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Page No 673:

Question 1:

Calculate the mean for the following distribution :

x: 5 6 7 8 9
f: 4 8 14 11 3

Answer:

Given:

First of all prepare the frequency table in such a way that its first column consist of the values of the variate and the second column the corresponding frequencies.

Thereafter multiply the frequency of each row with corresponding values of variable to obtain third column containing.

Then, sum of all entries in the column second and denoted by and in the third column to obtain.

We know that mean,
                                   

Hence, mean

Page No 673:

Question 2:

Find the missing value of p for the following distribution whose mean is 12.58

x: 5 8 10 12 p 20 25
f: 2 5 8 22 7 4 2

Answer:

Given:

Mean

First of all prepare the frequency table in such a way that its first column consist of the values of the variate and the second column the corresponding frequencies.

Thereafter multiply the frequency of each row with corresponding values of variable to obtain third column containing.

Then, sum of all entries in the column second and denoted by and in the third column to obtain.

We know that mean,

By using cross multiplication method

Hence, p

Page No 673:

Question 3:

The following table gives the number of boys of a particular age in a class of 40 students. Calculate the mean age of the students

Age (in years): 15 16 17 18 19 20
No. of students: 3 8 10 10 5 4

Answer:

Given:

First of all prepare the frequency table in such a way that its first column consist of the values of the variate and the second column the corresponding frequencies.

Thereafter multiply the frequency of each row with corresponding values of variable to obtain third column containing.

Then, sum of all entries in the column second and denoted by and in the third column to obtain.

We know that mean,
                               

Hence, the mean age of the students

Page No 673:

Question 4:

Five coins were simultaneously tossed 1000 times and at each toss the number of heads were observed. The number of tosses during which 0, 1, 2, 3, 4 and 5 heads were obtained are shown in the table below. Find the mean number of heads per toss.

No. of heads per toss No. of tosses
0
1
2
3
4
5
38
144
342
287
164
25
Total 1000

Answer:

Given:

First of all prepare the frequency table in such a way that its first column consist of the numnber of heads per tosses and the second column the corresponding number of tosses .

Thereafter multiply the frequency of each row with corresponding values of variable to obtain third column containing.

Then, sum of all entries in the column second and denote by and in the third column to obtain.

We know that mean,
                                    =24701000=2.47

Hence, the mean number of heads per toss is 2.47.



Page No 674:

Question 5:

The arithmetic mean of the following data is 14. Find the value of k.

xi: 5 10 15 20 25
fi: 7 k 8 4 5

Answer:

Given:

Mean

First of all prepare the frequency table in such a way that its first column consist of the values of the variate and the second column the corresponding frequencies.

Thereafter multiply the frequency of each row with corresponding values of variable to obtain third column containing.

Then, sum of all entries in the column second and denoted by and in the third column to obtain.

We know that mean,

By using cross multiplication method,

Hence, k = 6

Page No 674:

Question 6:

The arithmetic mean of the following data is 25, find the value of k.

xi: 5 15 25 35 45
fi: 3 k 3 6 2

Answer:

Given:

Mean

First of all prepare the frequency table in such a way that its first column consist of the values of the variate and the second column the corresponding frequencies.

Thereafter multiply the frequency of each row with corresponding values of variable to obtain third column containing.

Then, sum of all entries in the column second and denoted by and in the third column to obtain.

We know that mean,

By using cross multiplication method,

Hence, k = 4

Page No 674:

Question 7:

If the mean of the following data is 18.75. Find the value of p.

xi: 10 15 p 25 30
fi: 5 10 7 8 2

Answer:

Given:

Mean

First of all prepare the frequency table in such a way that its first column consist of the values of the variate and the second column the corresponding frequencies.

Thereafter multiply the frequency of each row with corresponding values of variable to obtain third column containing.

Then, sum of all entries in the column second and denoted by and in the third column to obtain.

We know that mean,

By using cross multiplication method,

Hence, p = 20

Page No 674:

Question 8:

Find the value of p, if the mean of the following distribution is 20.

x: 15 17 19 20+p 23
f: 2 3 4 5p 6

Answer:

Given:

Mean

First of all prepare the frequency table in such a way that its first column consist of the values of the variate and the second column the corresponding frequencies.

Thereafter multiply the frequency of each row with corresponding values of variable to obtain third column containing.

Then, sum of all entries in the column second and denoted by and in the third column to obtain.

We know that mean,
20=295+5p20+p15+5p

By using cross multiplication method,

5p2+100p+295=300+100p5p2=300-295=5p2=1p=1

Hence, p

Page No 674:

Question 9:

Find the missing frequencies in the following frequency distribution if it is known that the mean of the distribution is 50.

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

Answer:

Given:

Mean

First of all prepare the frequency table in such a way that its first column consists of the values of the variate and the second column the corresponding frequencies.

Thereafter multiply the frequency of each row with corresponding values of variable to obtain third column containing.

