RS Aggarwal 2017 Solutions for Class 9 Math Chapter 14 Statistics are provided here with simple step-by-step explanations. These solutions for Statistics are extremely popular among class 9 students for Math Statistics Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RS Aggarwal 2017 Book of class 9 Math Chapter 14 are provided here for you for free. You will also love the ad-free experience on Meritnation’s RS Aggarwal 2017 Solutions. All RS Aggarwal 2017 Solutions for class 9 Math are prepared by experts and are 100% accurate.

Page No 546:

Question 1:

Define statistics as a subject.

Answer:

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

Page No 546:

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.

Page No 546:

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.

Page No 546:

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 Upper limit+Lower limit2.
(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.

Page No 546:

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

Page No 546:

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

Page No 546:

Question 7:

Construct a frequency table for the following ages (in years) of 30 students using equal class intervals, one of them being 912, 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

Page No 546:

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

Page No 546:

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

Page No 546:

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 ÷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



Page No 547:

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

Page No 547:

Question 12:

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

Ages (in years) 1020 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

Page No 547:

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

Page No 547:

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

Page No 547:

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



Page No 554:

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:

Page No 554:

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:

Page No 554:

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:

Page No 554:

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:

Page No 554:

Question 5:

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

Year 200102 2002−03 2003−04 2004−05 2005−06
No. of students 800 975 1100 1400 1625

Answer:

Page No 554:

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:



Page No 555:

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:

Page No 555:

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:

Page No 555:

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:

Page No 555:

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:

Page No 555:

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:



Page No 555:

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:



Page No 556:

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 = 60+35+75+50+605 = 2805= 56



Page No 565:

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:

Page No 565:

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:



Page No 566:

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:

Page No 566:

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:

Page No 566:

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:

Page No 566:

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:

Page No 566:

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 =Minimum class sizeClass size of the class ×Its frequency 

We have the following table:
 

 Class Interval Frequency Adjusted Frequency
10-14 5 44×5=5
14-20 6 46×6=4
20-32 9 412×9=3
32-52 25 420×25=5
52-80 21 428×21=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:

Page No 566:

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.



Page No 567:

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:

Page No 567:

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:

Page No 567:

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:

 

Page No 567:

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.

 



Page No 571:

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 =Sum of observationsNumber of observations

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

1+2+3+4+5+6+7+88=368=4.5 

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

1+3+5+7+9+11+13+15+17+1910= 10010=10

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

 2+3+5+7+115= 285= 5.6

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

2+4+6+8+10+126=426= 7

(v) The first seven multiples of 5 are 5, 10, 15, 20, 25, 30 and 35.
Mean of these numbers:
 
5+10+15+20+25+30+357=1407=20

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

1+2+4+5+10+206= 426=7



Page No 572:

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 =Sum of observationsNumber of observations
 
= 2+4+3+4+2+0+3+5+1+610
=3010=3

Page No 572:

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 =Sum of observationsNumber of observations
 
=105+216+322+167+273+405+3467= 3467= 262

Page No 572:

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=35.5 + 30.8+27.3+32.1+23.8+29.96= 179.46= 29.9°F

Page No 572:

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 =Sum of observationsNumber of observations
 
=64+36+47+23+0+19+81+93+72+35+3+112= 47412= 39.5

Page No 572:

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 =Sum of observationsNumber of observations

Thus, we have:

13 = 7+9+11+13+x+21613 = 61+x678 = 61+xx = 17

Page No 572:

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 =Sum of observationsNumber of observations

Thus, we have:

35= x1+x2+.........+x2424

 840=x1+x2+......+x24  .......i



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

=x1+3+x2+3+..........+x24+324=(x1+x2+.........+x24) + 24×324= 840+7224      [From (i)]= 91224= 38

Page No 572:

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 =Sum of observationsNumber of observations

Thus, we have:

 43 = x1+x2+......+x2020

 860=x1+x2+......+x20  ......i


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

∴ New Mean = (x1-6)+(x2-6)+........+(x20-6)20

=(x1+x2+..........+x20)-(20×6)20= 860-12020      [From (i)]= 37

Page No 572:

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 =Sum of observationsNumber of observations

Thus, we have:

27 = x1+x2+...........+x1515

 x1+x2+.........+x15=405.........i

After multiplication, the numbers become 4x1, 4x2,...4x15
∴ New Mean = 4x1+4x2+......+4x1515

=4(x1+x2+.......+x15)15= 4×40515   [From (i)]=162015= 108

Page No 572:

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 =Sum of observationsNumber of observations

Thus, we have:

