Rs Aggarwal 2017 Solutions for Class 9 MATH Chapter 14

Question 1:

Define statistics as a subject.

Answer:

Statistics is the science which deals with the collection, presentation, analysis and interpretation of numerical data.

Question 2:

Define some fundamental characteristics of statistics.

Answer:

The fundamental characteristics of data (statistics) are as follows:
(i) Numerical facts alone constitute data.
(ii) Qualitative characteristics like intelligence and poverty, which cannot be measured numerically, do not form data.
(iii) Data are aggregate of facts. A single observation does not form data.
(iv) Data collected for a definite purpose may not be suited for another purpose.
(v) Data in different experiments are comparable.

Question 3:

What are primary data and secondary data? Which of the two is more reliable and why?

Answer:

Primary data: The data collected by the investigator himself with a definite plan in mind are known as primary data.
Secondary data: The data collected by someone other than the investigator are known as secondary data.

Primary data are highly reliable and relevant because they are collected by the investigator himself with a definite plan in mind, whereas secondary data are collected with a purpose different from that of the investigator and may not be fully relevant to the investigation.

Question 4:

Explain the meaning of each of the following terms:
(i) Variate
(ii) Class interval
(iii) Class size
(iv) Class mark
(v) Class limit
(vi) True class limits

Answer:

(i) Variate : Any character which is capable of taking several different values is called a variant or a variable.
(ii) Class interval : Each group into which the raw data is condensed is called class interval .
(iii) Class size: The difference between the true upper limit and the true lower limit of a class is called its class size.
(iv) Class mark of a class: The class mark is given by $(\frac{Upper limit + Lower limit}{2})$ .
(v) Class limit: Each class is bounded by two figures, which are called class limits.
(vi) True class limits: In the exclusive form, the upper and lower limits of a class are respectively known as true upper limit and true lower limit.
In the inclusive form of frequency distribution, the true lower limit of a class is obtained by subtracting 0.5 from the lower limit and the true upper limit of the class is obtained by adding 0.5 to the upper limit.
(vii) Frequency of a class: It is a number of data values that fall in the range specified by that class.
(viii) Cumulative frequency of a class: The cumulative frequency corresponding to that class is the sum of all frequencies up to and including that class.

Question 5:

Following data gives the number of children in 40 families:
1,2,6,5,1,5,1,3,2,6,2,3,4,2,0,4,4,3,2,2,0,0,1,2,2,4,3,2,1,0,5,1,2,4,3,4,1,6,2,2.
Represent it in the form of a frequency distribution, taking classes 0−2, 1−4, etc.

Answer:

The minimum observation is 0 and the maximum observation is 8.
Therefore, classes of the same size covering the given data are 0-2, 2-4, 4-6 and 6-8. .

Frequency distribution table:

Class	Tally mark	Frequency
0-2		11
2-4		17
4-6		9
6-8		3

Question 6:

The marks obtained by 40 students of a class in an examination are given below.
3,20,13,1,21,13,3,23,16,13,18,12,5,12,5,24,9,2,7,18,20,3,10,12,7,18,2,5,7,10,16,8,16,17,8,23,24,6,23, 15.
Present the data in the form of a frequency distribution using equal class size, one such class being 10−15 (15 not included).

Answer:

The minimum observation is 0 and the maximum observation is 25.
Therefore, classes of the same size covering the given data are 0-5, 5-10, 10-15, 15-20 and 20-25.
Frequency distribution table:

Class	Tally mark	Frequency
0-5		6
5-10		10
10-15		8
15-20		8
20-25		8

Question 7:

Construct a frequency table for the following ages (in years) of 30 students using equal class intervals, one of them being 9−12, where 12 is not included.
18,12,7,6,11,15,21,9,8,13,15,17,22,19,14,21,23,8,12,17,15,6,18,23,22,16,9,21,11,16.

Answer:

The minimum observation is 6 and the maximum observation is 24.
Therefore, classes of the same size covering the given data are 6-9, 9-12, 12-15, 15-18, 18-21 and 21-24.
Frequency distribution table:

Class	Tally mark	Frequency
6-9		5
9-12		4
12-15		4
15-18		7
18-21		3
21-24		7

Question 8:

Construct a frequency table with equal class intervals from the following data on the monthly wages (in rupees) of 28 labourers working in a factory, taking one of the class intervals as 210−230 (230 not included).
220,268,258,242,210,268,272,242,311,290,300,320,319,304,302,318,306,292,254,278,210,240,280,316,306,215,256,236.

Answer:

The minimum observation is 210 and the maximum observation is 330.
Therefore, classes of the same size covering the given data are 210-230, 230-250,250-270,270-290,290-310 and 310-330.
Frequency distribution table:

Class	Tally mark	Frequency
210-230		4
230-250		4
250-270		5
270-290		3
290-310		7
310-330		5

Question 9:

The weights (in grams) of 40 oranges picked at random from a basket are as follows:
40,50,60,65,45,55,30,90,75,85,70,85,75,80,100,110,70,55,30,35,45,70,80,85,95,70,60,70,75,100,65,60,40,100,75,110,30,45,84.
Construct a frequency table as well as a cumulative frequency table.

Answer:

The minimum observation is 30 and the maximum observation is 120.

Frequency distribution table:

Class	Tally mark	Frequency
30-40		4
40-50		6
50-60		3
60-70		5
70-80		9
80-90		6
90-100		2
100-110		3
110-120		2

Cumulative frequency table:

Class	Tally mark	Frequency	Cumulative frequency
30-40		4	4
40-50		6	10
50-60		3	13
60-70		5	18
70-80		9	27
80-90		6	33
90-100		2	35
100-110		3	38
110-120		2	40

Question 10:

The electricity bills (in rupees) of 40 houses in a locality are given below:
116,127,107,100,80,82,91,101,65,95,87,81,105,129,92,75,89,78,87,81,59,52,65,101,115,108,95,65,98,62,84,76,63,128,121,61,118,108,116,130.
Construct a grouped frequency table.

Answer:

The minimum observation is 800 and the maximum observation is 900.
So, range = 900 − 800 = 100
Class size = 10
Now, 100 $\div$ 10=10
i.e., we should have 10 classes each of size 10.
Therefore, classes of the same size covering the above data are 800-810, 810-820, 820-830, 830-840, 840-850, 850-860, 860-870, 870-880, 880-890 and 890-900.

Frequency distribution table:

Class	Tally mark	Frequency
800-810		3
810-820		2
820-830		1
830-840		8
840-850		5
850-860		1
860-870		3
870-880		1
880-890		1
890-900		5

Question 11:

The electricity bills (in rupees) of 40 houses in a locality are given below:
116,127,107,100,80,82,91,101,65,95,87,81,105,129,92,75,89,78,87,81,59,52,65,101,115,108,95,65,98,62,84,76,63,128,121,61,118,108,116,130.
Construct a grouped frequency table.

Answer:

The grouped frequency distribution table for the given data can be represented as follows:

Class	Tally mark	Frequency
50-60		2
60-70		6
70-80		3
80-90		8
90-100		5
100-110		7
110-120		4
120-130		4
130-140		1

Question 12:

Following are the ages (in years) of 360 patients, getting medical treatment in a hospital:

Ages (in years)	10−20	20−30	30−40	40−50	50−60	60−70
Number of patients	90	50	60	80	50	30

Construct the cumulative frequency table for the above data.

Answer:

The cumulative frequency table can be presented as given below:

Age (in years )	No. of patients	Cumulative frequency
10-20	90	90
20-30	50	140
30-40	60	200
40-50	80	280
50-60	50	330
60-70	30	360

Question 13:

Present the following as an ordinary grouped frequency table:

Marks (below)	10	20	30	40	50	60
Number of students	5	12	32	40	45	48

Answer:

The grouped frequency table can be presented as given below:

Marks	No. of students
0-10	5
10-20	7
20-30	20
30-40	8
40-50	5
50-60	3

Question 14:

Given below is a cumulative frequency table:

Marks	Number of students
Below 10	17
Below 20	22
Below 30	29
Below 40	37
Below 50	50
Below 60	60

Extract a frequency table from the above.

Answer:

The frequency table can be presented as given below:

Marks	Number of students
0-10	17
10-20	5
20-30	7
30-40	8
40-50	13
50-60	10

Question 15:

Make a frequency table from the following:

Marks obtained	Number of students
More than 60	0
More than 50	16
More than 40	40
More than 30	75
More than 20	87
More than 10	92
More than 0	100

Answer:

The frequency table can be presented as below:

Class	Frequency
0-10	8
10-20	5
20-30	12
30-40	35
40-50	24
50-60	16

Question 1:

The following table shows the number of students participating in various games in a school.

Game	Cricket	Football	Basketball	Tennis
Number of students	27	36	18	12

Draw a bar graph to represent the above data.

Answer:

Question 2:

On a certain day, the temperature in a city was recorded as under:

Time	5 a. m.	8 a. m.	11 a. m.	3 p. m.	6 p. m.
Temperature (in °C)	20	24	26	22	18

Illustrate the data by a bar graph.

Answer:

Question 3:

The approximate velocities of some vehicles are given below:

Name of vehicle	Bicycle	Scooter	Car	Bus	Train
Velocity (in km/hr)	27	36	18	12	80

Draw a bar graph to represent the above data.

Answer:

Question 4:

The following table shows the favourite sports of 250 students of a school. Represent the data by a bar graph.

Sports	Cricket	Football	Tennis	Badminton	Swimming
No. of students	75	35	50	25	65

Answer:

Question 5:

Given below is a table which shows the yearwise strength of a school. Represent this data by a bar graph.

Year	2001−02	2002−03	2003−04	2004−05	2005−06
No. of students	800	975	1100	1400	1625

Answer:

Question 6:

The following table shows the number of scooter produced by a company during six consecutive years. Draw a bar graph to represent this data.

Year	1999	2000	2001	2002	2004
No. of students	11000	14000	12500	17500	24000

Answer:

Question 7:

The birth rate per thousand in five countries over a period of time is shown below:

Country	China	India	Germany	UK	Sweden
Birth rate per thousand	42	35	14	28	21

Represent the above data by a bar graph.

Answer:

Question 8:

The following table shows the interest paid by India (in thousand crore rupees) on external debts during the period 1998−99 to 2002−03.
Represent the data by a bar graph.

Year	1998−99	1999−2000	2000−01	2001−02	2002−03
Intersect (in thousand crore rupees)	70	84	98	106	120

Answer:

Question 9:

The air distances of four cities from Delhi (in km) are given below:

City	Kolkata	Mumbai	Chennai	Hyderabad
Distance from Delhi (in km)	1340	1100	1700	1220

Draw a bar graph to represent the above data.

Answer:

Question 10:

The following table shows the life expectancy (average age to which people live) in various countries in a particular year.
Represent this data by a bar graph.

Country	Japan	India	Britain	Ethiopia	Cambodia
Life expectancy (in years)	76	57	70	43	36

Answer:

Question 11:

Gold prices on 4 consecutive Tuesday were as under:

Week	First	Second	Third	Fourth
Rate per 10 g (in Rs)	7250	7500	7600	7850

Draw a bar graph to show this information.

Answer:

Question 12:

Various modes of transport used by 1850 students of a school are given below.

School bus	Private bus	Bicycle	Rickshaw	By foot
640	360	490	210	150

Draw a bar graph to represent the above data.

Answer:

Question 13:

Look at the bar graph given below.

Read it carefully and answer the following questions.
(i) What information does the bar graph give?
(ii) In which subject is the student very good?
(iii) In which subject is he poor?
(iv) What is the average of his marks?

Answer:

(i) The bar graph shows the marks obtained by a student in various subjects in an examination.

(ii) The student scores very good in mathematics, as the height of the corresponding bar is the highest.

(iii) The student scores bad in Hindi, as the height of the corresponding bar is the lowest.

(iv) Average marks = $\frac{60 + 35 + 75 + 50 + 60}{5} = \frac{280}{5} = 56$

Question 1:

The daily wages of 50 workers in a factory are given below:

Daily wages (in rupees)	140−180	180−220	220−260	260−300	300−340	340−380
Number of workers	16	9	12	2	7	4

Construct a histogram to represent the above frequency distribution.

Answer:

The given frequency distribution is in exclusive form.
We will represent the class intervals [daily wages (in rupees)] along the x-axis & the corresponding frequencies [number of workers] along the y-axis.
The scale is as follows:
On x-axis: 1 big division = 40 rupees
On y-axis: 1 big division = 2 workers
Because the scale on the x-axis starts at 140, a kink, i.e., a break, is indicated near the origin to signify that the graph is drawn with a scale beginning at 140
and not at the origin.
We will construct rectangles with the class intervals as bases and the corresponding frequencies as heights.
Thus, we will obtain the following histogram:

Question 2:

The following table shows the average daily earnings of 40 general stores in a market, during a certain week.

Daily earnings (in rupees)	600−650	650−700	700−750	750−800	800−850	850−900
Number of stores	6	9	2	7	11	5

Draw a histogram to represent the above data.

Answer:

The given frequency distribution is in exclusive form.
We will represent the class intervals [daily earnings (in rupees)] along the x-axis & the corresponding frequencies [number of stores] along the y-axis.
The scale is as follows:
On x-axis: 1 big division = 50 rupees
On y-axis: 1 big division = 1 store
Because the scale on the x-axis starts at 600, a kink, i.e., a break is indicated near the origin to signify that the graph is drawn with a scale beginning at 600 and not at the origin.
We will construct rectangles with the class intervals as bases and the corresponding frequencies as heights.
Thus, we will obtain the following histogram:

Question 3:

The heights of 75 students in a school are given below:

Height (in cm)	130−136	136−142	142−148	148−154	154−160	160−166
Number of students	9	12	18	23	10	3

Draw a histogram to represent the above data.

Answer:

The given frequency distribution is in exclusive form.
We will represent the class intervals [heights (in cm)] along the x-axis & the corresponding frequencies [number of students ] along the y-axis.
The scale is as follows:
On x-axis: 1 big division = 6 cm
On y-axis: 1 big division = 2 students
Because the scale on the x-axis starts at 130, a kink, i.e., a break, is indicated near the origin to signify that the graph is drawn with a scale beginning at 130
and not at the origin.
We will construct rectangles with the class intervals as bases and the corresponding frequencies as heights.
Thus, we will obtain the following histogram:

Question 4:

Draw a histogram for the frequency distribution of the following data.

Class interval	8−13	13−18	18−23	23−28	28−33	33−38	38−43
Frequency	320	780	160	540	260	100	80

Answer:

The given frequency distribution is in exclusive form.
We will represent the class intervals along the x-axis & the corresponding frequencies along the y-axis.
The scale is as follows:
On x-axis: 1 big division = 5 units
On y-axis: 1 big division = 50 units
Because the scale on the x-axis starts at 8, a kink, i.e., a break, is indicated near the origin to signify that the graph is drawn with a scale beginning at 8
and not at the origin.
We will construct rectangles with the class intervals as bases and the corresponding frequencies as heights.
Thus, we will obtain the following histogram:

Question 5:

Construct a histogram for the following frequency distribution.

Class interval	5−12	13−20	21−28	29−36	37−44	45−52
Frequency	6	15	24	18	4	9

Answer:

The given frequency distribution is in inclusive form.
So, we will convert it into exclusive form, as shown below:

Class Interval	Frequency
4.5–12.5	6
12.5–20.5	15
20.5–28.5	24
28.5–36.5	18
36.5–44.5	4
44.5–52.5	9

We will mark class intervals along the x-axis and frequencies along the y-axis.
The scale is as follows:
On x-axis: 1 big division = 8 units
On y-axis: 1 big division = 2 units
Because the scale on the x-axis starts at 4.5, a kink, i.e., a break, is indicated near the origin to signify that the graph is drawn with a scale beginning at 4.5
and not at the origin.
We will construct rectangles with class intervals as bases and the corresponding frequencies as heights.
Thus, we will obtain the histogram as shown below:

Question 6:

The following table shows the number of illiterate persons in the age group (10−58 years) in. a town:

Age group (in years)	10−16	17−23	24−30	31−37	38−44	45−51	52−58
Number of illiterate persons	175	325	100	150	250	400	525

Draw a histogram to represent the above data.