Then, sum of all entries in the column second and denoted by and in the third column to obtain.

Now,

..... (1)

We know that mean,

By using cross multiplication method,


             ..... (2)

Putting the value of from equation (1) in (2), we get

Therefore,

Putting the value of in equation (1), we get

Hence,



Page No 679:

Question 1:

The number of telephone calls received at an exchange per interval for 250 successive one-minute intervals are given in the following frequency table:
 

No. of calls (x): 0 1 2 3 4 5 6
No. of intervals (f): 15 24 29 49 54 43 39

Compute the mean number of calls per intervals.

Answer:

Let the assume mean be.

We know that mean,

Here, we have .

Putting the values in the formula, we get


   

Hence, the mean number of calls per interval is 3.54.

Page No 679:

Question 2:

Five coins were simultaneously tossed 1000 times, and at each toss the number of heads was observed. The number of tosses during which 0, 1, 2, 3, 4 and 5 heads were obtained are shown in the table below: Find the mean number of heads per toss
 

No. of heads per toss (x): 0 1 2 3 4 5
No. of tosses (f): 38 144 342 287 164 25

Answer:

Let the assume mean be.

We know that mean

Now, we have

Putting the values above in formula, we have


Hence, the mean number of heads per toss is 2.47.

Page No 679:

Question 3:

The marks obtained out of 50, by 102 students in a Physics test are given in the frequency table below:
 

Marks (x): 15 20 22 24 25 30 33 38 45
Frequency (f): 5 8 11 20 23 18 13 3 1

Find the average number of marks.

Answer:

Let the assume mean be.

We know that mean,

Now, we have .

Putting the values in the above formula, we get



Hence, the average number of marks is 26.08.

Page No 679:

Question 4:

The number of students absent in a class were recorded every day for 120 days and the information is given in the following frequency table:
 

No. of students absent (x): 0 1 2 3 4 5 6 7
No. of days (f): 1 4 10 50 34 15 4 2

Find the mean number of students absent per day.

Answer:

Let the assume mean be.

We know that mean,

Now, we have .

Putting the values in the above formula,


    
  

Hence, the mean number of students absent per day is approximately 3.53.



Page No 686:

Question 1:

A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of a plants in 20 houses in a locality. Find the mean number of plants per house.
 

Number of plants: 0-2 2-4 4-6 6-8 8-10 10-12 12-14
Number of houses: 1 2 1 5 6 2 3

Which method did you use for finding the mean, and why?

Answer:

We may prepare the table as shown:

We know that mean,
                                  
Hence, mean = 8.1

Direct method is easier than other methods. Therefore, we used direct method.

Page No 686:

Question 2:

Thirty women were examined in a hospital by a doctor and the number of heart beats per minute recorded and summarised as follows. Find the mean heat beats per minute for these women, choosing a suitable method.

Number of heart beats per minute: 65-58 68-71 71-74 74-77 77-80 80-83 83-86
Number of women: 2 4 3 8 7 4 2

Answer:

Let the assumed mean be A = 75.5 and h = 3.

No. of Heart Beats per Min: Mid Value xi No. of Women fi: di=xi-A ui=1h×di fiui
65-68 66.5 2 -9 -3 -6
68-71 69.5 4 -6 -2 -8
71-74 72.5 3 -3 -1 -3
74-77 75.5 = A 8 0 0 0
77-80 78.5 7 3 1 7
80-83 81.5 4 6 2 8
83-86 84.5 2 9 3 6
    fi=30     fiui=4

We know that mean,

Now, we have.

Putting the values in the above formula, we have


  

Hence, the mean heart beats per minute for women is 75.9.

Page No 686:

Question 3:

Find the mean of each of the following frequency distributions :
 

Class interval: 3 - 5 5 - 7 7 - 9 9 - 11 11 - 13
Frequency: 5 10 10 7 8
​

Answer:

Class Frequency(fi) Class Mark(xi) fixi
3 - 5 5 4 20
5 - 7 10 6 60
7 - 9 10 8 80
9 - 11 7 10 70
11 - 13 8 12 96
  fi=40   fixi = 326

Mean, x=fixifi=32640=8.15
Thus, mean = 8.15

Page No 686:

Question 4:

The following distribution shows the daily pocket allowance given to the children of a multistorey building. The average pocket allowance is Rs 18.00. Find the missing frequency.
 

Class: 11−13 13−15 15−17 17−19 19−21 21−23 23−25
Frequency: 7 6 9 13 5 4

Answer:

Given: Mean = 18

Suppose the missing frequency is x.

Let the assumed mean A = 18 and h = 2.

We know that mean,

Now, we have,,.

Putting the values in the above formula, we have

18=18+2x-20x+442x-20x+44=0x-20=0x=20

Thus, the missing frequency is 20.

Page No 686:

Question 5:

If the mean of the following distribution is 27, find the value of p.
 

Class: 0−10 10−20 20−30 30−40 40−50
Frequency: 8 p 12 13 10

Answer:

Given: Mean = 27

Let the assumed mean A = 25 and h = 10.