40 = x1+x2+....+x1212


 x1+x2+....+x12=480   .....i

After division, the numbers become:

x18 , x28 , ........ , x128

New mean = x18+x28+....+x12812= x1+x2+...+x1212×8= x1+x2+...+x1296= 48096   From (i)=5    

Page No 572:

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,...x20.
We know:
Mean =Sum of observationsNumber of observations

Thus, we have:

18 = x1+x2+.......+x2020

x1+x2+ ..... +x20=360 .....i



New numbers are:

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

New Mean:

=x1+3+x2+3+........+x10+3+x11+........+x2020=(x1+x2+......+x10)+3×10+x11+........+x2020 = (x1+x2+.......+x20)+ 3020= 360+3020   From (i)=19.5 

Page No 572:

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 =Sum of observationsNumber of observations

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

48 = 51+45+49+46+44+x6288 = 235+xx = 53

Therefore, the sixth boy weighs 53 kg.

Page No 572:

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 =Sum of observationsNumber of observations

Thus, we have:

39 = x1+x2+...+x5050x
 x1+x2+ ....+x50=1950  ......i


Also, a score of 43 was misread as 23.
New Mean = x1+x2+...+x50-23 +4350= 1950-23 +4350               using i= 197050= 39.4



Page No 573:

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 =Sum of observationsNumber of observations

Thus, we have:64 = x1+x2+...+x100100

 x1+x2+ ....... +x100=6400   



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

Correct Mean = x1+x2+.......+x100-(26+9) +(36+90)100
=6400-35+126100=6491100= 64.91 

Page No 573:

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 =Sum of observationsNumber of observations

Thus, we have:

23 = x1+x2+...+x66
x1+x2.......x6=138.....................(i)

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

20 = x1+x2+...+x55
x1+x2......+x5=100....................(ii)

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

138 = x1+x2+...+x5+x6138 = 100+x6     ...(i) x6=38

Thus, the excluded number is 38



Page No 577:

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:

340+180+260+56+275+307+x7=2261418+x7=2261418+x=1582x=164

∴ Marks obtained by the 7th student = 164

Page No 577:

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 = (46.5×34) 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:

Sum of the weights of 34 students+Weight of the teacher35=471581+x35=471581+x=1645x=64

Therefore, the weight of the teacher is 64 kg.

Page No 577:

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 = 41×36 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: Sum of the weights of 36 students-x35=40.81476-x35=40.81476-x=1428x=48

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

Page No 577:

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 = 40×39 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:

Sum of the weights of 39 students+x40=39.81560+x40=39.81560+x=1592x=32

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

Page No 577:

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 7650×20=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:

Sum of the monthly salaries of 20 workers+x21=8200153000+x21=8200153000+x=172200x=19200

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

Page No 577:

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 = 9000×10=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 =New sum of the monthly wages of 10 persons10
                                       =8910010=Rs 8910

Page No 577:

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 = (330×7) litres=2310 litres
Sum of the monthly consumption of petrol for the next 5 months = 270×5 litres=1350 litres

Average consumption of petrol per month during the whole year=2310+135012                                                                                                  =366012                                                                                                   =305 litres

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

Page No 577:

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 = 15×18=270
Sum of the remaining 10 numbers = 10×13=130
Thus, we have:
Mean of 25 numbers=270+13025=16

Page No 577:

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 = (25×51) kg=1275 kg 
Now,
Let the sum of weights of the remaining 35 students be x kg.
Thus, we have:


Sum of the weights of 25 students+Sum of the weights of the remaining 35 students60=Mean weight of 60 students1275+x60=52.751275+x=3165x=1890Mean weight of the remaining 35 students=189035=54 kg



Page No 578:

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=Sum of the weights of initial 10 oarsmen-58+Weight of the new man10x+1.5=10x-58+Weight of the new man10

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

Page No 578:

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 = 35×8=280
Let the excluded number be x.
Now,
New mean = 35 - 3 = 32
Thus, we have:

Sum of 8 numbers-x7=32280-x7=32280-x=224x=56

Therefore, the excluded number is 56.

Page No 578:

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 = (150×60)=9000
New sum = [9000 - (52 + 8) + (152 + 88)] = 9180

Correct mean = New sumTotal items=9180150=61.2

Therefore, the correct mean is 61.2.

Page No 578:

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 = 31×60=1860
Mean of the first 16 results = 58
Sum of the first 16 results = 58×16=928
Mean of the last 16 results = 62
Sum of the last 16 results = 62×16=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

Page No 578:

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 = 42×11 = 462
Mean of the first 6 numbers = 37
Sum of the first 6 numbers = 37×6 = 222
Mean of the last 6 numbers = 46
Sum of the last 6 numbers = 46×6 = 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.