Answer:

The given frequency distribution is inclusive form.
So, we will convert it into exclusive form, as shown below:

Age (in years)	Number of Illiterate Persons
9.5-16.5	175
16.5-23.5	325
23.5-30.5	100
30.5-37.5	150
37.5-44.5	250
44.5-51.5	400
51.5-58.5	525

We will mark the age groups (in years) along the x-axis & frequencies (number of illiterate persons) along the y-axis.
The scale is as follows:
On x-axis: 1 big division = 7 years
On y-axis: 1 big division = 50 persons
Because the scale on the x-axis starts at 9.5, a kink, i.e., a break, is indicated near the origin to signify that the graph is drawn with a scale beginning at 9.5
and not at the origin.
We will construct rectangles with class intervals (age) as bases and the corresponding frequencies (number of illiterate persons) as
heights.
Thus, we obtain the histogram, as shown below:

Question 7:

Draw a histogram to represent the following data.

Clas interval	10−14	14−20	20−32	32−52	52−80
Frequency	5	6	9	25	21

Answer:

In the given frequency distribution, class sizes are different.
So, we calculate the adjusted frequency for each class.
The minimum class size is 4.
Adjusted frequency of a class = $\frac{Minimum class size}{Class size of the class} \times Its frequency$

We have the following table:

Class Interval	Frequency	Adjusted Frequency
10-14	5	$\frac{4}{4} \times 5 = 5$
14-20	6	$\frac{4}{6} \times 6 = 4$
20-32	9	$\frac{4}{12} \times 9 = 3$
32-52	25	$\frac{4}{20} \times 25 = 5$
52-80	21	$\frac{4}{28} \times 21 = 3$

We mark the class intervals along the x-axis and the corresponding adjusted frequencies along the y-axis.
We have chosen the scale as follows:
On the x- axis,
1 big division = 5 units
On the y-axis,
1 big division = 1 unit
We draw rectangles with class intervals as the bases and the corresponding adjusted frequencies as the heights.
Thus, we obtain the following histogram:

Question 8:

In a study of diabetic patients in a village, the following observations were noted.

Age in years	10−20	20−30	30−40	40−50	50−60	60−70
Number of patients	2	5	12	19	9	4

Represent the above data by a frequency polygon.

Answer:

We take two imagined classes—one at the beginning (0–10) and other at the end (70–80)—each with frequency zero.
With these two classes, we have the following frequency table:

Age in Years	Class Mark	Frequency (Number of Patients)
0–10	5	0
10–20	15	2
20–30	25	5
30–40	35	12
40–50	45	19
50–60	55	9
60–70	65	4
70–80	75	0

Now, we plot the following points on a graph paper:
A(5, 0), B(15, 2), C(25, 5), D(35, 12), E(45, 19), F(55, 9), G(65, 4) and H(75, 0)
Join these points with line segments AB, BC, CD, DE, EF, FG, GH ,HI and IJ to obtain the required frequency polygon.

Question 9:

The ages (in years) of 360 patients treated in a hospital on a particular day are given below.

Age in years	10−20	20−30	30−40	40−50	50−60	60−70
Number of patients	90	40	60	20	120	30

Draw a histogram and a frequency polygon on the same graph to represent the above data.

Answer:

We represent the class intervals along the x-axis and the corresponding frequencies along the y-axis.
We construct rectangles with class intervals as bases and respective frequencies as heights.
We have the scale as follows:
On the x-axis:
1 big division = 10 years
On the y-axis:
1 big division = 2 patients
Thus, we obtain the histogram, as shown below.
We join the midpoints of the tops of adjacent rectangles by line segments.
Also, we take the imagined classes 0–10 and 70–80, each with frequency 0. The class marks of these classes are 5 and 75, respectively.
So, we plot the points A(5, 0) and B(75, 0). We join A with the midpoint of the top of the first rectangle and B with the midpoint of the top of the last rectangle.
Thus, we obtain a complete frequency polygon, as shown below:

Question 10:

Draw a histogram and the frequency polygon from the following data.

Class interval	20−25	25−30	30−35	35−40	40−45	45−50
Frequency	30	24	52	28	46	10

Answer:

We represent the class intervals along the x-axis and the corresponding frequencies along the y-axis.
We construct rectangles with class intervals as bases and respective frequencies as heights.
We have the scale as follows:
On the x-axis, 1 big division = 5 units.
On the y-axis, 1 big division = 5 units.
Because the scale on the x-axis starts at 15, a kink, i.e., a break, is indicated near the origin to signify that the graph is drawn with a scale beginning at 15
and not at the origin.
Thus, we obtain the histogram, as shown below.
We join the midpoints of the tops of adjacent rectangles by line segments.
Also, we take the imagined classes 15–20 and 50–55, each with frequency 0. The class marks of these classes are 17.5 and 52.5, respectively.
So, we plot the points A( 17.5, 0) and B(52.5, 0). We join A with the midpoint of the top of the first rectangle and B with the midpoint of the top of the last rectangle.
Thus, we obtain a complete frequency polygon, as shown below:

Question 11:

Draw a histogram for the following data.

Class interval	600−640	640−680	680−720	720−760	760−800	800−840
Frequency	18	45	153	288	171	63

Using this histogram, draw the frequency polygon on the same graph.

Answer:

We represent the class intervals along the x-axis and the corresponding frequencies along the y-axis.
We construct rectangles with class intervals as bases and respective frequencies as heights.
We have the scale as follows:
On the x-axis:
1 big division = 40 units
On the y-axis:
1 big division = 20 units
Thus, we obtain the histogram, as shown below:
We join the midpoints of the tops of adjacent rectangles by line segments.
Also, we take the imagined classes 560–600 and 840–880, each with frequency 0.
The class marks of these classes are 580 and 860, respectively.
Because the scale on the x-axis starts at 560, a kink; i.e., a break, is indicated near the origin to signify that the graph is drawn with a scale beginning at 560
and not at the origin.
So, we plot the points A( 580, 0) and B(860, 0). We join A with the midpoint of the top of the first rectangle and join B with the midpoint of the top of the last rectangle.
Thus, we obtain a complete frequency polygon, as shown below:

Question 12:

Draw a frequency polygon for the following frequency distribution.

Class interval	1−10	11−20	21−30	31−40	41−50	51−60
Frequency	8	3	6	12	2	7

Answer:

Though the given frequency table is in inclusive form, class marks in case of inclusive and exclusive forms are the same.
We take the imagined classes ( $-$ 9)–0 at the beginning and 61–70 at the end, each with frequency zero.
Thus, we have:

Class Interval	Class Mark	Frequency
$-$ 9–0	–4.5	0
1–10	5.5	8
11–20	15.5	3
21–30	25.5	6
31–40	35.5	12
41–50	45.5	2
51–60	55.5	7
61–70	65.5	0

Along the x-axis, we mark –4.5, 5.5, 15.5, 25.5, 35.5, 45.5, 55.5 and 65.5.
Along the y-axis, we mark 0, 8, 3, 6, 12, 2, 7 and 0.
We have chosen the scale as follows :
On the x-axis, 1 big division = 10 units.
On the y-axis, 1 big division = 1 unit.
We plot the points A(–4.5,0), B(5.5, 8), C(15.5, 3), D(25.5, 6), E(35.5, 12), F(45.5, 2), G(55.5, 7) and H(65.5, 0).
We draw line segments AB, BC, CD, DE, EF, FG, GH to obtain the required frequency polygon, as shown below.

Question 1:

Find the arithmetic mean of
(i) the first eight natural numbers
(ii) the first ten odd numbers
(iii) the first five prime numbers
(iv) the first six even numbers
(v) the first seven multiplies of 5
(vi) all the factors of 20

Answer:

We know:

$Mean = \frac{Sum of observa tions}{Number of observations}$

(i) The first eight natural numbers are 1, 2, 3, 4, 5, 6, 7 and 8.
Mean of these numbers:

$\frac{1 + 2 + 3 + 4 + 5 + 6 + 7 + 8}{8} = \frac{36}{8} = 4.5$

(ii) The first ten odd numbers are 1, 3, 5, 7, 9, 11, 13, 15, 17 and 19.
Mean of these numbers:

$\frac{1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19}{10} = \frac{100}{10} = 10$

(iii) The first five prime numbers are 2, 3, 5, 7 and 11.
Mean of these numbers:

$\frac{2 + 3 + 5 + 7 + 11}{5} = \frac{28}{5} = 5.6$

(iv) The first six even numbers are 2, 4, 6, 8, 10 and 12.
Mean of these numbers:

$\frac{2 + 4 + 6 + 8 + 10 + 12}{6} = \frac{42}{6} = 7$

(v) The first seven multiples of 5 are 5, 10, 15, 20, 25, 30 and 35.
Mean of these numbers:

$\frac{5 + 10 + 15 + 20 + 25 + 30 + 35}{7} = \frac{140}{7} = 20$

(vi) The factors of 20 are 1, 2, 4, 5, 10 and 20.
Mean of these numbers:

$\frac{1 + 2 + 4 + 5 + 10 + 20}{6} = \frac{42}{6} = 7$

Question 2:

The number of children in 10 families of a locality are
2, 4, 3, 4, 2, 0, 3, 5, 1, 6.
Find the mean number of children per family.

Answer:

Numbers of children in 10 families = 2, 4, 3, 4, 2, 0, 3, 5, 1 and 6.
Thus, we have:

$Mean = \frac{Sum of observa tions}{Number of observations}$

$= \frac{2 + 4 + 3 + 4 + 2 + 0 + 3 + 5 + 1 + 6}{10}$
$= \frac{30}{10} = 3$

Question 3:

The following are the number of books issued in a school library during a week:
105, 216, 322, 167, 273, 405 and 346.
Find the average number of books issued per day.

Answer:

Numbers of books issued in the school library: 105, 216, 322, 167, 273, 405 and 346
Thus, we have:

$Mean = \frac{Sum of observa tions}{Number of observations}$

$= \frac{105 + 216 + 322 + 167 + 273 + 405 + 346}{7} = \frac{346}{7} = 262$

Question 4:

The daily minimum temperature recorded (in degree F) at a place during a week was as under:

Monday	Tuesday	Wednesday	Thursday	Friday	Saturday
35.5	30.8	27.3	32.1	23.8	29.9

Find the mean temperature.

Answer:

Daily minimum temperatures = 35.5, 30.8, 27.3, 32.1, 23.8 and 29.9
Thus, we have:

$Mean temperature = \frac{35.5 + 30.8 + 27.3 + 32.1 + 23.8 + 29.9}{6} = \frac{179.4}{6} = 29.9 ° F$

Question 5:

The percentages of marks obtained by 12 students of a class in mathematics are
64, 36, 47, 23, 0, 19, 81, 93, 72, 35, 3, 1.
Find the mean percentage of marks.

Answer:

Percentages of marks obtained: 64, 36, 47, 23, 0, 19, 81, 93, 72, 35, 3, 1
Now,

$Mean = \frac{Sum of observa tions}{Number of observations}$

$= \frac{64 + 36 + 47 + 23 + 0 + 19 + 81 + 93 + 72 + 35 + 3 + 1}{12} = \frac{474}{12} = 39.5$

Question 6:

If the arithmetic mean of 7, 9, 11, 13, x, 21 is 13, find the value of x.

Answer:

Observations: 7, 9, 11, 13, x, 21
Given:
Mean = 13
We know:

$Mean = \frac{Sum of observa tions}{Number of observations}$

Thus, we have:

$13 = \frac{7 + 9 + 11 + 13 + x + 21}{6} \Rightarrow 13 = \frac{61 + x}{6} \Rightarrow 78 = 61 + x \Rightarrow x = 17$

Question 7:

The mean of 24 numbers is 35. If 3 is added to each number, what will be the new mean?

Answer:

Let the numbers be x₁, x₂,...x₂₄.

We know:

$Mean = \frac{Sum of observa tions}{Number of observations}$

Thus, we have:

$35 = \frac{x_{1} + x_{2} + . . . . . . . . . + x_{24}}{24}$ _{$\Rightarrow 840 = x_{1} + x_{2} + . . . . . . + x_{24} . . . . . . . (i)$}

After addition, the new numbers become (x₁+3), (x₂+3),...(x₂₄+3).
New mean:

$= \frac{(x_{1} + 3) + (x_{2} + 3) + . . . . . . . . . . + (x_{24} + 3)}{24} = \frac{(x_{1} + x_{2} + . . . . . . . . . + x_{24}) + (24 \times 3)}{24} = \frac{840 + 72}{24} [From (i)] = \frac{912}{24} = 38$

Question 8:

The mean of 20 numbers is 43. If 6 is subtracted from each of the numbers, what will be the new mean?

Answer:

Let the numbers be x₁, x₂,...x₂₀.
We know:
$Mean = \frac{Sum of observa tions}{Number of observations}$

Thus, we have:

$43 = \frac{x_{1} + x_{2} + . . . . . . + x_{20}}{20}$ _{$\Rightarrow 860 = x_{1} + x_{2} + . . . . . . + x_{20} . . . . . . (i)$}

Numbers after subtraction: (x₁ $-$ 6), (x₂ $-$ 6),...(x₂₀ $-$ 6)

∴ New Mean = $\frac{(x_{1} - 6) + (x_{2} - 6) + . . . . . . . . + (x_{20} - 6)}{20}$

$= \frac{(x_{1} + x_{2} + . . . . . . . . . . + x_{20}) - (20 \times 6)}{20} = \frac{860 - 120}{20} [From (i)] = 37$

Question 9:

The mean of 15 numbers is 27. If each number is multiplied by 4, what will be the mean of the new numbers?

Answer:

Let the numbers be x₁, x₂,...x₁₅
We know:

$Mean = \frac{Sum of observa tions}{Number of observations}$

Thus, we have:

$27 = \frac{x_{1} + x_{2} + . . . . . . . . . . . + x_{15}}{15}$

$\Rightarrow x_{1} + x_{2} + . . . . . . . . . + x_{15} = 405 . . . . . . . . . (i)$

After multiplication, the numbers become 4x₁, 4x₂,...4x₁₅
∴ New Mean = $\frac{4 x_{1} + 4 x_{2} + . . . . . . + 4 x_{15}}{15}$

$= \frac{4 (x_{1} + x_{2} + . . . . . . . + x_{15})}{15} = \frac{4 \times 405}{15} [From (i)] = \frac{1620}{15} = 108$

Question 10:

The mean of 12 numbers is 40. If each number is divided by 8, what will be the mean of the new numbers?

Answer:

Let the numbers be x₁, x₂,...x₁₂.
We know:

$Mean = \frac{Sum of observa tions}{Number of observations}$

Thus, we have:

$40 = \frac{x_{1} + x_{2} + . . . . + x_{12}}{12}$

$\Rightarrow x_{1} + x_{2} + . . . . + x_{12} = 480 . . . . . (i)$

After division, the numbers become:

$\frac{x_{1}}{8}, \frac{x_{2}}{8}, . . . . . . . ., \frac{x_{12}}{8}$

$∴ New mean = \frac{(\frac{x_{1}}{8}) + (\frac{x_{2}}{8}) + . . . . + (\frac{x_{12}}{8})}{12} = \frac{x_{1} + x_{2} + . . . + x_{12}}{12 \times 8} = \frac{x_{1} + x_{2} + . . . + x_{12}}{96} = \frac{480}{96} [From (i)] = 5$

Question 11:

The mean of 12 numbers is 40. If each number is divided by 8, what will be the mean of the new numbers?