Page No 578:

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 = (52×25) kg = 1300 kg
Mean weight of the first 13 students = 48 kg
Sum of the weights of the first 13 students = (48×13) kg = 624 kg
Mean weight of the last 13 students = 55 kg
Sum of the weights of the last 13 students = (55×13) 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.

Page No 578:

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 = 80×25 = 2000
Mean score of another 55 observations = 60
Sum of the scores of another 55 observations = 60×55 = 3300

Mean score of the whole set of observations=Sum of the scores of 25 observations+Sum of the scores of another 55 observationsTotal number of observations
     
                                                       =2000+330080=530080=66.25

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

Page No 578:

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=36+44+75+x4 50=155+x4
155+x=200x=200-155=45

∴ Marks scored by Arun in science = 45

Page No 578:

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 (5400×25) = Rs 135000
Mean salary of the other 30 workers = Rs 5700
Sum of the monthly salaries of the other 30 workers = Rs (5700×30) = Rs 171000
Let the sum of the salaries of the remaining 30 workers be Rs x.
Mean monthly salary paid to 75 workers=135000+171000+x75 5680=306000+x75426000=306000+xx=120000Therefore, the sum of the salaries of the remaining 20 workers is Rs 120000.Mean salary of the remaining 20 workers=Sum of the salaries of 20 workers20                                                        =12000020=Rs 6000  Hence, the mean salary of the remaining 20 workers is Rs 6000.

Page No 578:

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 =Distance SpeedTime taken by the ship to travel from the starting point to the island =x15 hTime taken by the ship to travel from the island to the starting point=x10 hAverage speed=Total distance travelledTotal time taken=x+xx15+x10 =2x2x+3x30 =2x5x30 =605=12 km/h

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

Page No 578:

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 ×10) kg = 400 kg
Thus, we have:

Average weight of students in the class=Total weight of girls in the class+Total weight of boys in the classTotal number of students in the class44=400+Total weight of boys in the class50
Total weight of boys+400 =2200Total weight of boys =1800 kg
Average weight of boys =180040=45 kg



Page No 584:

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 (xi) No. of Workers (fi)     (fi)(xi)
90 12 1080
110 14 1540
120 13 1560
130 11 1430
150 10 1500
  fi = 60 fi xi = 7110

Thus, we have:

Mean = fixixi

        = 711060 = Rs 118.50

Page No 584:

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 (xi) No. of Workers (fi)    (fi)(xi)
60 4 240
63 3 189
66 2 132
69 2 138
72 1 72
  fi = 12 fi xi = 771

Thus, we have:

Mean = fixixi
= 77112= 64.25 kg



Page No 585:

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 (xi) Frequency (fi)    (fi)(xi)
15 3 45
16 8 128
17 9 153
18 11 198
19 6 114
20 3 60
  fi = 40 fi xi = 698

Thus, we have:
Mean = fixixi
= 69840= 17.45 years

Page No 585:

Question 4:

Find the mean of the following frequency distribution:

Variable (xi) 10 30 50 70 89
Frequency (fi) 7 8 10 15 10

Answer:

We will make the following table:
 

Variable (xi) Frequency (fi)     (fi)(xi)
10 7 70
30 8 240
50 10 500
70 15 1050
89 10 890
  fi = 50 fi xi = 2750

Thus, we have:

Mean = fixixi
= 275050 = 55

Page No 585:

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:
 

(xi) (fi)     (fi)(xi)
3 6 18
5 8 40
7 15 105
9 p 9p
11 8 88
13 4 52
  fi = 41 + p fi xi = 303 + 9p

We know:

Mean = fixixi
Given:
Mean = 8
Thus, we have:

8 = 303+9p41+p328+8p = 303+9pp = 25

Page No 585:

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:
 

(xi) (fi)      (fi)(xi)
15 8 120
20 7 140
25 p 25p
30 14 420
35 15 525
40 6 240
  fi = 50 + p fi xi = 1445 + 25p

Thus, we have:
Mean = fixixi
 28.25 = 1445+25p50+p28.2550+p = 1445+25p1412.5 + 28.25p = 1445+25p3.25 p = 32.5 p =10 

Page No 585:

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:
 

 (xi) (fi)     (fi)(xi)
8 12 96
12 16 192
15 20 300
p 24 24p
20 16 320
25 8 200
30 4 120
  fi = 100 fi xi = 1228 + 24p

Thus, we have:

Mean = fixixi
16.6 = 1228+24p10016.6×100 = 1228+24p1660 = 1228+24p
24p=432p=18

Page No 585:

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:
 