Answer:

Let the numbers be x₁, x₂_,...x_20.
We know:
$Mean = \frac{Sum of observa tions}{Number of observations}$

Thus, we have:

$18 = \frac{x_{1} + x_{2} + . . . . . . . + x_{20}}{20}$ _{$\Rightarrow x_{1} + x_{2} + . . . . . + x_{20} = 360 . . . . . (i)$}

New numbers are:

(x₁ + 3), (x₂ + 3),...(x₁₀ + 3), x₁₁,...x₂₀

New Mean:

$= \frac{(x_{1} + 3) + (x_{2} + 3) + . . . . . . . . + (x_{10} + 3) + x_{11} + . . . . . . . . + x_{20}}{20} = \frac{(x_{1} + x_{2} + . . . . . . + x_{10}) + (3 \times 10) + x_{11} + . . . . . . . . + x_{20}}{20} = \frac{(x_{1} + x_{2} + . . . . . . . + x_{20}) + 30}{20} = \frac{360 + 30}{20} [From (i)] = 19.5$

Question 12:

The mean weight of 6 boys in a group is 48 kg. The individual weights of five of them are 51 kg, 45 kg, 49 kg, 46 kg and 44 kg. Find the weight of the sixth boy.

Answer:

The individual weights of five boys are 51 kg, 45 kg, 49 kg, 46 kg and 44 kg.
Now,
Let the weight of the sixth boy be x kg.
We know:
$Mean = \frac{Sum of observa tions}{Number of observations}$

Also,
Given mean = 48 kg
Thus, we have:

$\Rightarrow 48 = \frac{51 + 45 + 49 + 46 + 44 + x}{6} \Rightarrow 288 = 235 + x \Rightarrow x = 53$

Therefore, the sixth boy weighs 53 kg.

Question 13:

The mean of the marks scored by 50 students was found to be 39. Later on it was discovered that a score of 43 was misread as 23. Find the correct mean.

Answer:

Let the marks scored by 50 students be x₁, x₂,...x₅₀.
Mean = 39
We know:
$Mean = \frac{Sum of observa tions}{Number of observations}$

Thus, we have:

$39 = \frac{x_{1} + x_{2} + . . . + x_{50}}{50} \Rightarrow x$ _{$\Rightarrow x_{1} + x_{2} + . . . . + x_{50} = 1950 . . . . . . (i)$}

Also, a score of 43 was misread as 23.
$∴ New Mean = \frac{(x_{1} + x_{2} + . . . + x_{50}) - 23 + 43}{50} = \frac{1950 - 23 + 43}{50} [using (i)] = \frac{1970}{50} = 39.4$

Question 14:

The mean of 100 items was found to be 64. Later on it was discovered that two items were misread as 26 and 9 instead of 36 and 90 respectively. Find the correct mean.

Answer:

Let the items be x₁, x₂,...x₁₀₀
Given:
Mean = 64
We know:

$Mean = \frac{Sum of observa tions}{Number of observations}$

$Thus, we have : 64 = \frac{x_{1} + x_{2} + . . . + x_{100}}{100}$ _{$\Rightarrow x_{1} + x_{2} + . . . . . . . + x_{100} = 6400$}

Because items 36 and 90 were misread as 26 and 9, we have:

Correct Mean = $\frac{(x_{1} + x_{2} + . . . . . . . + x_{100}) - (26 + 9) + (36 + 90)}{100}$
$= \frac{6400 - 35 + 126}{100} = \frac{6491}{100} = 64.91$

Question 15:

The mean of six numbers is 23. If one of the numbers is excluded, the mean of the remaining numbers is 20. Find the excluded number.

Answer:

Let the numbers be x₁, x₂,..., x₆.
Mean = 23
We know:
$Mean = \frac{Sum of observa tions}{Number of observations}$

Thus, we have:

$23 = \frac{x_{1} + x_{2} + . . . + x_{6}}{6}$
$x_{1} + x_{2} . . . . . . . x_{6} = 138$ .....................(i)

If one number, say, x₆, is excluded, then we have:

$20 = \frac{x_{1} + x_{2} + . . . + x_{5}}{5}$
$x_{1} + x_{2} . . . . . . + x_{5} = 100 . . . . . . . . . . . . . . . . . . . . (ii)$

Using (i) and (ii), we get:

$138 = x_{1} + x_{2} + . . . + x_{5} + x_{6} \Rightarrow 138 = 100 + x_{6} . . . (i) \Rightarrow x_{6} = 38$

Thus, the excluded number is 38

Question 1:

The mean mark obtained by 7 students in a group is 226. If the marks obtained by six of them are 340, 180, 260, 56, 275 and 307 respectively, find the marks obtained by the seventh student.

Answer:

Given:
Marks obtained by 6 students: 340, 180, 260, 56, 275 and 307
Now,
Let the marks obtained by the 7th student be x.
We know:
Mean marks obtained by 7 students = 226
Thus, we have:

$\frac{340 + 180 + 260 + 56 + 275 + 307 + x}{7} = 226 \Rightarrow \frac{1418 + x}{7} = 226 \Rightarrow 1418 + x = 1582 \Rightarrow x = 164$

∴ Marks obtained by the 7th student = 164

Question 2:

The mean weight of a class of 34 students is 46.5 kg. If the weight of the teacher is included, the mean is rises by 500 g. Find the weight of the teacher.

Answer:

Mean weight of 34 students = 46.5 kg
Sum of the weights of 34 students = $(46.5 \times 34) kg = 1581$ kg
Increase in the mean weight when the weight of the teacher is included = 500 g = 0.5 kg
∴ New mean weight = (46.5 + 0.5) kg = 47 kg
Now,
Let the weight of the teacher be x kg.
Thus, we have:

$\frac{Sum of the weights of 34 students + Weight of the teacher}{35} = 47 \Rightarrow \frac{1581 + x}{35} = 47 \Rightarrow 1581 + x = 1645 \Rightarrow x = 64$

Therefore, the weight of the teacher is 64 kg.

Question 3:

The mean weight of a class of 36 students is 41 kg. If one of the students leaves the class then the mean is decreased by 200 g. Find the weight of the student who left.

Answer:

Mean weight of 36 students = 41 kg
Sum of the weights of 36 students = $(41 \times 36) kg = 1476 kg$
Decrease in the mean when one of the students left the class = 200 g = 0.2 kg
Mean weight of 35 students = (41 $-$ 0.2) kg = 40.8 kg
Now,
Let the weight of the student who left the class be x kg.

$Thus, we have : \frac{Sum of the weights of 36 students - x}{35} = 40.8 \Rightarrow \frac{1476 - x}{35} = 40.8 \Rightarrow 1476 - x = 1428 \Rightarrow x = 48$

Hence, the weight of the student who left the class is 48 kg.

Question 4:

The average weight of a class of 39 students is 40 kg. When a new student is admitted to the class, the average decreases by 200 g. Find the weight of the new student.

Answer:

Average weight of 39 students = 40 kg
Sum of the weights of 39 students = $(40 \times 39) kg = 1560 kg$
Decrease in the average when new student is admitted in the class = 200 g = 0.2 kg
∴ New average weight = (40 $-$ 0.2) kg = 39.8 kg
Now,
Let the weight of the new student be x kg.
Thus, we have:

$\frac{Sum of the weights of 39 students + x}{40} = 39.8 \Rightarrow \frac{1560 + x}{40} = 39.8 \Rightarrow 1560 + x = 1592 \Rightarrow x = 32$

Therefore, the weight of the new student is 32 kg.

Question 5:

The average monthly salary of 20 workers in an office is Rs 7650. If the manager's salary is added, the average salary becomes Rs 8200 per month. What is the manager's salary per month?

Answer:

Average monthly salary of 20 workers = Rs 7650
Sum of the monthly salaries of 20 workers = $Rs (7650 \times 20) = Rs 153000$
By adding the manager's monthly salary, we get:
Average salary = Rs 8200
Now,
Let the manager's monthly salary be Rs x.
Thus, we have:

$\frac{Sum of the monthly salaries of 20 workers + x}{21} = 8200 \Rightarrow \frac{153000 + x}{21} = 8200 \Rightarrow 153000 + x = 172200 \Rightarrow x = 19200$

Therefore, the manager's monthly salary is Rs 19200.

Question 6:

The average monthly wage of a group of 10 persons is Rs 9000. One member of the group, whose monthly wage is Rs 8100, leaves the group and is replaced by a new member whose monthly wages is Rs 7200. Find the new monthly average wage.

Answer:

Average monthly wages of 10 persons = Rs 9000
Sum of the monthly wages of 10 persons = $(9000 \times 10) = Rs 90000$
Sum of the monthly wages of 9 persons if one of the members left the group = Rs (90000 $-$ 8100) = Rs 81900
Sum of the monthly wages of 10 persons if one new member joined the group = Rs (81900 + 7200) = Rs 89100

$∴ New monthly average wages = \frac{New sum of the monthly wages of 10 persons}{10}$
$= \frac{89100}{10} = Rs 8910$

Question 7:

The average monthly consumption of petrol for a car for the first 7 months of a year is 330 litres, and for the next 5 months is 270 litres. What is the average consumption per month during the whole year?

Answer:

Average monthly consumption of petrol for the first 7 months = 330 litres
Average monthly consumption of petrol for the next 5 months = 270 litres
Sum of the monthly consumption of petrol for the first 7 months = $(330 \times 7) litres = 2310 litres$
Sum of the monthly consumption of petrol for the next 5 months = $(270 \times 5) litres = 1350 litres$

$∴ Average consumption of petrol per month during the whole year = \frac{2310 + 1350}{12} = \frac{3660}{12} = 305 litres$

Therefore, the average consumption of petrol per month during the whole year is 305 litres.

Question 8:

Find the mean of 25 numbers if the mean of 15 of them is 18 and the mean of the remaining numbers is 13.

Answer:

Mean of 15 numbers = 18
Mean of the remaining 10 numbers = 13
Sum of 15 numbers = $15 \times 18 = 270$
Sum of the remaining 10 numbers = $10 \times 13 = 130$
Thus, we have:
$Mean of 25 numbers = \frac{270 + 130}{25} = 16$

Question 9:

The mean weight of 60 students of a class is 52.75 kg. If the mean weight of 25 of them is 51 kg, find the mean weight of the remaining students.

Answer:

Mean weight of 60 students of the class= 52.75 kg
Mean weight of 25 of them = 51 kg
Sum of the weights of those 25 students = $(25 \times 51) kg = 1275 kg$
Now,
Let the sum of weights of the remaining 35 students be x kg.
Thus, we have:

$\frac{Sum of the weights of 25 students + Sum of the weights of the remaining 35 students}{60} = Mean weight of 60 students \Rightarrow \frac{1275 + x}{60} = 52.75 \Rightarrow 1275 + x = 3165 \Rightarrow x = 1890 ∴ Mean weight of the remaining 35 students = \frac{1890}{35} = 54 kg$

Question 10:

The average weight of 10 oarsmen in a boat is increased by 1.5 kg when one of the crew who weighs 58 kg is replaced by a new man. Find the weight of the new man.

Answer:

Let the average weight of 10 oarsmen be x kg.
Sum of the weights of 10 oarsmen = 10x kg
∴ New average weight = (x + 1.5) kg
Now, we have:
$New average weight = \frac{Sum of the weights of initial 10 oarsmen - 58 + Weight of the new man}{10} \Rightarrow x + 1.5 = \frac{10 x - 58 + Weight of the new man}{10}$

$\Rightarrow Weight of the new man + 10 x - 58 = 10 x + 15 \Rightarrow Weight of the new man - 58 = 15 \Rightarrow Weight of the new man = 15 + 58 = 73 kg$

Question 11:

The mean of 8 numbers is 35. If a number is excluded then the mean is reduced by 3. Find the excluded number.

Answer:

Mean of 8 numbers = 35
Sum of 8 numbers = $35 \times 8 = 280$
Let the excluded number be x.
Now,
New mean = 35 $-$ 3 = 32
Thus, we have:

$\frac{Sum of 8 numbers - x}{7} = 32 \Rightarrow \frac{280 - x}{7} = 32 \Rightarrow 280 - x = 224 \Rightarrow x = 56$

Therefore, the excluded number is 56.

Question 12:

The mean of 150 items was found to be 60. Later on, it was discovered that the values of two items were misread as 52 and 8 instead of 152 and 88 respectively. Find the correct men.

Answer:

Mean of 150 items = 60
Sum of 150 items = $(150 \times 60) = 9000$
New sum = [9000 $-$ (52 + 8) + (152 + 88)] = 9180

Correct mean = $\frac{New sum}{Total items} = \frac{9180}{150} = 61.2$

Therefore, the correct mean is 61.2.

Question 13:

The mean of 31 results is 60. If the mean of the first 16 results is 58 and that of the last 16 results is 62, find the 16th result.

Answer:

Mean of 31 results = 60
Sum of 31 results = $31 \times 60 = 1860$
Mean of the first 16 results = 58
Sum of the first 16 results = $58 \times 16 = 928$
Mean of the last 16 results = 62
Sum of the last 16 results = $62 \times 16 = 992$
Value of the 16th result = (Sum of the first 16 results + Sum of the last 16 results) $-$ Sum of 31 results
= (928 + 992) $-$ 1860
= 1920 $-$ 1860
= 60

Question 14:

The mean of 11 numbers is 42. If the mean of the first 6 numbers is 37 and that of the last 6 numbers is 46, find the 6th number.

Answer:

Mean of 11 numbers = 42
Sum of 11 numbers = 42 $\times$ 11 = 462
Mean of the first 6 numbers = 37
Sum of the first 6 numbers = 37 $\times$ 6 = 222
Mean of the last 6 numbers = 46
Sum of the last 6 numbers = 46 $\times$ 6 = 276
∴ 6th number = [(Sum of the first 6 numbers + Sum of the last 6 numbers) $-$ Sum of 11 numbers]
= [(222 + 276) $-$ 462]
= [498 $-$ 462]
= 36
Hence, the 6th number is 36.

Question 15:

The mean weight of 25 students of a class is 52 kg. If the mean weight of the first 13 students of the class is 48 kg and that of the last 13 students is 55 kg, find the weight of the 13th student.

Answer:

Mean weight of 25 students = 52 kg
Sum of the weights of 25 students = (52 $\times$ 25) kg = 1300 kg
Mean weight of the first 13 students = 48 kg
Sum of the weights of the first 13 students = (48 $\times$ 13) kg = 624 kg
Mean weight of the last 13 students = 55 kg
Sum of the weights of the last 13 students = (55 $\times$ 13) kg =715 kg
Weight of the 13th student = (Sum of the weights of the first 13 students + Sum of the weights of the last 13 students) $-$ Sum of the weights of 25 students
= [(624+715) $-$ 1300] kg
= 39 kg
Therefore, the weight of the 13th student is 39 kg.

Question 16:

The mean score of 25 observations is 80 and the mean score of another 55 observations is 60. Determine the mean score of the whole set of observations.

Answer:

Mean score of 25 observations = 80
Sum of the scores of 25 observations = 80 $\times$ 25 = 2000
Mean score of another 55 observations = 60
Sum of the scores of another 55 observations = 60 $\times$ 55 = 3300

$Mean score of the whole set of observations = \frac{Sum of the scores of 25 observations + Sum of the scores of another 55 observations}{Total number of observations}$

$= \frac{2000 + 3300}{80} = \frac{5300}{80} = 66.25$

Therefore, the mean score of the whole set of observations is 66.25.

Question 17:

Arun scored 36 marks in English, 44 marks in Hindi, 75 marks in mathematics and x marks in science. If he has secured an average of 50 marks, find the value of x.

Answer:

Marks scored by Arun in English = 36
Marks scored by Arun in Hindi = 44
Marks scored by Arun in mathematics = 75
Marks scored by Arun in science = x
Average marks = 50
Thus, we have:
$Average marks = \frac{36 + 44 + 75 + x}{4} \Rightarrow 50 = \frac{155 + x}{4}$
$\Rightarrow 155 + x = 200 \Rightarrow x = 200 - 155 = 45$

∴ Marks scored by Arun in science = 45

Question 18:

The mean monthly salary paid to 75 workers in a factory is Rs 5680. The mean salary of 25 of them is Rs 5400 and that of 30 others is Rs 5700. Find the mean salary of the remaining workers.

Answer:

Mean monthly salary paid to 75 workers = Rs 5680
Mean salary of 25 workers = Rs 5400
Sum of the monthly salaries paid to 25 workers = Rs (5400 $\times$ 25) = Rs 135000
Mean salary of the other 30 workers = Rs 5700
Sum of the monthly salaries of the other 30 workers = Rs (5700 $\times$ 30) = Rs 171000
Let the sum of the salaries of the remaining 30 workers be Rs x.
$Mean monthly salary paid to 75 workers = \frac{135000 + 171000 + x}{75} \Rightarrow 5680 = \frac{306000 + x}{75} \Rightarrow 426000 = 306000 + x \Rightarrow x = 120000 Therefore, the sum of the salaries of the remaining 20 workers is Rs 120000 . Mean salary of the remaining 20 workers = \frac{Sum of the salaries of 20 workers}{20} = \frac{120000}{20} = Rs 6000 Hence, the mean salary of the remaining 20 workers is Rs 6000 .$

Question 19:

A ship sails out to an island at the rate of 15 km/h and sails back to the starting point at 10 km/h. Find the average sailing speed for the whole journey.

Answer:

Let the distance from the starting point to the island be x km.
Speed of the ship sailing out to the island = 15 km/h
Speed of the ship sailing back to the starting point = 10 km/h
We know:
$Time = \frac{Distance}{Speed} Time taken by the ship to travel from the starting point to the island = \frac{x}{15} h Time taken by the ship to travel from the island to the starting point = \frac{x}{10} h Average speed = \frac{Total distance travelled}{Total time taken} = \frac{x + x}{\frac{x}{15} + \frac{x}{10}} = \frac{2 x}{\frac{2 x + 3 x}{30}} = \frac{2 x}{\frac{5 x}{30}} = \frac{60}{5} = 12 km / h$

Therefore, the average speed of the ship in the whole journey was 12 km/h.

Question 20:

There are 50 students in a class, of which 40 are boys. The average weight of the class is 44 kg an that of the girls is 40 kg. Find the average weight of the boys.

Answer:

Total students in the class = 50
Number of boys = 40
∴ Number of girls = (50 $-$ 40) = 10
Average weight of students in the class = 44 kg
Average weight of girls in the class = 40 kg
Sum of the weights of girls in the class = (40 $\times$ 10) kg = 400 kg
Thus, we have:

$Average weight of students in the class = \frac{Total weight of girls in the class + Total weight of boys in the class}{Total number of students in the class} \Rightarrow 44 = \frac{400 + Total weight of boys in the class}{50}$
$\Rightarrow Total weight of boys + 400 = 2200 \Rightarrow Total weight of boys = 1800 kg$
$Average weight of boys = \frac{1800}{40} = 45 kg$

Question 1:

Find the mean of daily wages of 60 workers in a factory as per data given below:

Daily wages (in Rs)	90	110	120	130	150
No. of workers	12	14	13	11	10

Answer:

Daily Wages (x_i)	No. of Workers (f_i)	(f_i)(x_i)
90	12	1080
110	14	1540
120	13	1560
130	11	1430
150	10	1500
	$\sum_{}^{} f_{i} =$ 60	$\sum_{}^{} f_{i} x_{i} =$ 7110

Thus, we have:

Mean = \frac{\sum_{}^{} f_{i} x_{i}}{\sum_{}^{} x_{i}}

= \frac{7110}{60} = Rs 118.50

Question 2:

The following table shows the weights of 12 workers in a factory:

Weight (in kg)	60	63	66	69	72
No. of workers	4	3	2	2	1

Find the mean weight of the workers.

Answer:

We will make the following table:

Weight (x_i)	No. of Workers (f_i)	(f_i)(x_i)
60	4	240
63	3	189
66	2	132
69	2	138
72	1	72
	$\sum_{}^{} f_{i} =$ 12	$\sum_{}^{} f_{i} x_{i} =$ 771

Thus, we have:

$Mean = \frac{{\sum f_{i} x_{i}}_{}^{}}{\sum_{}^{} x_{i}}$
= $\frac{771}{12} = 64.25 kg$

Question 3:

The following table shows the weights of 12 workers in a factory:

Weight (in kg)	60	63	66	69	72
No. of workers	4	3	2	2	1

Find the mean weight of the workers.

Answer:

We will make the following table:

Age (x_i)	Frequency (f_i)	(f_i)(x_i)
15	3	45
16	8	128
17	9	153
18	11	198
19	6	114
20	3	60
	$\sum_{}^{} f_{i} =$ 40	$\sum_{}^{} f_{i} x_{i} =$ 698

Thus, we have:

Mean = \frac{{\sum f_{i} x_{i}}_{}^{}}{\sum_{}^{} x_{i}}

= \frac{698}{40} = 17.45 years

Question 4:

Find the mean of the following frequency distribution:

Variable (x_i)	10	30	50	70	89
Frequency (f_i)	7	8	10	15	10

Answer:

We will make the following table:

Variable (x_i)	Frequency (f_i)	(f_i)(x_i)
10	7	70
30	8	240
50	10	500
70	15	1050
89	10	890
	$\sum_{}^{} f_{i} =$ 50	$\sum_{}^{} f_{i} x_{i} =$ 2750

Thus, we have:

Mean = \frac{{\sum f_{i} x_{i}}_{}^{}}{\sum_{}^{} x_{i}}

= \frac{2750}{50} = 55

Question 5:

If the mean of the following frequency distribution is 8, find the value of p.

x	3	5	7	9	11	13
f	6	8	15	p	8	4

Answer:

We will make the following table:

(x_i)	(f_i)	(f_i)(x_i)
3	6	18
5	8	40
7	15	105
9	p	9p
11	8	88
13	4	52
	$\sum_{}^{} f_{i} =$ 41 + p	$\sum_{}^{} f_{i} x_{i} =$ 303 + 9p

We know:

Mean = \frac{{\sum f_{i} x_{i}}_{}^{}}{\sum_{}^{} x_{i}}

Given:
Mean = 8
Thus, we have:

8 = \frac{303 + 9 p}{41 + p} \Rightarrow 328 + 8 p = 303 + 9 p \Rightarrow p = 25

Question 6:

Find the missing frequency p for the following frequency distribution whose mean is 28.25.

x	15	20	25	30	35	40
f	8	7	p	14	15	6

Answer:

We will prepare the following table:

(x_i)	(f_i)	(f_i)(x_i)
15	8	120
20	7	140
25	p	25p
30	14	420
35	15	525
40	6	240
	$\sum_{}^{} f_{i} =$ 50 + p	$\sum_{}^{} f_{i} x_{i} =$ 1445 + 25p

Thus, we have:

Mean = \frac{{\sum f_{i} x_{i}}_{}^{}}{\sum_{}^{} x_{i}}

\Rightarrow 28.25 = \frac{1445 + 25 p}{50 + p} \Rightarrow 28.25 (50 + p) = (1445 + 25 p) \Rightarrow 1412.5 + 28.25 p = 1445 + 25 p \Rightarrow 3.25 p = 32.5 \Rightarrow p = 10

Question 7:

Find the value of p for the following frequency distribution whose mean is 16.6

x	8	12	15	p	20	25	30
f	12	16	20	24	16	8	4

Answer:

We will make the following table:

(x_i)	(f_i)	(f_i)(x_i)
8	12	96
12	16	192
15	20	300
p	24	24p
20	16	320
25	8	200
30	4	120
	$\sum_{}^{} f_{i} =$ 100	$\sum_{}^{} f_{i} x_{i} =$ 1228 + 24p

Thus, we have:

Mean = \frac{{\sum f_{i} x_{i}}_{}^{}}{\sum_{}^{} x_{i}}

\Rightarrow 16.6 = \frac{(1228 + 24 p)}{100} \Rightarrow 16.6 \times 100 = (1228 + 24 p) \Rightarrow 1660 = 1228 + 24 p

\Rightarrow 24 p = 432 \Rightarrow p = 18

Question 8:

Find the missing frequencies in the following frequency distribution, whose mean is 50.

x	10	30	50	70	90	Total
f	17	f₁	32	f₂	19	120

Answer:

We will prepare the following table:

(x_i)	(f_i)	(f_i)(x_i)
10	17	170
30	f₁	30f₁
50	32	1600
70	f₂	70f₂
90	19	1710
	$\sum_{}^{} f_{i} =$ 120	$\sum_{}^{} f_{i} x_{i} =$ 3480 + 30f₁₊ 70f₂

Thus, we have:

M e a n = \frac{{\sum f_{i} x_{i}}_{}^{}}{\sum_{}^{} x_{i}}

\Rightarrow 50 = \frac{3480 + 30 f_{1} + 70 f_{2}}{120}

_{$\Rightarrow 6000 = 3480 + 30 f_{1} + 70 f_{2} \Rightarrow 30 f_{1} + 70 f_{2} = 2520 . . . . . (i)$}

Also,
Given:
17 + f₁ + 32 + f₂ + 19 = 120
⇒ 68 + f₁ + f₂ = 120
⇒ f₁+ f₂ = 52
or, f₂ = 52

-

f₁ ...(ii)
By putting the value of f₂ in (i), we get:
2520 = 30f₁ + 70(52

-

f₁)
⇒ 2520 = 30f₁+ 3640

-

70f₁
⇒ 40f₁= 1120
⇒ f₁= 28
Substituting the value in (ii), we get:
f₂ = 52

-

f₁ = 52

-

28 = 24

Question 9:

Use the assumed-mean method to find the mean weekly wages from the data given below.

Weekly wages (in Rs)	800	820	860	900	920	980	1000
No. of workers	7	14	19	25	20	10	5

Answer:

Let the assumed mean be A = 900.
Now, we will prepare the following table:

Weekly Wages (x_i)	No. of Workers (f_i)	d_i = (x_i $-$ A)=(x_i $-$ 900)	(f_i)(d_i)
800	7	$-$ 100	$-$ 700
820	14	$-$ 80	$-$ 1120
860	19	$-$ 40	$-$ 760
900	25	0	0
920	20	20	400
980	10	80	800
1000	5	100	500
	$\sum_{}^{} f_{i} =$ 100		$\sum_{}^{} f_{i} d i =$ $-$ 880

Thus, we have:

Mean = x = A + \frac{{\sum f_{i} d_{i}}_{}^{}}{\sum_{}^{} f_{i}}

= 900 + \frac{(- 880)}{100} = 900 - \frac{88}{10} = \frac{9000 - 88}{10} = \frac{8912}{10} = Rs 891.2

Question 10:

Use the assumed-mean method to find the mean height of the plants from the following frequency-distribution table.

Height in cm (x_i)	61	64	67	70	73
No. of plants (f_i)	5	18	42	27	8

Answer:

Let the assumed mean be A = 67.
Now, we will make the following table:

Height in cm (x_i)	No. of Plants (f_i)	d_i = (x_i $-$ A)= (x_i $-$ 67)	(f_i)(d_i)
61	5	$-$ 6	$-$ 30
64	18	$-$ 3	$-$ 54
67	42	0	0
70	27	3	81
73	8	6	48
	$\sum_{}^{} f_{i} =$ 100		$\sum_{}^{} f_{i} d i =$ 45

Thus, we have:

Mean = x = A + \frac{{\sum f_{i} d_{i}}_{}^{}}{\sum_{}^{} f_{i}}

= 67 + \frac{(45)}{100} = \frac{6700 + 45}{100} = \frac{6745}{100} = 67.45 cm

Question 11:

Use the step-deviation method to find the arithmetic mean from the following data.

x	18	19	20	21	22	23	24
f	170	320	530	700	230	140	110

Answer:

Here,
h = 1
Let the assumed mean be A = 21.
Now, we will prepare the following table:

(x_i)	(f_i)	u_i = $\frac{(x_{i} - 21)}{1}$	(u_i)(f_i)
18	170	$-$ 3	$-$ 510
19	320	$-$ 2	$-$ 640
20	530	$-$ 1	$-$ 530
21 = A	700	0	0
22	230	1	230
23	140	2	280
24	110	3	330
	$\sum_{}^{} f_{i} =$ 2200		$\sum_{}^{} f_{i} u_{i} =$ $-$ 840

Thus, we have:

Mean = x = A + (h \times \frac{{\sum f_{i} u_{i}}_{}^{}}{\sum_{}^{} f_{i}})

= 21 + [1 \times \frac{(- 840)}{2200}] = 21 - \frac{840}{2200} = \frac{46200 - 840}{2200} = \frac{45360}{2200} = 20.62

Question 12:

The table given below gives the distribution of villages and their heights from the sea level in a certain region.

Height (in metres)	200	600	1000	1400	1800	2200
No. of villages	142	265	560	271	89	16

Compute the mean height, using the step-deviation method.

Answer:

Here,
h = 400
Let the assumed mean be A = 1400.
We will prepare the following table:

(x_i)	(f_i)	u_i = $\frac{(x_{i} - 1400)}{400}$	(f_i)(u_i)
200	142	$-$ 3	$-$ 426
600	265	$-$ 2	$-$ 530
1000	560	$-$ 1	$-$ 560
1400 = A	271	0	0
1800	89	1	89
2200	16	2	32
	$\sum_{}^{} f_{i} =$ 1343		$\sum_{}^{} f_{i} u_{i} =$ $-$ 1395

We have:

Mean = x = A + (h \times \frac{{\sum f_{i} u_{i}}_{}^{}}{\sum_{}^{} f_{i}})

= 1400 + (400 \times \frac{- 1395}{1343}) = 1400 - \frac{558000}{1343} = \frac{1880200 - 558000}{1343} = \frac{1322200}{1343} = 984.51 m

Question 1:

Find the median of
(i) 2, 10, 9, 9, 5, 2, 3, 7, 11
(ii) 15, 6, 16, 8, 22, 21, 9, 18, 25
(iii) 20, 13, 18, 25, 6, 15, 21, 9, 16, 8, 22
(iv) 7, 4, 2, 5, 1, 4, 0, 10, 3, 8, 5, 9, 2

Answer:

(i) Arranging the numbers in ascending order, we get:
2, 2, 3, 5, 7, 9, 9, 10, 11
Here, n is 9, which is an odd number.
If n is an odd number, we have:
$Median = Value of (\frac{n + 1}{2}) th observation$
Now,
$Median = Value of (\frac{9 + 1}{2}) th observation = Value of the 5 th observation = 7$

(ii) Arranging the numbers in ascending order, we get:
6, 8, 9, 15, 16, 18, 21, 22, 25
Here, n is 9, which is an odd number.
If n is an odd number, we have:
$Median = Value of (\frac{n + 1}{2}) th observation$
Now,
$Median = Value of (\frac{9 + 1}{2}) th observation = Value of the 5 th observation = 16$

(iii) Arranging the numbers in ascending order, we get:
6, 8, 9, 13, 15, 16, 18, 20, 21, 22, 25
Here, n is 11, which is an odd number.
If n is an odd number, we have:
$Median = Value of (\frac{n + 1}{2}) th observation$
Now,
$Median = Value of (\frac{11 + 1}{2}) th observation = Value of the 6 th observation = 16$

(iv) Arranging the numbers in ascending order, we get:
0, 1, 2, 2, 3, 4, 4, 5, 5, 7, 8, 9, 10
Here, n is 13, which is an odd number.
If n is an odd number, we have:
$Median = Value of (\frac{n + 1}{2}) th observation$
Now,
$Median = Value of (\frac{13 + 1}{2}) th observation = Value of the 7 th observation = 4$

Question 2:

Find the median of
(i) 17, 19, 32, 10, 22, 21, 9, 35
(ii) 72, 63, 29, 51, 35, 60, 55, 91, 85, 82
(iii) 10, 75, 3, 15, 9, 47, 12, 48, 4, 81, 17, 27

Answer:

(i) Arranging the numbers in ascending order, we get:
9, 10, 17, 19, 21, 22, 32, 35
Here, n is 8, which is an even number.
If n is an even number, we have:
$Median = Mean of (\frac{n}{2}) th & (\frac{n}{2} + 1) th observations$
Now,
$Median = Mean of (\frac{8}{2}) th & (\frac{8}{2} + 1) th observations = Mean of the 4 th & 5 th observations = \frac{1}{2} (19 + 21) = 20$

(ii) Arranging the numbers in ascending order, we get:
29, 35, 51, 55, 60, 63, 72, 82, 85, 91
Here, n is 10, which is an even number.
If n is an even number, we have:
$Median = Mean of (\frac{n}{2}) th & (\frac{n}{2} + 1) th observations$
Now,
$Median = Mean of (\frac{10}{2}) th & (\frac{10}{2} + 1) th observations = Mean of the 5 th & 6 th observations = \frac{1}{2} (60 + 63) = 61.5$

(iii) Arranging the numbers in ascending order, we get:
3, 4, 9, 10, 12, 15, 17, 27, 47, 48, 75, 81
Here, n is 12, which is an even number.
If n is an even number, we have:
$Median = Mean of (\frac{n}{2}) th & (\frac{n}{2} + 1) th observations$
Now,
$Median = Mean of (\frac{12}{2}) th & (\frac{12}{2} + 1) th observations = Mean of the 6 th & 7 th observations = \frac{1}{2} (15 + 17) = 16$

Question 3:

The marks of 15 students in an examination are:
25, 19, 17, 24, 23, 29, 31, 40, 19, 20, 22, 26, 17, 35, 21.
Find the median score.