 (xi) (fi)     (fi)(xi)
10 17 170
30 f1 30f1
50 32 1600
70 f2 70f2
90 19 1710
  fi = 120 fi xi = 3480 + 30f1 + 70f2

Thus, we have:
Mean = fixixi

50 = 3480+30f1+70f2120

6000=3480+30f1+70f230f1+70f2 = 2520  .....   i



Also,
G
iven:
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

Page No 585:

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 (xi) No. of Workers (fi) di = (xi-A)=(xi-900)     (fi)(di)
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
  fi = 100   fi di = -880

Thus, we have:
Mean=  x = A +fidifi
= 900+-880100= 900-8810= 9000-8810= 891210 = Rs 891.2



Page No 586:

Question 10:

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

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

Answer:

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

Height in cm (xi) No. of Plants (fi) di = (xi-A)= (xi-67)        (fi)(di)
61 5 -6 -30
64 18 -3 -54
67 42 0 0
70 27 3 81
73 8 6 48
  fi = 100   fi di = 45

Thus, we have:
Mean=  x = A +fidifi
= 67+45100= 6700+45100= 6745100= 67.45 cm

Page No 586:

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:
 

(xi)  (fi) ui = xi-211      (ui)(fi)
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
  fi = 2200   fi ui = -840
 
Thus, we have:
Mean=  x = A +h×fiuifi
= 21+1×-8402200=  21-8402200= 46200-8402200=453602200= 20.62

Page No 586:

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:
 

(xi)  (fi) ui = xi-1400400     (fi)(ui)
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
  fi = 1343   fi ui = -1395
We have:
Mean=  x = A +h×fiuifi
= 1400+400×-13951343= 1400- 5580001343= 1880200-5580001343=13222001343= 984.51 m



Page No 590:

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 n+12th observation
Now,
Median=Value of 9+12th observation             =Value of the 5th 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 n+12th observation
Now,
Median=Value of 9+12th observation             =Value of the 5th 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 n+12th observation
Now,
Median=Value of 11+12th observation             =Value of the 6th 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 n+12th observation
Now,
Median=Value of 13+12th observation             =Value of the 7th observation             =4

Page No 590:

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 n2th & n2+1th observations
Now,
Median =Mean of 82th & 82+1th observations              =Mean of the 4th & 5th observations              =1219+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 n2th & n2+1th observations
Now,
Median =Mean of 102th & 102+1th observations              =Mean of the 5th & 6th observations              =1260+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 n2th & n2+1th observations
Now,
Median =Mean of 122th & 122+1th observations              =Mean of the 6th & 7th observations              =1215+17              =16

Page No 590:

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 n+12th observation
Thus, we have:
Median score=Value of 15+12 th observation                        =Value of the 8th observation                        =23

Page No 590:

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 n+12th observation
Median height =Value of 9+12th observation                               =Value of the 5th observation                       =147.3Hence, the median height is 147.3 cm. 

Page No 590:

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 n2th & n2+1th observations
Median weight=Mean of 82th & 82+1th observations                           =Mean of 4th & 5th observations                           =1213.4+14.3                          =13.85               

Hence, the median weight is 13.85 kg.



Page No 591:

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 n2th & n2+1th observations
Median age =Mean of 102th & 102+1th observations                                           = Mean of 5th & 6th observations                                           =1240+44                                           =42Hence, the median age is 42 years.

Page No 591:

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 n2th & n2+1th observations
Median=Mean of 102th & 102+1th observations                                  =Mean of 5th & 6th observations                                  =12x+1+x+3                                  =122x+4                                  =(x+2)Given: Median=24x+2=24x=22

Page No 591:

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 n+12th observation
Median weight =Weight of 41+12th student                       = Weight of the 21st studentHere, we see that the weight of the 20th to 28th students is 50 kg.This means that the weight of the 21st student is 50 kg. Hence, the median weight is 50 kg.   

Page No 591:

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 n+12th observation
Median=Value of 37+12th observation                                 =Value of 19th observationHere, we see that value of the 18th to 21st observations is 22.Therefore, the value of the 19th observation is 22.Median = 22

Page No 591:

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 n+12th observation
Median marks=Marks obtained by 43+12th student                     =Marks obtained by the 22nd student Here, we see that 25 marks are obtained by 11th to 26th students.Marks obtained by the 22nd student = 25Median marks = 25

Page No 591:

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 n2th & n2+1th observations
Median height=Mean of the heights of 502th and 502+1th students                     =Mean of the heights of the 25th and 26th studentsHere, we see that the height of the 25th student is 154 cm & the height of the 26th to 35th students is 155 cm.Median height = 12154+155 cm=154.5 cm