Answer:

Arranging the marks of 15 students in ascending order, we get:
17, 17, 19, 19, 20, 21, 22, 23, 24, 25, 26, 29, 31, 35, 40
Here, n is 15, which is an odd number.
We know:
$Median = Value of (\frac{n + 1}{2}) th observation$
Thus, we have:
$Median score = Value of (\frac{15 + 1}{2}) th observation = Value of the 8 th observation = 23$

Question 4:

The heights (in cm) of 9 girls are:
144.2, 148.5, 143.7, 149.6, 150, 146.5, 145, 147.3, 152.1.
Find the median height.

Answer:

On arranging the heights in ascending order, we get:
143.7, 144.2, 145, 146.5, 147.3, 148.5, 149.6, 150, 152.1
Here, n is 9, which is an odd number.
Thus, we have:
$Median = Value of (\frac{n + 1}{2}) th observation$
$Median height = Value of (\frac{9 + 1}{2}) th observation = Value of the 5 th observation = 147.3 Hence, the median height is 147.3 cm .$

Question 5:

The weights (in kg) of 8 children are:
13.4, 10.6, 12.7, 17.2, 14.3, 15, 16.5, 9.8.
Find the median weight.

Answer:

Arranging the weights (in kg) in ascending order, we have:
9.8, 10.6, 12.7, 13.4, 14.3, 15, 16.5, 17.2
Here, n is 8, which is an even number.
Thus, we have:
$Median = Mean of (\frac{n}{2}) th & (\frac{n}{2} + 1) th observations$
$Median weight = Mean of (\frac{8}{2}) th & (\frac{8}{2} + 1) th observations = Mean of 4 th & 5 th observations = \frac{1}{2} (13.4 + 14.3) = 13.85$

Hence, the median weight is 13.85 kg.

Question 6:

The ages (in years) of 10 teachers in a school are:
32, 44, 53, 47, 37, 54, 34, 36, 40, 50.
Find the median age.

Answer:

Arranging the ages (in years) in ascending order, we have:
32, 34, 36, 37, 40, 44, 47, 50, 53, 54
Here, n is 10, which is an even number.
Thus, we have:
$Median = Mean of (\frac{n}{2}) th & (\frac{n}{2} + 1) th observations$
$Median age = Mean of (\frac{10}{2}) th & (\frac{10}{2} + 1) th observations = Mean of 5 th & 6 th observations = \frac{1}{2} (40 + 44) = 42 Hence, the median age is 42 years .$

Question 7:

If 10, 13, 15, 18, x + 1, x + 3, 30, 32, 35, 41 are ten observation in an ascending order with median 24, find the value of x.

Answer:

10, 13, 15, 18, x+1, x+3, 30, 32, 35 and 41 are arranged in ascending order.
Median = 24
We have to find the value of x.
Here, n is 10, which is an even number.
Thus, we have:
$Median = Mean of (\frac{n}{2}) th & (\frac{n}{2} + 1) th observations$
$∴ Median = Mean of (\frac{10}{2}) th & (\frac{10}{2} + 1) th observations = Mean of 5 th & 6 th observations = \frac{1}{2} (x + 1 + x + 3) = \frac{1}{2} (2 x + 4) = (x + 2) Given : Median = 24 \Rightarrow x + 2 = 24 \Rightarrow x = 22$

Question 8:

Find the median weight for the following data.

Weight (in kg)	45	46	48	50	52	54	55
Number of students	8	5	6	9	7	4	2

Answer:

The following data is in ascending order:

Weight (in kg)	Frequency	Cumulative Frequency
45	8	8
46	5	13
48	6	19
50	9	28
52	7	35
54	4	39
55	2	41

Here, n is 41, which is an odd number.
Thus, we have:

Median = Value of (\frac{n + 1}{2}) th observation

Median weight = Weight of (\frac{41 + 1}{2}) th student = Weight of the 21 st student Here, we see that the weight of the 20 th to 28 th students is 50 kg . This means that the weight of the 21 st student is 50 kg . Hence, the median weight is 50 kg .

Question 9:

Find the median for the following frequency distribution.

Variate	17	20	22	15	30	25
Frequency	5	9	4	3	10	6

Answer:

We will first arrange the given data in ascending order as follows:

Variate	Frequency
15	3
17	5
20	9
22	4
25	6
30	10

Now, we will make the cumulative frequency table as follows:

Variate	Frequency	Cumulative Frequency
15	3	3
17	5	8
20	9	17
22	4	21
25	6	27
30	10	37

Here, n is 37, which is an odd number.
Thus, we have:

Median = Value of (\frac{n + 1}{2}) th observation

Median = Value of (\frac{37 + 1}{2}) th observation = Value of 19 th observation Here, we see that value of the 18 th to 21 st observations is 22 . Therefore, the value of the 19 th observation is 22 . ∴ Median = 22

Question 10:

Calculate the median for the following data.

Marks	20	9	25	50	40	80
Number of students	6	4	16	7	8	2

Answer:

We will first arrange the given data in ascending order as follows:

Marks	Number of Students
9	4
20	6
25	16
40	8
50	7
80	2

Now, we will prepare the cumulative frequency table as follows:

Marks	Frequency	Cumulative Frequency
9	4	4
20	6	10
25	16	26
40	8	34
50	7	41
80	2	43

Here, n is 43, which is an odd number.
Thus, we have:

Median = Value of (\frac{n + 1}{2}) th observation

Median marks = Marks obtained by (\frac{43 + 1}{2}) th student = Marks obtained by the 22 nd student Here, we see that 25 marks are obtained by 11 th to 26 th students . Marks obtained by the 22 nd student = 25 Median marks = 25

Question 11:

The heights (in cm) of 50 students of a class are given below:

Height (in cm)	156	154	155	151	157	152	153
Number of students	8	4	10	6	7	3	12

Find the median height.

Answer:

We will first arrange the given data in ascending order as follows:

Height (in cm)	Number of Students
151	6
152	3
153	12
154	4
155	10
156	8
157	7

Now, we will prepare the cumulative frequency table as follows:

Height (in cm)	Frequency	Cumulative Frequency
151	6	6
152	3	9
153	12	21
154	4	25
155	10	35
156	8	43
157	7	50

Here, n is 50, which is an even number.
Thus, we have:

Median = Mean of (\frac{n}{2}) th & (\frac{n}{2} + 1) th observations

Median height = Mean of the heights of (\frac{50}{2}) th and (\frac{50}{2} + 1) th students = Mean of the heights of the 25 th and 26 th students Here, we see that the height of the 25 th student is 154 cm & the height of the 26 th to 35 th students is 155 cm . Median height = \frac{1}{2} (154 + 155) cm = 154.5 cm

Question 12:

Find the median for the following data.

Variate	23	26	20	30	28	25	18	16
Frequency	4	6	13	5	11	4	8	9

Answer:

We will first arrange the given data in ascending order.

Variate	Frequency
16	9
18	8
20	13
23	4
25	4
26	6
28	11
30	5

Now, we will prepare a cumulative frequency table.

Variate	Frequency	Cumulative Frequency
16	9	9
18	8	17
20	13	30
23	4	34
25	4	38
26	6	44
28	11	55
30	5	60

Here, n is 60, which is an even number.
Now,

Median = Mean of (\frac{n}{2}) th & (\frac{n}{2} + 1) th observations

Median = Mean of (\frac{60}{2}) th & (\frac{60}{2} + 1) th observations = Mean of 30 th & 31 st observations Value of the 30 th observation is 20 & value of 31 st to 34 th observations is 23 . ∴ Median = \frac{1}{2} (20 + 23) = 21.5

Question 1:

Find the mode of the following items.
0, 6, 5, 1, 6, 4, 3, 0, 2, 6, 5, 6

Answer:

On arranging the items in ascending order, we get:
0, 0, 1, 2, 3, 4, 5, 5, 6, 6, 6, 6
Clearly, 6 occurs maximum number of times.
∴ Mode = 6

Question 2:

Determine the mode of the following values of a variable.
23, 15, 25, 40, 27, 25, 22, 25, 20

Answer:

On arranging the values in ascending order, we get:
15, 20, 22, 23, 25, 25, 25, 27, 40
Clearly, 25 occurs maximum number of times.
∴ Mode = 25

Question 3:

Calculate the mode of the following sizes of shoes sold by a shop on a particular day.
5, 9, 8, 6, 9, 4, 3, 9, 1, 6, 3, 9, 7, 1, 2, 5, 9

Answer:

On arranging the shoe sizes in ascending order, we get:
1, 1, 2, 3, 3, 4, 5, 5, 6, 6, 7, 8, 9, 9, 9, 9, 9
Clearly, 9 occurs maximum number of times.
∴ Mode = 9

Question 4:

A cricket player scored the following runs in 12 one-day matches:
50, 30, 9, 32, 60, 50, 28, 50, 19, 50, 27, 35.
Find his modal score.

Answer:

On arranging the runs in ascending order, we get:
9, 19, 27, 28, 30, 32, 35, 50, 50, 50, 50, 60
Clearly, 50 occurs maximum number of times.
∴ Modal score = 50

Question 5:

Calculate the mode of each of the following using the empirical formula.
17, 10, 12, 11, 10, 15, 14, 11, 12, 13

Answer:

We have:
17, 10, 12, 11, 10, 15, 14, 11, 12, 13

We know:

$Mean = \frac{Sum of all observations}{Number of observations} = \frac{17 + 10 + 12 + 11 + 10 + 15 + 14 + 11 + 12 + 13}{10} = \frac{125}{10} = 12.5 Here, the number of observations is even . Also, on arranging the numbers in ascending order, we get : 10, 10, 11, 11, 12, 12, 13, 14, 15, 17 Thus, we have : Median = \frac{1}{2} [{(\frac{n}{2})}^{th} term + {(\frac{n}{2} + 1)}^{th} term] = \frac{1}{2} [{(\frac{10}{2})}^{th} term + {(\frac{10}{2} + 1)}^{th} term] = \frac{1}{2} [{(5)}^{th} term + {(6)}^{th} term] = \frac{1}{2} [12 + 12] = 12$

By the empirical formula, we have:
Mode = 3 Median $-$ 2 Mean
Or,
Mode = 3(12) $-$ 2(12.5) = 11

Question 6:

Calculate the mode of each of the following using the empirical formula.

Marks	10	11	12	13	14	16	19	20
Number of students	3	5	4	5	2	3	2	1

Answer:

Marks (x_i)	No. of Students (f_i)	Cumulative Frequency	(x_i)(f_i)
10	3	3	30
11	5	8	55
12	4	12	48
13	5	17	65
14	2	19	28
16	3	22	48
19	2	24	38
20	1	25	20
	n = 25		$\sum_{}^{} f_{i} x_{i} =$ 332

The number of observations is 25, which is an odd number . Thus, we have : Median = [{(\frac{n + 1}{2})}^{th} term] = [{(\frac{26}{2})}^{th} term]

^{$= {(13)}^{th} term = 13$}

Also,

Mean = \frac{\sum_{}^{} f_{i} x_{i}}{\sum_{}^{} f_{i}} = \frac{332}{25} = 13.28

By the empirical formula, we have:
Mode = 3 Median

-

2 Mean
Or,
Mode = 3(13)

-

2(13.28) = 12.44

Question 7:

Calculate the mode of each of the following using the empirical formula.

Item (x)	5	7	9	12	14	17	19	21
Frequency (f)	6	5	3	6	5	3	2	4

Answer:

Item (x_i)	Frequency (f_i)	Cumulative Frequency	(x_i)(f_i)
5	6	6	30
7	5	11	35
9	3	14	27
12	6	20	72
14	5	25	70
17	3	28	51
19	2	30	38
21	4	34	84
	n = 34		$\sum_{}^{} f_{i} x_{i} =$ 407

The number of observations is 34, which is an even number . Thus, we have : Median = \frac{1}{2} [{(\frac{n}{2})}^{th} term + {(\frac{n}{2} + 1)}^{th} term] = \frac{1}{2} [{(\frac{34}{2})}^{th} term + {(\frac{34}{2} + 1)}^{th} term] = \frac{1}{2} [{(17)}^{th} term + {(18)}^{th} term] = \frac{1}{2} [12 + 12] = 12

Also,

Mean = \frac{\sum_{}^{} f_{i} x_{i}}{\sum_{}^{} f_{i}} = \frac{407}{34} = 11.97

By the empirical formula, we have:
Mode = 3 Median

-

2 Mean
Or,
Mode = 3(12)

-

2(11.97) = 12.06

Question 8:

Calculate the mode of each of the following using the empirical formula.

x	18	20	25	30	34	38	40
f	6	7	3	7	7	5	5

Answer:

(x_i)	(f_i)	Cumulative Frequency	(x_i)(f_i)
18	6	6	108
20	7	13	140
25	3	16	75
30	7	23	210
34	7	30	238
38	5	35	190
40	5	40	200
	n = 40		$\sum_{}^{} f_{i} x_{i} =$ 1161

The number of observations is 40, which is an even number . Thus, we have : Median = \frac{1}{2} [{(\frac{n}{2})}^{th} term + {(\frac{n}{2} + 1)}^{th} term] = \frac{1}{2} [{(\frac{40}{2})}^{th} term + {(\frac{40}{2} + 1)}^{th} term] = \frac{1}{2} [20 th term + 21 st term] = \frac{1}{2} [30 + 30] = 30

Also,

Mean = \frac{\sum_{}^{} f_{i} x_{i}}{\sum_{}^{} f_{i}} = \frac{1161}{40} = 29.025 \approx 29

By the empirical formula, we have:
Mode = 3 Median

-

2 Mean
Or,
Mode = 3(30)

-

2(29) = 32

Question 9:

The table given below shows the weights (in kg) of 50 persons:

Weight (in kg)	42	47	52	57	62	67	72
Number of persons	3	8	6	8	11	5	9

Find the mean, median and mode.