Page No 591:

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 n2th & n2+1th observations
Median=Mean of 602th & 602+1th observations           =Mean of 30th & 31st observationsValue of the 30th observation is 20 & value of 31st to 34th observations is 23.Median=1220+23 =21.5          



Page No 594:

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

Page No 594:

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

Page No 594:

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

Page No 594:

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

Page No 594:

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=Sum of all observationsNumber of observations= 17+10+12+11+10+15+14+11+12+1310= 12510= 12.5Here, 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, 17Thus, we have:Median = 12n2th term +n2+1th term= 12102th term +102+1th term=125th term +6th term=1212+12 =12

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

Page No 594:

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 (xi) No. of Students (fi) Cumulative Frequency      (xi)(fi)
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   fi xi = 332

The number of observations is 25, which is an odd number.Thus, we have:Median =n+12th term= 262th term 

=13th term=13



Also,
Mean = fixifi= 33225= 13.28

By the empirical formula, we have:
Mode = 3 Median - 2 Mean
Or,
Mode = 3(13) - 2(13.28) = 12.44

Page No 594:

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 (xi) Frequency (fi) Cumulative Frequency      (xi)(fi)
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   fi xi = 407

The number of observations is 34, which is an even number.Thus, we have:Median = 12n2th term +n2+1th term= 12342th term +342+1th term=  1217th term +18th term= 1212+12=12

Also,
Mean  = fixifi= 40734= 11.97

By the empirical formula, we have:
Mode = 3 Median - 2 Mean
Or,
Mode = 3(12) - 2(11.97) = 12.06

Page No 594:

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:

 (xi) (fi) Cumulative Frequency      (xi)(fi)
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   fi xi = 1161

The number of observations is 40, which is an even number.Thus, we have:Median = 12n2th term +n2+1th term= 12402th term +402+1th term=  1220th term +21st term= 1230+30=30

Also,
Mean  = fixifi= 116140= 29.025 29

By the empirical formula, we have:
Mode = 3 Median - 2 Mean
Or,
Mode = 3(30) - 2(29) = 32

Page No 594:

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 (xi) Number of Persons (fi) Cumulative Frequency      (xi)(fi)
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   fi xi = 2935

The number of observations is 50, which is an even number.Thus, we have:Median=12n2th term+n2+1th term=12502th term+502+1th term=1225th term+26th term=1257+62=59.5

Also,

Mean=fixifi=293550=58.7

By the empirical formula, we have:
Mode = 3 Median - 2 Mean
Or,
Mode = 3(59.5) - 2(58.7) = 61.1



Page No 595:

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 (xi) Number of Students (fi) Cumulative Frequency      (xi)(fi)
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   fi xi = 1992

The number of observations is 80, which is an even number.Thus, we have:Median=12n2th term+n2+1th term=12802th term+802+1th term=1240th term+41st term=1228+28=28

Also,
Mean=fixifi=199280=24.9

By the empirical formula, we have:
Mode = 3 Median - 2 Mean
Or,
Mode = 3(28) - 2(24.9) = 34.2

Page No 595:

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 (xi) Number of Persons (fi) Cumulative Frequency      (xi)(fi)
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   fi xi = 2650

The number of observations is 106, which is an even number.Thus, we have:Median= 12n2th term +n2+1th term= 121062th term +1062+1th term=  1253rd term +54th term= 1225+25=25

Also,
Mean  = fixifi= 2650106= 25

By the empirical formula, we have:
Mode = 3 Median - 2 Mean
Or,
Mode = 3(25) - 2(25) = 25

Page No 595:

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 (xi) Number of Students (fi) Cumulative Frequency      (xi)(fi)
47 4 4 188
50 3 7 150
53 2 9 106
56 2 11 112
60 4 15 240
  n = 15   fi xi = 796

The number of observations is 15, which an odd number.Thus, we have:Median=n+12th term= 15+12th term= 8th term= 53

Also,
Mean=fixifi=79615=53.06

By the empirical formula, we have:
Mode = 3 Median - 2 Mean
Or,
Mode = 3(53) - 2(53.06) = 52.88



Page No 597:

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

Page No 597:

Question 2:

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

Answer:

(b) 110

Class mark = Upper limit+Lower limit2=120+1002=110



Page No 598:

Question 3:

In the class intervals 1020, 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) 2030
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 2030.

Page No 598:

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=20-52=352=17.5Upper limit =20+52=452=22.5

Thus, the required class is 17.5-22.5.

Page No 598:

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=x+x+62=2x+62=x+3x+3=10x=7

Thus, the lower limit is 7.

Page No 598:

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) mu
(d) m + u

Answer:

(a) 2m - u

Given:
Mid value = m
Upper limit = u

We know:
Lower limit+Upper limit2=Mid value Lower limit +u2=mLower limit+u=2mLower limit=2m-u

Page No 598:

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 × 5 = 35

Page No 598:

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+(m + L)2
(b) L + m + L2
(c) 2m - L
(d) m - 2L

Answer:

(c) 2m-L

Mid value=Lower limit+Upper limit2m=L+U2U=2m-LUpper class boundary of the class=2m-L

Page No 598:

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

x+x+102=422x+102=422x+10=842x=74x=37Thus, we have:Lower limit=37 Upper limit=37+10=47

Page No 598:

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=Sum of all observationsTotal number of observations 11=x+x+2+x+4+x+6+x+8511=5x+2055x+20=555x=35x=7

Page No 598:

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) 1013
(b) 1023
(c) 1113
(d) 1123

Answer:

(c) 1113
Mean of 5 observations = 9
We know:
Mean=Sum of observationsTotal number of observations9=x+x+3+x+5+x+7+x+1059=5x+2555x+25=455x=20x=4Therefore, 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=9+11+143=343=1113

Page No 598:

Question 12:

If x is the mean of x1, x2, x3, ..., xn, then i=1n(xi - x) = ?
(a) −1
(b) 0
(c) 1
(d) n − 1

Answer:

(b) 0

If x  is the mean of x1, x2, x3, x4,...xn, then we have:i=1nxi=x Or,i=1nxi-x =0



Page No 599:

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 x1, x2,...xn.
Hence, mean = x1+x2+....+xnn


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

New Mean = x1+5+x2+5+....+xn+5n

           = x1+x2+....+xn+5nn

           = x1+x2+....+xnn+5New mean = mean + 5          

Hence, mean is increased by 5

Page No 599:

Question 14:

Let x be the mean of x1, x2, ..., xn and y be the mean of y1, y2, ..., yn.
If z is the mean of x1, x2, ..., xn, y1, y2, ... , yn, then z = ?
(a) (x + y)
(b) 12(x + y)
(c) 1n(x + y)
(d) 12n(x + y)

Answer:

(b) 12(x + y)

z¯=(x1+x2+...+xn)+(y1+y2+...+yn)2n

Given:x¯ = x1+x2+.....xnnx1+x2+......+xn=nx¯andy ¯=y1+y2+......+ynny1+y2+......+yn=n y¯ z¯ = n x¯ + n y¯2n          =12x¯+y¯

Page No 599:

Question 15:

If x is the mean of x1, x2, ..., xn, then for a  0, the mean of ax1, ax2, ..., axn, x1a,x2a, ...,xna is
(a) a+1ax

(b) a+1ax2

(c) a+1axn

(d) a+1ax2n

Answer:

(b) a+1ax2
Required mean =(ax1+ax2+...+axn)+x1a+x2a+...+xna2n                             =12a(x1+x2+...+xn)n+1a(x1+x2+...+xn)n                                                          =12ax¯+1ax¯                                 x¯=x1+x2+...+xnn                                                       =a+1ax¯2

Page No 599:

Question 16:

If x1, x2, ..., xn are the means of n groups with n1, n2, ..., nn number of observations respectively, then the mean x of all the groups taken together is
(a) i=1nnixi

(b) i=1n2nnixi

(c) i=1nnixii=1nni

(d) i=1nnixi2n

Answer:


(c) i=1nnixii=1nni


Sum of all terms=n1x¯1+n2x¯2+...nnx¯nNumber of terms=(n1+n2+...nn)Mean=i=1nnix¯ii=1nni

Page No 599:

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=Sum of all observations Total number of observations =51+45+49+46+44+x6=235+x6Given:Mean=48 kg 235+x6=48235+x=288x=53Hence, the weight of the 6th boy is 53 kg.

Page No 599:

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 = 39×50=1950
Correct sum = (1950 + 43 - 23) = 1970
Mean=197050=39.4

Page No 599:

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 = 64×100=6400
Correct sum = (6400 + 36 + 90 - 26 - 9) = 6491
Correct mean=6491100=64.91

Page No 599:

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 = 100×50=5000
It is given that one of the observations, 50, is replaced by 150.
∴ New sum = (5000 - 50 + 150) = 5100
And,
Resulting mean=5100100=51



Page No 600:

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

Page No 600:

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 x1, x2,.......x50.Each number is to be subtacted from 53.Mean of (53-x1), (53-x2),...(53-x50)= (53-x1)+(53-x2)+....+(53-x50)50                                                              = (53×50)-(x1+x2+....+x50)50(53×50)-(x1+x2+....+x50)50=-3.5(53×50)-(x1+x2+....+x50)=-175(x1+x2+....+x50)= 175+2650(x1+x2+....+x50)=2825Mean=282550=56.5

Page No 600:

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×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=303+9p41+pGiven:Mean=8303+9p41+p=8303+9p=328+8pp=25

Page No 600:

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 n+12th observation Median score=Value of 11+12th term                    =Value of 6th term                     =29

Page No 600:

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 n2th observation & n2+1th observationMedian weight=Mean of the weights of 102th student & 102+1th student                     =Mean of the weights of 5th student & 6th student                     =1240+44 =42Hence, the median weight is 42 kg.