Answer:

Weight (x_i)	Number of Persons (f_i)	Cumulative Frequency	(x_i)(f_i)
42	3	3	126
47	8	11	376
52	6	17	312
57	8	25	456
62	11	36	682
67	5	41	335
72	9	50	648
	n = 50		$\sum_{}^{} f_{i} x_{i} =$ 2935

The number of observations is 50, which is an even number . Thus, we have : Median = \frac{1}{2} [{(\frac{n}{2})}^{th} term + {(\frac{n}{2} + 1)}^{th} term] = \frac{1}{2} [{(\frac{50}{2})}^{th} term + {(\frac{50}{2} + 1)}^{th} term] = \frac{1}{2} [{(25)}^{th} term + {(26)}^{th} term] = \frac{1}{2} [57 + 62] = 59.5

Also,

Mean = \frac{\sum_{}^{} f_{i} x_{i}}{\sum_{}^{} f_{i}} = \frac{2935}{50} = 58.7

By the empirical formula, we have:
Mode = 3 Median

-

2 Mean
Or,
Mode = 3(59.5)

-

2(58.7) = 61.1

Question 10:

The marks obtained by 80 students in test are given below:

Marks	4	12	20	28	36	44
Number of students	8	10	16	24	15	7

Find the modal marks.

Answer:

Marks (x_i)	Number of Students (f_i)	Cumulative Frequency	(x_i)(f_i)
4	8	8	32
12	10	18	120
20	16	34	320
28	24	58	672
36	15	73	540
44	7	80	308
	n = 80		$\sum_{}^{} f_{i} x_{i} =$ 1992

The number of observations is 80, which is an even number . Thus, we have : Median = \frac{1}{2} [{(\frac{n}{2})}^{th} term + {(\frac{n}{2} + 1)}^{th} term] = \frac{1}{2} [{(\frac{80}{2})}^{th} term + {(\frac{80}{2} + 1)}^{th} term] = \frac{1}{2} [{(40)}^{th} term + {(41)}^{st} term] = \frac{1}{2} [28 + 28] = 28

Also,

Mean = \frac{\sum_{}^{} f_{i} x_{i}}{\sum_{}^{} f_{i}} = \frac{1992}{80} = 24.9

By the empirical formula, we have:
Mode = 3 Median

-

2 Mean
Or,
Mode = 3(28)

-

2(24.9) = 34.2

Question 11:

The ages of the employees of a company are given below:

Age (in years)	19	21	23	25	27	29	31
Number of persons	13	15	16	18	16	15	13

Find the mean, median and mode for the above data.

Answer:

Age (x_i)	Number of Persons (f_i)	Cumulative Frequency	(x_i)(f_i)
19	13	13	247
21	15	28	315
23	16	44	368
25	18	62	450
27	16	78	432
29	15	93	435
31	13	106	403
	n = 106		$\sum_{}^{} f_{i} x_{i} =$ 2650

The number of observations is 106, which is an even number . Thus, we have : Median = \frac{1}{2} [{(\frac{n}{2})}^{th} term + {(\frac{n}{2} + 1)}^{th} term] = \frac{1}{2} [{(\frac{106}{2})}^{th} term + {(\frac{106}{2} + 1)}^{th} term] = \frac{1}{2} [{(53)}^{rd} term + {(54)}^{th} term] = \frac{1}{2} [25 + 25] = 25

Also,

Mean = \frac{\sum_{}^{} f_{i} x_{i}}{\sum_{}^{} f_{i}} = \frac{2650}{106} = 25

By the empirical formula, we have:
Mode = 3 Median

-

2 Mean
Or,
Mode = 3(25)

-

2(25) = 25

Question 12:

The following table shows the weights of 12 students:

Weight (in kg)	47	50	53	56	60
Number of students	4	3	2	2	4

Find the mean, median and mode for the above data.

Answer:

Weight (x_i)	Number of Students (f_i)	Cumulative Frequency	(x_i)(f_i)
47	4	4	188
50	3	7	150
53	2	9	106
56	2	11	112
60	4	15	240
	n = 15		$\sum_{}^{} f_{i} x_{i} =$ 796

The number of observations is 15, which an odd number . Thus, we have : Median = [{(\frac{n + 1}{2})}^{th} term] = [{(\frac{15 + 1}{2})}^{th} term] = [8^{th} term] = 53

Also,

Mean = \frac{\sum_{}^{} f_{i} x_{i}}{\sum_{}^{} f_{i}} = \frac{796}{15} = 53.06

By the empirical formula, we have:
Mode = 3 Median

-

2 Mean
Or,
Mode = 3(53)

-

2(53.06) = 52.88

Question 1:

The range of the data
12, 25, 15, 18, 17, 20, 22, 6, 16, 11, 8, 19, 10, 30, 20, 32 is
(a) 10
(b) 15
(c) 18
(d) 26

Answer:

(d) 26

We have:
Maximum value = 32
Minimum value = 6
We know:
Range = Maximum value $-$ Minimum value
=32 $-$ 6
=26

Question 2:

The class mark of the class 100−120 is
(a) 100
(b) 110
(c) 115
(d) 120

Answer:

(b) 110

Class mark = $\frac{Upper limit + Lower limit}{2} = \frac{120 + 100}{2} = 110$

Question 3:

In the class intervals 10−20, 20−30, the number 20 is included in
(a) 10−20
(b) 20−30
(c) in each of 10−20 and 20−30
(d) in none of 10−20 and 20−30

Answer:

(b) 20−30
This is the continuous form of frequency distribution. Here, the upper limit of each class is excluded, while the lower limit is included. So, the number 20 is included in the class interval 20−30.

Question 4:

The class marks of a frequency distribution are 15, 20, 25, 30, .. . The class corresponding to the class marks 20 is
(a) 12.5−17.5
(b) 17.5−22.5
(c) 18.5−21.5
(d) 19.5−20.5

Answer:

(b) 17.5 $-$ 22.5

We are given frequency distribution 15, 20, 25, 30,...
Class size = 20 $-$ 15 = 5
Class marks = 20
Now,
$Lower limit = (20 - \frac{5}{2}) = \frac{35}{2} = 17.5 Upper limit = (20 + \frac{5}{2}) = \frac{45}{2} = 22.5$

Thus, the required class is 17.5 $-$ 22.5.

Question 5:

In a frequency distribution, the mid-value of a class is 10 and width of each class is 6. The lower limit of the class is
(a) 6
(b) 7
(c) 8
(d) 12

Answer:

(b) 7

Given:
Mid value of the class = 10
Width of each class = 6
Now,
Let the lower limit be x.
We know:
Upper limit = Lower limit + Class size
= x + 6
Also,
$Mid value = \frac{x + x + 6}{2} = \frac{2 x + 6}{2} = x + 3 \Rightarrow x + 3 = 10 \Rightarrow x = 7$

Thus, the lower limit is 7.

Question 6:

Let m be the midpoint and u be the upper class limit of a class in a continuous frequency distribution. The lower class limit of the class is
(a) 2m − u
(b) 2m + u
(c) m − u
(d) m + u

Answer:

(a) 2m $-$ u

Given:
Mid value = m
Upper limit = u

We know:
$\frac{Lower limit + Upper limit}{2} = Mid value \Rightarrow \frac{Lower limit + u}{2} = m \Rightarrow Lower limit + u = 2 m \Rightarrow Lower limit = 2 m - u$

Question 7:

The width of each of the five continuous classes in a frequency distribution is 5 and the lower class limit of the lowest class is 10. The upper class limit of the highest class is
(a) 45
(b) 25
(c) 35
(d) 40

Answer:

(c) 35

We have:
Class width = 5
Lower class limit of the lowest class = 10
Now,
Upper class limit of the highest class = 10 + 5 $\times$ 5 = 35

Question 8:

Let L be the lower class boundary of a class in a frequency distribution and m be the midpoint of the class. Which one of the following is the upper class boundary of the class?
(a) $m + \frac{(m + L)}{2}$
(b) $L + \frac{m + L}{2}$
(c) $2 m - L$
(d) $m - 2 L$

Answer:

(c) 2m $-$ L

$Mid value = \frac{Lower limit + Upper limit}{2} \Rightarrow m = \frac{L + U}{2} \Rightarrow U = 2 m - L ∴ Upper class boundary of the class = 2 m - L$

Question 9:

The mid-value of a class interval is 42 and the class size is 10. The lower and upper limits are
(a) 37−47
(b) 37.5−47.5
(c) 36.5−47.5
(d) 36.5−46.5

Answer:

(a) 37–47

Let the lower limit be x.
Here,
Class size = 10
∴ Upper limit = Class size + Lower limit
Upper limit = (x + 10)
Mid value of the class interval = 42

$∴ \frac{x + x + 10}{2} = 42 \Rightarrow \frac{2 x + 10}{2} = 42 \Rightarrow 2 x + 10 = 84 \Rightarrow 2 x = 74 \Rightarrow x = 37 Thus, we have : Lower limit = 37 Upper limit = 37 + 10 = 47$

Question 10:

If the mean of five observations x, x + 2, x + 4, x + 6 and x + 8 is 11, then the value of x is
(a) 5
(b) 6
(c) 7
(d) 8

Answer:

(c) 7

Mean of 5 observations = 11
We know:
$Mean = \frac{Sum of all observations}{Total number of observations} \Rightarrow 11 = \frac{x + x + 2 + x + 4 + x + 6 + x + 8}{5} \Rightarrow 11 = \frac{5 x + 20}{5} \Rightarrow 5 x + 20 = 55 \Rightarrow 5 x = 35 \Rightarrow x = 7$

Question 11:

If the mean of x, x + 3, x + 5, x + 7, x + 10 is 9, the mean of the last three observations is
(a) $10 \frac{1}{3}$
(b) $10 \frac{2}{3}$
(c) $11 \frac{1}{3}$
(d) $11 \frac{2}{3}$

Answer:

(c) 11 $\frac{1}{3}$
Mean of 5 observations = 9
We know:
$Mean = \frac{Sum of observations}{Total number of observations} \Rightarrow 9 = \frac{x + x + 3 + x + 5 + x + 7 + x + 10}{5} \Rightarrow 9 = \frac{5 x + 25}{5} \Rightarrow 5 x + 25 = 45 \Rightarrow 5 x = 20 \Rightarrow x = 4 Therefore, the last three observations are (4 + 5), (4 + 7) and (4 + 10), i . e ., 9, 11 and 14 . Now, Mean of the last three terms = \frac{9 + 11 + 14}{3} = \frac{34}{3} = 11 \frac{1}{3}$

Question 12:

If $x$ is the mean of $x_{1}, x_{2}, x_{3}, . . ., x_{n},$ then $\sum_{i = 1}^{n} (x_{i} - x) = ?$
(a) −1
(b) 0
(c) 1
(d) n − 1

Answer:

(b) 0

$I f x is the mean of x_{1}, x_{2}, x_{3}, x_{4, . . .} x_{n,} then we have : \sum_{i = 1}^{n} x_{i} = x O r, \sum_{i = 1}^{n} x_{i} - x = 0$

Question 13:

If each observation of a data is increased by 5, then their mean
(a) remains the same
(b) becomes 5 times the original mean
(c) is decreased by 5
(d) is increased by 5

Answer:

(d) is increased by 5

$Let the numbers be x$ ₁, x₂_,...x_n.
$Hence, mean = \frac{x_{1} + x_{2} + . . . . + x_{n}}{n}$

Now the new numbers after increasing every number by 5 : (x₁+5) , (x₂+5)...,(x_n+5)

$New Mean = \frac{(x_{1} + 5) + (x_{2} + 5) + . . . . + (x_{n} + 5)}{n}$

$= \frac{x_{1} + x_{2} + . . . . + x_{n} + 5 n}{n}$

$= \frac{x_{1} + x_{2} + . . . . + x_{n}}{n} + 5 ∴ New mean = mean + 5$

Hence, mean is increased by 5

Question 14:

Let $x$ be the mean of $x_{1}, x_{2}, . . ., x_{n}$ and $y$ be the mean of $y_{1}, y_{2}, . . ., y_{n}$ .
If $z$ is the mean of $x_{1}, x_{2}, . . ., x_{n}, y_{1}, y_{2}, . . ., y_{n},$ then $z = ?$
(a) $(x + y)$
(b) $\frac{1}{2} (x + y)$
(c) $\frac{1}{n} (x + y)$
(d) $\frac{1}{2 n} (x + y)$

Answer:

(b) $\frac{1}{2} (x + y)$

$\bar{z} = \frac{(x_{1} + x_{2} + . . . + x_{n}) + (y_{1} + y_{2} + . . . + y_{n})}{2 n}$

$Given : \bar{x} = \frac{x_{1} + x_{2} + . . . . . x_{n}}{n} \Rightarrow x_{1} + x_{2} + . . . . . . + x_{n} = n \bar{x} and \bar{y} = \frac{y_{1} + y_{2} + . . . . . . + y_{n}}{n} \Rightarrow y_{1} + y_{2} + . . . . . . + y_{n} = n \bar{y} ∴ \bar{z} = \frac{n \bar{x} + n \bar{y}}{2 n} = \frac{1}{2} (\bar{x} + \bar{y})$

Question 15:

If $x$ is the mean of $x_{1}, x_{2}, . . ., x_{n},$ then for $a \neq 0$ , the mean of $a x_{1,} a x_{2}, . . ., a x_{n}, \frac{x_{1}}{a}, \frac{x_{2}}{a}, . . ., \frac{x_{n}}{a}$ is
(a) $(a + \frac{1}{a}) x$

(b) $(a + \frac{1}{a}) \frac{x}{2}$

(c) $(a + \frac{1}{a}) \frac{x}{n}$

(d) $\frac{(a + \frac{1}{a}) x}{2 n}$

Answer:

(b) $(a + \frac{1}{a}) \frac{x}{2}$
$Required mean = \frac{(a x_{1} + a x_{2} + . . . + a x_{n}) + (\frac{x_{1}}{a} + \frac{x_{2}}{a} + . . . + \frac{x_{n}}{a})}{2 n} = \frac{1}{2} \{\frac{a (x_{1} + x_{2} + . . . + x_{n})}{n} + \frac{\frac{1}{a} (x_{1} + x_{2} + . . . + x_{n})}{n}\} = \frac{1}{2} \{a \bar{x} + \frac{1}{a} \bar{x}\} (\bar{x} = \frac{x_{1} + x_{2} + . . . + x_{n}}{n}) = \{a + \frac{1}{a}\} \frac{\bar{x}}{2}$

Question 16:

If $x_{1}, x_{2}, . . ., x_{n}$ are the means of n groups with $n_{1}, n_{2}, . . ., n_{n}$ number of observations respectively, then the mean $x$ of all the groups taken together is
(a) $\sum_{i = 1}^{n} n_{i} x_{i}$

(b) $\sum_{\frac{i = 1}{n^{2}}}^{n} n_{i} x_{i}$

(c) $\frac{\sum_{i = 1}^{n} n_{i} x_{i}}{\sum_{i = 1}^{n} n_{i}}$

(d) $\frac{\sum_{i = 1}^{n} n_{i} x_{i}}{2 n}$

Answer:

$(c) \frac{\sum_{i = 1}^{n} n_{i} x_{i}}{\sum_{i = 1}^{n} n_{i}}$

$Sum of all terms = n_{1} {\bar{x}}_{1} + n_{2} {\bar{x}}_{2} + . . . n_{n} {\bar{x}}_{n} Number of terms = (n_{1} + n_{2} + . . . n_{n}) ∴ Mean = \frac{\sum_{i = 1}^{n} n_{i} {\bar{x}}_{i}}{\sum_{i = 1}^{n} n_{i}}$

Question 17:

The mean weight of six boys in a groups is 48 kg. The individual weights of five of them are 51 kg, 45 kg, 49 kg, 46 kg and 44 kg. The weight of the 6th boy is
(a) 52 kg
(b) 52.8 kg
(c) 53 kg
(d) 47 kg

Answer:

(c) 53 kg

Mean weight of six boys = 48 kg
Let the weight of the 6th boy be x kg.

$We know : Mean = \frac{Sum of all observations}{Total number of observations} = \frac{51 + 45 + 49 + 46 + 44 + x}{6} = \frac{235 + x}{6} Given : Mean = 48 kg \Rightarrow \frac{235 + x}{6} = 48 \Rightarrow 235 + x = 288 \Rightarrow x = 53 Hence, the weight of the 6 th boy is 53 kg .$

Question 18:

The mean of the marks scored by 50 students was found to be 39. Latter on it was discovered that a core of 43 was misread as 23. The correct mean is
(a) 38.6
(b) 39.4
(c) 39.8
(d) 39.2

Answer:

(b) 39.4

Mean of the marks scored by 50 students = 39
Sum of the marks scored by 50 students = $(39 \times 50) = 1950$
Correct sum = (1950 + 43 $-$ 23) = 1970
$∴ Mean = \frac{1970}{50} = 39.4$

Question 19:

The mean of 100 items was found to be 64. Later on it was discovered that two items were misread as 26 and 9 instead of 36 and 90 respectively. The correct mean is
(a) 64.86
(b) 65.31
(c) 64.91
(d) 64.61

Answer:

(c) 64.91

Mean of 100 items = 64
Sum of 100 items = $64 \times 100 = 6400$
Correct sum = (6400 + 36 + 90 $-$ 26 $-$ 9) = 6491
$Correct mean = \frac{6491}{100} = 64.91$

Question 20:

The mean of 100 observations is 50. If one of the observations 50 is replaced by 150, the resulting mean will be
(a) 50.5
(b) 51
(c) 51.5
(d) 52

Answer:

(b) 51

Mean of 100 observations = 50
Sum of 100 observations = $100 \times 50 = 5000$
It is given that one of the observations, 50, is replaced by 150.
∴ New sum = (5000 $-$ 50 + 150) = 5100
And,
$Resulting mean = \frac{5100}{100} = 51$

Question 21:

The mean of 25 observations is 36. Out of these observations, the mean of first 13 is 32 and that of the last 13 is 40. The 13th observations is
(a) 23
(b) 36
(c) 38
(d) 40

Answer:

(b) 36

Mean of 25 observations = 36
Sum of 25 observations = $36 \times 25 = 900$
Mean of the first 13 observations = 35
Sum of the first 13 observations = 32 $\times$ 13 = 416
Mean of the last 13 observations = 40
Sum of the last 13 observations = 40 $\times$ 13 = 520
∴ 13th observation = (Sum of the first 13 observations + Sum of the last 13 observations) $-$ Sum of 25 observations
= 416 + 520 $-$ 900
= 36

Question 22:

There are 50 numbers. Each number is subtracted from 53 and the mean of the numbers so obtained is found to be −3.5. The mean of the given numbers is
(a) 46.5
(b) 49.5
(c) 53.5
(d) 56.5

Answer:

(d) 56.5
Numbers = 50
$Let the numbers be x_{1}, x_{2}, . . . . . . . x_{50} . Each number is to be subtacted from 53 . Mean of (53 - x_{1}), (53 - x_{2}), . . . (53 - x_{50}) = \frac{(53 - x_{1}) + (53 - x_{2}) + . . . . + (53 - x_{50})}{50} = \frac{(53 \times 50) - (x_{1} + x_{2} + . . . . + x_{50})}{50} ∴ \frac{(53 \times 50) - (x_{1} + x_{2} + . . . . + x_{50})}{50} = - 3.5 \Rightarrow (53 \times 50) - (x_{1} + x_{2} + . . . . + x_{50}) = - 175 \Rightarrow (x_{1} + x_{2} + . . . . + x_{50}) = 175 + 2650 \Rightarrow (x_{1} + x_{2} + . . . . + x_{50}) = 2825 Mean = \frac{2825}{50} = 56.5$

Question 23:

The mean of the following data is 8.

x	3	5	7	9	11	13
y	6	8	15	p	8	4

The value of p is
(a) 23
(b) 24
(c) 25
(d) 21

Answer:

(c) 25

x	_y	_{x $\times$ y}
3	6	18
5	8	40
7	15	105
9	p	9p
11	8	88
13	4	52
Total	41 + p	303 + 9p

Now, Mean = \frac{303 + 9 p}{41 + p} Given : Mean = 8 ∴ \frac{303 + 9 p}{41 + p} = 8 \Rightarrow 303 + 9 p = 328 + 8 p \Rightarrow p = 25

Question 24:

The runs scored by 11 members of a cricket team are
15, 34, 56, 27, 43, 29, 31, 13, 50, 20, 0
The median score is
(a) 27
(b) 29
(c) 31
(d) 20

Answer:

(b) 29

Arranging the weight of 10 students in ascending order, we have:
0, 13, 15, 20, 27, 29, 31, 34, 43, 50, 56
Here, n is 11, which is an odd number.
Thus, we have:
$Median = Value of (\frac{n + 1}{2}) th observation Median score = Value of (\frac{11 + 1}{2}) th term = Value of 6 th term = 29$

Question 25:

The weight of 10 students (in kgs) are
55, 40, 35, 52, 60, 38, 36, 45, 31, 44
The median weight is
(a) 40 kg
(b) 41 kg
(c) 42 kg
(d) 44 kg

Answer:

(c) 42 kg

Arranging the numbers in ascending order, we have:
31, 35, 36, 38, 40, 44, 45, 52, 55, 60
Here, n is 10, which is an even number.
Thus, we have:
$Median = Mean of (\frac{n}{2}) th observation & (\frac{n}{2} + 1) th observation Median weight = Mean of the weights of (\frac{10}{2}) th student & (\frac{10}{2} + 1) th student = Mean of the weights of 5 th student & 6 th student = \frac{1}{2} (40 + 44) = 42 Hence, the median weight is 42 kg .$

Question 26:

The median of the numbers 4, 4, 5, 7, 6, 7, 7, 12, 3 is
(a) 4
(b) 5
(c) 6
(d) 7

Answer:

(c) 6

We will arrange the given data in ascending order as:
3, 4, 4, 5, 6, 7, 7, 7, 12
Here, n is 9, which is an odd number.
Thus, we have:
$Median = Value of \frac{1}{2} (n + 1) th term Median score = \frac{1}{2} (9 + 1) th term = 5 th term = 6$

Question 27:

The median of the numbers 84, 78, 54, 56, 68, 22, 34, 45, 39, 54 is
(a) 45
(b) 49.5
(c) 54
(d) 56

Answer:

(c) 54

We will arrange the data in ascending order as:
22, 34, 39, 45, 54, 54, 56, 68, 78, 84
Here, n is 10, which is an even number.
Thus, we have:
$Median = Mean of (\frac{n}{2}) th & (\frac{n}{2} + 1) th observations = \frac{1}{2} (5 th observation + 6 th observation) = \frac{1}{2} (54 + 54) = 54$

Question 28:

Mode of the data 15, 17, 15, 19, 14, 18, 15, 14, 16, 15, 14, 20, 19, 14, 15 is
(a) 14
(b) 15
(c) 16
(d) 17

Answer:

(b) 15

Here, 14 occurs 4 times, 15 occurs 5 times, 16 occurs 1 time, 18 occurs 1 time, 19 occurs 1 time and 20 occurs 1 time. Therefore, the mode, which is the most occurring item, is 15.

Question 29:

For drawing a frequency polygon of a continuous frequency distribution, we plot the points whose ordinates are the frequencies of the respective classes and abscissae are respectively
(a) upper limits of the classes
(b) lower limits of the classes
(c) class marks of the classes
(d) upper limits of preceding classes

Answer:

(c) class marks of the classes

To draw a frequency polygon of the continuous frequency distribution, we plot the class marks of the classes on the x-axis.

Question 30:

The marks obtained by 17 students of a class in a test (out of 100) are given below:
90, 79, 76, 82, 46, 64, 72, 49, 68, 66,48, 91, 82, 100, 96, 65, 84
The range of the data is
(a) 46
(b) 54
(c) 90
(d) 100

Answer:

(b) 54
We know:
Range = Maximum marks $-$ Minimum marks
Here,
Maximum marks = 100
Minimum marks = 46
∴ Range = 100 $-$ 46 = 54

Question 31:

The class mark of the class 130−150 is
(a) 130
(b) 135
(c) 140
(d) 145

Answer:

(c) 140

$Class mark = \frac{Lower limit + Upper limit}{2} = \frac{130 + 150}{2} = 140$

Question 32:

The mean of five numbers is 30. If one number is excluded, their mean becomes 28. The excluded number is
(a) 28
(b) 30
(c) 35
(d) 38

Answer:

(d) 38
Mean of 5 numbers = 30
Sum of 5 numbers = 30 $\times$ 5 = 150
By excluding one number, the mean becomes 28.
Now,
Sum of 4 numbers = 28 $\times$ 4 = 112
Excluded number = (150 $-$ 112) = 38

Question 33:

The median of the data arranged in ascending order 8, 9, 12, 18, (x + 2), (x + 4), 30, 31, 34, 39 is 24. The value of x is
(a) 22
(b) 21
(c) 20
(d) 24

Answer:

(b) 21

The given data is in ascending order.
Here, n is 10, which is an even number.
Thus, we have:
$Median = Mean of (\frac{n}{2}) th & (\frac{n}{2} + 1) th observations = \frac{1}{2} (5 th observation + 6 th observation) = \frac{1}{2} (x + 2 + x + 4) = (x + 3) = 24 Also, x + 3 = 24 \Rightarrow x = 21$

Question 34:

Assertion: The mean of 15 numbers is 25. If 6 is subtracted from each number, the mean of new numbers is 19.
Reason: Mode = 3(median) − 2(mean).
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true and Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Answer:

(b) Both Assertion (A) and Reason (R) are true and Reason (R) is not a correct explanation of Assertion (A).

Assertion (A):
Mean of 15 numbers = 25

$Let the numbers be x_{1}, x_{2}, . . . x_{15} . Given : \frac{x_{1} + x_{2} + . . . + x_{15}}{15} = 25$
$\Rightarrow x_{1} + x_{2} + . . . + x_{15} = 375$
When 6 is subtracted from each number, we get:
$\frac{(x_{1} - 6) + (x_{2} - 6) + . . . . . + (x_{15} - 6)}{15} = \frac{(x_{1} + x_{2} + . . . . + x_{15}) - 6 \times 15}{15} = \frac{375 - 90}{15} = \frac{285}{15} = 19$

This means that Assertion (A) is true.

Reason (R):
Mode = 3(Median) $-$ 2(Mean)
This is the empirical formula.
Hence, Reason (R) is true.

But Reason (R) is not the correct explanation of Assertion (A).

Question 35:

Assertion: Median of 51, 70, 65, 82, 60, 68, 62, 95, 55, 64, 58, 75, 80, 85, 90 is 68.
Reason: When n observations are arranged in an ascending order and n is odd, then $median = value of \frac{1}{2} (n + 1)$ th observation.
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true and Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Answer:

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).

Assertion:

We will arrange the numbers in ascending order as:
51, 55, 58, 60, 62, 64, 65, 68, 70, 75, 80, 82, 85, 90, 95

$Here, n is 15, which is an odd number . Median = Value of (\frac{n + 1}{2}) th observation = Value of 8 th observation = 68 This means that Assertion is correct .$

Reason is true. It is also the correct explanation of Assertion.

Question 36:

The mode of the data 2, 3, 9, 16, 9, 3, 9 is 16.

Answer:

False

Mode is the most frequently occurring observation. Therefore, in the given data, 9 is the mode.

Question 37:

The median of 3, 14, 18, 20, 5 is 18.

Answer:

False

We will arrange the given numbers in ascending order as:
3, 5, 14, 18, 20
Here, n is 5, which is an odd number.

$Thus, we have : Median = Value of (\frac{n + 1}{2}) th term = Value of the third term = 14$

Question 38:

The median of $1, 3, 2, 5, 8, 6, 1, 4, 7, 9 = \frac{1}{2} (5 th term + 6 th term) = \frac{1}{2} (8 + 6) = 7$ .

Answer:

We will arrange the given numbers in ascending order as:
1, 1, 2, 3, 4, 5, 6, 7, 8, 9
Here, n is 10, which is an even number.
Thus, we have:
$Median = Mean of (\frac{n}{2}) th & (\frac{n}{2} + 1) th term s = \frac{1}{2} (5 th term + 6 th term) = \frac{1}{2} (4 + 5) = \frac{9}{2} = 4.5$

Therefore, the given statement is false.

Question 39:

Match the following columns:

Column I	Column II
(a) The mean of first 10 odd number =	(p) 11.2
(b) The mean of first 10 even numbers =	(q) 10
(c) The mean of first 10 prime numbers =	(r) 11
(d) The mean of first 10 composite numbers =	(s) 12.9

(a) .......,
(b) .......,
(c) .......,
(d) .......,

Answer:

$Mean = \frac{Sum of all observations}{Total number of observations}$

(a) The first 10 odd numbers are 1, 3, 5, 7, 9, 11, 13, 15, 17 and 19.

$Mean = \frac{1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19}{10} = \frac{100}{10} = 10$

(b) The first 10 even numbers are 2, 4, 6, 8, 10, 12, 14, 16, 18 and 20.

$Mean = \frac{2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 18 + 20}{10} = \frac{110}{10} = 11$

(c) The first 10 prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23 and 29.

$Mean = \frac{2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 + 29}{10} = \frac{129}{10} = 12.9$

(d) The first 10 composite numbers are 4, 6, 8, 9, 10, 12, 14, 15, 16 and 18.

$Mean = \frac{4 + 6 + 8 + 9 + 10 + 12 + 14 + 15 + 16 + 18}{10} = \frac{112}{10} = 11.2$

Hence, the correct answers are:

$(a) \leftrightarrow (q) (b) \leftrightarrow (r) (c) \leftrightarrow (s) (d) \leftrightarrow (p)$

Question 40:

The class marks of a frequency distribution are 47, 52, 57, 62, 67, 72, 77. Determine the (i) class size (ii) class limits with respect to the class mark 52 (iii) true class limits for class mark 52.

Answer:

Class marks of the frequency distribution: 47, 52, 57, 62, 67, 72, 77
(i) Class size = Length of each class
= 52 $-$ 47
= 5
(ii) $\frac{Class size}{2} = \frac{5}{2} = 2.5$
Class limits with respect to the class mark 52 are (52 $-$ 2.5) and (52 + 2.5), i.e., 49.5 and 54.5.

(iii) Classes are in the exclusive form and in the exclusive form, the upper and lower limits of a class are known as true upper limit and true lower limit, respectively.

Therefore, the true class limits are 49.5 and 54.5.

Question 41:

Which is false?
(a) If n is odd, then $median = value of (\frac{n + 1}{2}) th$ item.
(b) If n is even, then $median = \frac{1}{2} \times \{value of \frac{n}{2} th item + value of (\frac{n}{2} + 1) th item\}$ .
(c) Mode is the item which occurs most often.
(d) $Mode = \frac{1}{2} (mean + median)$ .

Answer:

(a) True
(b) True
(c) True
(d) False
Correct formula for mode:
Mode = 3(Median) $-$ 2(Mean)

Question 42:

Which is false?
(a) If $x$ is the mean of $x_{1}, x_{2}, . . ., x_{n}$ , then $\sum_{i = 1}^{n} (x_{i} - x) = 0$ .
(b) If the mean of $x_{1}, x_{2}, . . ., x_{n}$ is $x$ , then the mean of $(x_{1} + a), (x_{2} + a), . . ., (x_{n} + a) i s (x + a)$ .
(c) If the mean of $x_{1}, x_{2}, . . ., x_{n}$ is $x$ and $a \neq 0$ , then the mean of $a x_{1}, a x_{2}, . . ., a x_{n}$ is $a x$ .
(d) If M is the median of $x_{1}, x_{2}, . . ., x_{n}$ and $a \neq 0$ , then aM is the median of $a x_{1}, a x_{2}, . . . a x_{n}$ .