Page No 600:

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 12(n+1)th term Median score=12(9+1)th term=5th term=6

Page No 600:

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 n2th & n2+1th observations            =125th observation+6th observation           =1254+54=54

Page No 600:

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.

Page No 600:

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.

Page No 600:

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



Page No 601:

Question 31:

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

Answer:

(c) 140

Class mark =Lower limit+Upper limit 2=130+1502=140

Page No 601:

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

Page No 601:

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 n2th & n2+1th observations            =125th observation+6th observation            =12(x+2+x+4)=(x+3)            =24Also,x+3=24x=21

Page No 601:

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 x1, x2,...x15.Given: x1+x2+...+x1515=25
 x1 + x2 + ... + x15 = 375
When
 6 is subtracted from each number, we get:
x1 - 6 + x2 - 6 + ..... + x15 - 615 = x1 +x2 + ....+ x15  - 6×1515= 375 - 9015=28515= 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).

Page No 601:

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 12(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 n+12th observation            =Value of 8th observation            =68This means that Assertion is correct.

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

Page No 601:

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.

Page No 601:

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 n+12th term            =Value of the third term            =14         



Page No 602:

Question 38:

The median of 1, 3, 2, 5, 8, 6, 1, 4, 7, 9=12 (5th term + 6th term) = 12(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 n2th & n2+1th terms            =125th term+6th term                        =124+5=92=4.5   

Therefore, the given statement is false.

Page No 602:

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 =Sum of all observationsTotal number of observations 

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

Mean =1+3+5+7+9+11+13+15+17+1910=10010=10

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

Mean =2+4+6+8+10+12+14+16+18+2010=11010=11

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

Mean =2+3+5+7+11+13+17+19+23+2910=12910=12.9

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

Mean =4+6+8+9+10+12+14+15+16+1810=11210=11.2

Hence, the correct answers are:

(a)(q)(b)(r)(c)(s)(d)(p)

Page No 602:

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) Class size 2=52=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.

Page No 602:

Question 41:

Which is false?
(a) If n is odd, then median = value of n + 12th  item.
(b) If n is even, then median=12×value of n2th item + value of n2+1th item.
(c) Mode is the item which occurs most often.
(d) Mode=12 (mean + median).

Answer:

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

Page No 602:

Question 42:

Which is false?
(a) If x is the mean of x1, x2, ..., xn, then i=1n(xi - x) = 0.
(b) If the mean of x1, x2, ..., xn is x, then the mean of (x1 + a), (x2 + a), ..., (xn + a) is (x + a).
(c) If the mean of x1, x2, ... , xn is x and a  0, then the mean of ax1, ax2, ..., axn is ax.
(d) If M is the median of x1, x2, ..., xn and a  0, then aM is the median of ax1, ax2, ... axn.

Answer:

(a)

If x is the mean of x1, x2,..., xn, then we have:Mean =Sum of all observations Total number of observations x= x1+x2+...+xnn
 x1 + x2 + ..... + xn = nx ........1

We have:xi - x = x1 - x + x2 - x+.......+xn - x                   =  x1 +x2 + ....+ xn - nx                   = nx - nx                    From 1                    = 0

Hence, (a) is correct.

(b) 
Now, the numbers are (x1+a), (x2+a),...(xn+a).Mean =x1+a+x2+a+....xn+an         =(x1+x2.....+xn)+nan         =(x1+x2.....+xn)n+a         =x+a

Hence, (b) is correct.

(c)
Now, the numbers are ax1, ax2,.....axn.Mean =ax1+ax2+....axnn         =a(x1+x2+....xn)n         =axHence, (c) is correct. 

(d) M is the median of x1, x2,...xn.
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=12(2+3)=2.5Now, we will multiply the numbers by 5.The new numbers are 5, 10, 15 and 20.Median =12(10+15)=12.5=5×2.5Now, 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=2Now, multiply the numbers by 5.The new numbers are 5, 10 and 15.Median =10=5×2From the above examples, we can say that if M is the median of x1, x2,...xn and a0, then aM is the median of ax1, ax2,...axn.