Answer:

(a)

$If x is the mean of x_{1}, x_{2}, . . ., x_{n,} then we have : Mean = \frac{Sum of all observations}{Total number of observations} x = \frac{x_{1} + x_{2} + . . . + x_{n}}{n}$
$\Rightarrow x_{1} + x_{2} + . . . . . + x_{n} = n x . . . . . . . . (1)$

$We have : \sum (x_{i} - x) = (x_{1} - x) + (x_{2} - x) + . . . . . . . + (x_{n} - x) = (x_{1} + x_{2} + . . . . + x_{n}) - n x = n x - n x [From (1)] = 0$
Hence, (a) is correct.

(b)
$Now, the numbers are (x_{1} + a), (x_{2} + a), . . . (x_{n} + a) . Mean = \frac{x_{1} + a + x_{2} + a + . . . . x_{n} + a}{n} = \frac{(x_{1} + x_{2} . . . . . + x_{n}) + n a}{n} = \frac{(x_{1} + x_{2} . . . . . + x_{n})}{n} + a = x + a$

Hence, (b) is correct.

(c)
$Now, the numbers are a x_{1}, a x_{2}, . . . . . a x_{n} . Mean = \frac{a x_{1} + a x_{2} + . . . . a x_{n}}{n} = \frac{a (x_{1} + x_{2} + . . . . x_{n})}{n} = a x Hence, (c) is correct .$

(d) M is the median of x₁, x₂,...x_n.
We know that median is the middle term.
Hence, on multiplying the numbers by a, aM would be the median.
Examples:
$Let the numbers be 1, 2, 3 and 4 . Here, n is 4, which is even . ∴ Median = \frac{1}{2} (2 + 3) = 2.5 Now, we will multiply the numbers by 5 . The new numbers are 5, 10, 15 and 20 . Median = \frac{1}{2} (10 + 15) = 12.5 = 5 \times 2.5 Now, considering the case when the number of terms is odd . Let the numbers be 1, 2 and 3 . Here, n is 3, which is odd . ∴ Median = 2 Now, multiply the numbers by 5 . The new numbers are 5, 10 a n d 15 . Median = 10 = 5 \times 2 From the above examples, w e can say that if M is the median of x_{1}, x_{2, . . .} x_{n} and a \neq 0, then a M is the median of a x_{1}, a x_{2}, . . . a x_{n} .$

Hence, (d) is correct.

Question 43:

Which is false?
(a) If the mean of 4, 6, x, 8, 10, 13 is 8, then x = 7.
(b) If the median of the following array 59, 62, 65, x, x + 2, 72, 85, 99 is 67, then x = 66.
(c) If the mode of 1, 3, 5, 7, 5, 2, 7, 5, 9, 3, p, 11 is 5, then the value of p is 7.
(d) If the mean of 10 observations is 15 and that of other 15 observations is 18, then the mean of all the 25 observations is 16.8.

Answer:

$(a) Mean = \frac{Sum of all observations}{Total number of observations} = \frac{4 + 6 + x + 8 + 10 + 13}{6} We have : Mean = 8 \Rightarrow \frac{4 + 6 + x + 8 + 10 + 13}{6} = 8 \Rightarrow 41 + x = 48 \Rightarrow x = 7 Hence, (a) is true .$

$(b) Here, n = 8, which is an even number . ∴ Median = Mean of (\frac{n}{2}) th term & (\frac{n}{2} + 1) th term = \frac{1}{2} (4 th term + 5 th term) = \frac{1}{2} (x + x + 2) = (x + 1) Now, x + 1 = 67 \Rightarrow x = 66 Hence, (b) is true .$

(c) When p = 7, mode cannot be 5.
Hence, (c) is false.

(d) Mean of 10 observations = 15
Sum of these 10 observations = 15 $\times$ 10 = 150
Mean of other 15 observations = 18
Sum of these 15 observations = 18 $\times$ 15 = 270
Now,

$Mean = \frac{Sum of all observations}{Total number of observations} = \frac{150 + 270}{25} = 16.8 Hence, (d) is true .$

Question 1:

Look at the table given below:

Marks	0−10	11−20	21−30	31−40
No. of students	6	9	11	4

The true lower limit of the class 21−30 is
(a) 21
(b) 20
(c) 20.5
(d) 21.5

Answer:

(c) 20.5

To find the true lower limits of the classes, we will draw a continuous frequency distribution table, as given below.

Marks	No. of Students
$-$ 0.5−10.5	6
10.5−20.5	9
20.5−30.5	11
30.5−40.5	4

Thus, the true lower limit of the class 21−30 is 20.5

Question 2:

Look at the table given below:

Marks	0−10	10−20	20−30	30−40
No. of students	8	11	7	3

The true upper limit of the class 10−20 is
(a) 19.5
(b) 20
(c) 20.5
(d) none of these

Answer:

(b) 20

Because the data are in the continuous frequency distribution form, the true upper limit of the class 10−20 is 20.

Question 3:

Look at the table given below:

Marks	0−10	11−20	21−30	31−40
No. of students	7	8	10	5

What is the class size of the class 11−20 in this table?
(a) 9
(b) 15.5
(c) 10
(d) 4.5

Answer:

(c) 10

We know:
Class size = True upper limit $-$ True lower limit
We form the exclusive frequency distribution table as:

Marks	No. of Students
$-$ 0.5−10.5	7
10.5−20.5	8
20.5−30.5	10
30.5−40.5	5

∴ Class size = 20.5

-

10.5 = 10

Question 4:

What is the class mark of class 21−30 in table of Q. 3?
(a) 4.5
(b) 9
(c) 25.5
(d) 26

Answer:

(c) 25.5

Class mark = $\frac{Upper limit + Lower limit}{2}$
= $\frac{21 + 30}{2} = 25.5$

Question 5:

If the mean of five observations x, x + 2, x + 4, x + 6 and x + 8 is 11, find the value of x.

Answer:

We know:

$Mean = \frac{Sum of observations}{Number of observations} \Rightarrow 11 = \frac{x + x + 2 + x + 4 + x + 6 + x + 8}{5} \Rightarrow 55 = 5 x + 20 \Rightarrow 35 = 5 x \Rightarrow x = 7$

Question 6:

The points scored by a kabaddi team in a series of matches are as follows:
8, 24, 10, 14, 5, 15, 7, 2, 17, 27, 10, 7, 48, 8, 18, 28.
Find the median of the points scored by the team.

Answer:

Arranging the scores in ascending order, we have:
2, 5, 7, 7, 8, 8, 10, 10, 14, 15, 17, 18, 24, 28, 48

$The number of observations is 16, which is an even number . Thus, we have : Median = \frac{1}{2} [{\frac{n}{2}}^{th} term + {(\frac{n}{2} + 1)}^{th} term] = \frac{1}{2} [{\frac{16}{2}}^{th} term + {(\frac{16}{2} + 1)}^{th} term] = \frac{1}{2} [8^{th} term + {(9)}^{th} term] = \frac{1}{2} [10 + 14] = \frac{1}{2} \times 24 = 12$

Question 7:

The following table shows the number of students participating in various games in a school:

Game	Cricket	Football	Basket ball	Tennis
No. of students	27	36	18	12

Draw a bar graph to represent the above data.

Answer:

The histogram is given below:-

Question 8:

The heights of five players are 148 cm, 154 cm, 153 cm, 140 cm and 150 cm respectively. Find the mean height per player.

Answer:

We know:
$Mean = \frac{Sum of observations}{Number of observations} = \frac{148 + 154 + 153 + 140 + 150}{5} = \frac{745}{5} = 149 cm$

Thus, the mean height of players is 149 cm.

Question 9:

The marks obtained by 12 students of a class in a test are
36, 27, 5, 19, 34, 23, 37, 23, 16, 23, 20, 38.
Find the modal marks.

Answer:

Arranging the marks in ascending order, we have:
5, 16, 19, 20, 23, 23, 23, 34, 36, 37, 38
Because 23 occurs the maximum number of times, the mode is 23.

Question 10:

The class marks of a frequency distribution are
26, 31, 36, 41, 46, 51.
Find the true class limits.

Answer:

We have,
Class size = (31 $-$ 26) = 5

Class mark = 26
Lower limit = $(26 - \frac{5}{2}) = 23.5$
Upper limit = $(26 + \frac{5}{2}) = 28.5$

Class mark = 31
Lower limit = $(31 - \frac{5}{2}) = 28.5$
Upper limit = $(31 + \frac{5}{2}) = 33.5$

Class mark = 36
Lower limit = $(36 - \frac{5}{2}) = 33.5$
Upper limit = $(36 + \frac{5}{2}) = 38.5$

Class mark = 41
Lower limit = $(41 - \frac{5}{2}) = 38.5$
Upper limit = $(41 + \frac{5}{2}) = 43.5$

Class mark = 46
Lower limit = $(46 - \frac{5}{2}) = 43.5$
Upper limit = $(46 + \frac{5}{2}) = 48.5$

Class mark = 51
Lower limit = $(51 - \frac{5}{2}) = 48.5$
Upper limit = $(51 + \frac{5}{2}) = 53.5$

Thus, the true class limits are 23.5, 28.5, 33.5, 38.5 , 43.5, 48.5, 53.5

Question 11:

The mean of the following frequency distribution is 8. Find the value of p.

x	3	5	7	9	11	13
f	6	8	15	p	8	4

Answer:

(x_i)	(f_i)	(f_i)(x_i)
3	6	18
5	8	40
7	15	105
9	p	9p
11	8	88
13	4	52
	$\sum_{}^{} f_{i}$ = 41 + p	$\sum_{}^{} f_{i} x_{i} =$ 303 + 9p

We have:

$Mean = \frac{\sum_{}^{} x_{i} f_{i}}{\sum_{}^{} f_{i}}$

$\Rightarrow 8 = \frac{303 + 9 p}{41 + p} \Rightarrow 328 + 8 p = 303 + 9 p \Rightarrow 25 = p$

Question 12:

If 10, 13, 15, 18, x + 1, x + 3, 30, 32, 35, 41 are ten observations in an ascending order with median 24, find the value of x.

Answer:

Here, the observations are in ascending order.
Also,
n = 10 (even number)
Thus, we have:
$Median = \frac{1}{2} [{(\frac{n}{2})}^{th} term + {(\frac{n}{2} + 1)}^{th} term] \Rightarrow 24 = \frac{1}{2} [{(\frac{10}{2})}^{th} term + {(\frac{10}{2} + 1)}^{th} term] \Rightarrow 24 = \frac{1}{2} [5^{th} term + {(6)}^{th} term] \Rightarrow 24 = \frac{1}{2} [x + 1 + x + 3] \Rightarrow 48 = 2 x + 4 \Rightarrow 44 = 2 x \Rightarrow x = 22$

Question 13:

Calculate the mode of the following using empirical formula:
17, 10, 12, 11, 10, 15, 14, 11, 12, 13.

Answer:

We know:

$Mean = \frac{Sum of observations}{Number of observations} = \frac{17 + 10 + 12 + 11 + 10 + 15 + 14 + 11 + 12 + 13}{10} = \frac{125}{10} = 12.5$

$Arranging the observations in ascending form, we get : 10, 10, 11, 11, 12, 12, 13, 14, 15, 17$

$The number of observations is 10, which is an even number . Thus, we have : Median = \frac{1}{2} [{(\frac{n}{2})}^{th} term + {(\frac{n}{2} + 1)}^{th} term] = \frac{1}{2} [{(\frac{10}{2})}^{th} term + {(\frac{10}{2} + 1)}^{th} term] = \frac{1}{2} [5^{th} term + {(6)}^{th} term] = \frac{1}{2} [12 + 12] = 12$

On applying the empirical formula, we have:
Mode = 3(Median) $-$ 2(Mean)
or, Mode = 3(12) $-$ 2(12.5)
= 11

Question 14:

Find the median of the following frequency distribution:

Variate	3	6	10	12	7	15
Frequency	3	4	2	8	12	10

Answer:

The following table is in ascending order:

Variate	Frequency	Cumulative Frequency
3	3	3
6	4	7
7	13	20
10	2	22
12	8	30
15	10	40
	n = 40

The number of observations is 40, which is an even number . Thus, we have : Median = \frac{1}{2} [{(\frac{n}{2})}^{th} term + {(\frac{n}{2} + 1)}^{th} term] = \frac{1}{2} [{(\frac{40}{2})}^{th} term + {(\frac{40}{2} + 1)}^{th} term] = \frac{1}{2} [20^{th} term + {(21)}^{th} term] = \frac{1}{2} [7 + 10] = 8.5

Question 15:

The mean of six numbers is 23. If one of the numbers is excluded, the mean of the remaining numbers is 20. Find the excluded number.

Answer:

Let the six numbers be x₁, x₂,..., x₆_.
Given:
Mean = 23
We know:
$Mean = \frac{Sum of observations}{Number of observations} \Rightarrow 23 = \frac{x_{1} + x_{2} + . . . . + x_{6}}{6}$

$x_{1} + x_{2} + . . . x_{6} = 132 . . . . . . . . . (i)$

Let the number to be excluded be x₆.
Then, we have:
_{$20 = \frac{x_{1} + x_{2} + . . . . + x_{5}}{5}$ _{$x_{1} + x_{2} . . . . . . x_{5} = 100 . . . . . . . . (i)$}}

Using (i) in (ii), we get:
100 + x₆ = 138
or, x₆ = 38

Question 16:

Fill in the blanks in the following table:

Marks	Frequency	Cumulative frequency
0−5	3	3
5−10	5	......
10−15	8	......
15−20	4	......

Answer:

Marks	Frequency	Cumulative Frequency
0−5	3	3
5−10	5	8
10−15	8	16
15−20	4	20

Question 17:

In a city, the weekly observations made on the cost of living index are given below.

Cost of living index	Number of weeks
140−150	5
150−160	10
160−170	20
170−180	9
180−190	6
190−200	2

Represent the above information in the form of a histogram.

Answer:

The histogram is shown below:-

Question 18:

The mean of the marks scored by 50 students was found to be 39. Later on, it was discovered that a score was 43 was misread as 23. Find the correct mean.

Answer:

Mean of marks of 50 students = 39
Sum of marks of 50 students = $39 \times 50 = 1950$
Correct sum of marks of 50 students = $1950 - 23 + 43 = 1970$
Correct mean = $\frac{1970}{50} = 39.4$

Question 19:

The following table shows the weights of 12 workers in a factory.

Weight (in kg)	60	63	66	69	72
No. of workers	4	3	2	2	1

Find the mean weight of the workers.

Answer:

Weight (x_i)	No. of Workers (f_i)	(f_i)(x_i )
60	4	240
63	3	189
66	2	132
69	2	138
72	1	72
	$\sum_{}^{} f_{i}$ = 12	$\sum_{}^{} f_{i} x_{i} =$ 771

Thus, we have:

Mean = \frac{\sum_{}^{} x_{i} f_{i}}{\sum_{}^{} f_{i}}

= \frac{771}{12} = 64.25

Thus, the mean weight of the workers is 64.25 kg.

Question 20:

The heights (in cm) of 50 students of a class are given below.

Height (in cm)	156	154	155	151	157	152	153
No. of students	8	4	10	6	7	3	12

Find the median height.

Answer:

The following table represents heights of students in ascending order:

Height (cm)	No. of Students	Cumulative Frequency
151	6	6
152	3	9
153	12	21
154	4	25
155	10	35
156	8	43
157	7	50
	n = 50

Here, the number of observations is 50, which is an even number . Thus, we have : Median = \frac{1}{2} [{(\frac{n}{2})}^{th} term + {(\frac{n}{2} + 1)}^{th} term] = \frac{1}{2} [{(\frac{50}{2})}^{th} term + {(\frac{50}{2} + 1)}^{th} term] = \frac{1}{2} [25^{th} term + {(26)}^{th} term] = \frac{1}{2} [155 + 155] = 155 cm

RS Aggarwal 2017 Solutions for Class 9 Math Chapter 14 - Statistics

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