Hence, (d) is correct.



Page No 603:

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=Sum of all observations Total number of observations              =4+6+x+8+10+136We have:Mean=84+6+x+8+10+136=841+x=48x=7Hence, (a) is true. 

(b) Here, n= 8, which is an even number. Median =Mean of n2th term & n2+1th term                              =124th term+5th term                             =12(x+x+2)                             =(x+1)Now, x+1=67x=66Hence, (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 = 15×10 = 150
Mean of other 15 observations = 18
Sum of these 15 observations = 18×15 = 270
Now,

Mean =Sum of all observations Total number of observations          =150+27025         =16.8Hence, (d) is true. 



Page No 608:

Question 1:

Look at the table given below:

Marks 010 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
-0.510.5 6
10.520.5 9
20.530.5 11
30.540.5 4

Thus, the true lower limit of the class 2130 is 20.5

Page No 608:

Question 2:

Look at the table given below:

Marks 010 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 1020 is 20.

Page No 608:

Question 3:

Look at the table given below:

Marks 010 11−20 21−30 31−40
No. of students 7 8 10 5
What is the class size of the class 1120 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
-0.510.5 7
10.520.5 8
20.530.5 10
30.540.5 5


∴ Class size = 20.5 - 10.5 = 10

Page No 608:

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 =  Upper limit+Lower limit2
                  = 21+302=25.5

Page No 608:

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 = Sum of observationsNumber of observations11= x+x+2+x+4+x+6+x+8555 = 5x +2035  =5xx=7

Page No 608:

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 = 12n2th term +n2+1th term = 12162th term +162+1th term=  128th term +9th term= 1210+14=12×24=12

Page No 608:

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:-



Page No 609:

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  = Sum of observationsNumber of observations            = 148+154+153+140+1505            =7455            = 149 cm

Thus, the mean height of players is 149 cm.

Page No 609:

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.

Page No 609:

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 = 26-52=23.5
Upper limit = 26+52 = 28.5

Class mark = 31
Lower limit = 31-52=28.5
Upper limit = 31+52 = 33.5

Class mark = 36
Lower limit = 36-52=33.5
Upper limit = 36+52 = 38.5

Class mark = 41
Lower limit = 41-52=38.5
Upper limit = 41+52 = 43.5

Class mark = 46
Lower limit = 46-52=43.5
Upper limit = 46+52 = 48.5

Class mark = 51
Lower limit = 51-52=48.5
Upper limit = 51+52 = 53.5

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

Page No 609:

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:

(xi) (fi)     (fi)(xi)
3 6 18
5 8 40
7 15 105
9 p 9p
11 8 88
13 4 52
  fi = 41 + p fi xi = 303 + 9p

We have:

Mean =xififi

8=303+9p41+p328+8p=303+9p25=p

Page No 609:

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 = 12n2th term +n2+1th term24= 12102th term +102+1th term24=  125th term +6th term24= 12x+1+x+348 = 2x+444 = 2xx=22

Page No 609:

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 =Sum of observationsNumber of observations          = 17+10+12+11+10+15+14+11+12+1310          = 12510=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 = 12n2th term +n2+1th term = 12102th term +102+1th term=  125th term +6th term= 1212+12=12

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

Page No 609:

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 = 12n2th term +n2+1th term = 12402th term +402+1th term=  1220th term +21th term= 127+10=8.5

Page No 609:

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 = Sum of observationsNumber of observations23 = x1+x2+....+x66

x1+x2+...x6=132 .........(i)

Let the number to be excluded be x6.
Then, we have:
20 = x1+x2+....+x55
x1+x2......x5=100........(i)


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

Page No 609:

Question 16:

Fill in the blanks in the following table:

Marks Frequency Cumulative frequency
05 3 3
510 5 ......
10−15 8 ......
15−20 4 ......

Answer:

Marks Frequency Cumulative Frequency
05 3 3
510 5 8
10−15 8 16
15−20 4 20



Page No 610:

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:-

Page No 610:

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 = 39×50=1950
Correct sum of marks of 50 students = 1950-23+43=1970
Correct mean = 197050=39.4

Page No 610:

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 (xi) No. of Workers (fi)     (fi)(xi )
60 4 240
63 3 189
66 2 132
69 2 138
72 1 72
  fi= 12 fi xi = 771

Thus, we have:

Mean =xififi

        = 77112= 64.25 

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

Page No 610:

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 = 12n2th term +n2+1th term = 12502th term +502+1th term=  1225th term +26th term= 12[155+155]=155 cm



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