We know that mean,

Now, we have,,

Putting the values in the above formula, we have

 
    

Thus, the value of p is 7.

Page No 686:

Question 6:

The table below shows the daily expenditure on food of 25 household in a locality

Daily expenditure (in Rs): 100−150 150−200 200−250 250−300 300−350
Number of households: 4 5 12 2 2

Find the mean daily expenditure on food by a suitable method.

Answer:

Let the assumed mean a = 225 and h = 50.
 

Daily expenditure (in Rs) fi xi di = xi − 225 fiui
100−150 4 125 − 100 − 2 − 8
150−200 5 175 − 50 − 1 − 5
200−250 12 225 0 0 0
250−300 2 275 50 1 2
300−350 2 325 100 2 4
Total fi=25       fiui=− 7

Now,

Therefore, mean daily expenditure on food is Rs 211.

Page No 686:

Question 7:

A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.
(i)

Number of days: 0 − 6 6 − 12 12 − 18 18 − 24 24 − 30 30 − 36 36 − 42
Number of students: 10 11 7 4 4 3 1

(ii)
Number of days: 0 − 6 6 − 10 10 − 14 14 − 20 20 − 28 28 − 38 38 − 40
Number of students: 11 10 7 4 4 3 1

Answer:

(i) Let the assume mean = 17.

We know that mean, 

Now, we have.

Putting the values in the above formula, we have


    

Hence, the mean number of days a student was absent is 12.475.

(ii)

Number of Days Mid-Value (xi) Number of Students (fi) fixi
0−6 3 11 33
6−10 8 10 80
10−14 12 7 84
14−20 17 4 68
20−28 24 4 96
28−38 33 3 99
38−40 39 1 39
    fi=40 fixi=499


We know

Mean=fixifi
∴ Mean number of days a student was absent=fixifi=49940=12.475

Hence, the mean number of days a student was absent is 12.475.

Page No 686:

Question 8:

The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.

Literacy rate (in %): 45−55 55−65 65−75 75−85 85−95
Number of cities: 3 10 11 8 3

Answer:

Let the assumed mean A = 70 and h = 10.

We know that mean,

Now, we have

Putting the values in the above formula, we have
 

   

Hence, the mean literacy rate is approximately 69.43%.

Page No 686:

Question 9:

 The following is the cumulative frequency distribution ( of less than type ) of 1000 persons each of age 20 years and above . Determine the mean age .

Age below (in years) :    30      40      50      60    70    80 

Number of persons:       100     220   350  750   950   1000  

Answer:


Let the assumed mean A = 55 and h = 10.

Marks Mid-Value Frequency ui=xi-5510 fiui
20–30 25 100 −3 −300
30–40 35 120 −2 −240
40–50 45 130 −1 −130
50–60 55 400 0 0
60–70 65 200 1 200
70–80 75 50 2 100
    N = 1000   fiui=-370

We know that mean,
Now, we have N=fi=1000, h=10, A=55, fiui=-370
X¯=55+1011000×-370=55-3.7=51.3 years



Page No 687:

Question 10:

The marks obtained by 110 students in an examination are given below:
 

Marks: 30 – 35 35 – 40 40 – 45 45 – 50 50 – 55 55 – 60 60 – 65
Frequency: 14 16 28 23 18 8 3

Find the mean marks of the students.

Answer:


Let the assumed mean, A be 47.5.

Here, h = 5

Marks Mid-Values(xi) Frequency (fi) di=xi-47.5 ui=xi-47.55 fiui
30–35 32.5 14 −15 −3 −42
35–40 37.5 16 −10 −2 −32
40–45 42.5 28 −5 −1 −28
45–50 47.5 23 0 0 0
50–55 52.5 18 5 1 18
55–60 57.5 8 10 2 16
60–65 62.5 3 15 3 9
    fi=110     fiui=-59

We know

Mean =A+h×fiuifi
∴ Mean marks of the students

=47.5+5×-59110=47.5-2.68=44.82

Hence, the mean marks of the students is 44.82.

Page No 687:

Question 11:

Find the mean of each of the following frequency distributions :
 

Classes: 25−29 30−34 35−39 40−44 45−49 50−54 55−59
Frequency: 14 22 16 6 5 3 4

Answer:

The given series is an inclusive series. Firstly, make it an exclusive series.

Class Interval Mid-Value(xi) Frequency(fi) di=xi-A   =xi-42 ui=dih=di5 fiui
24.5-29.5 27 14 −15 −3 −42
29.5-34.5 32 22 −10 −2 −44
34.5-39.5 37 16 −5 −1 −16
39.5-44.5 A = 42 6 0 0 0
44.5-49.5 47 5 5 1 5
49.5-54.5 52 3 10 2 6
54.5-59.5 57 4 15 3 12
    fi=70     fiui=-79

Let the assumed mean be A = 42 and h = 5.

We know that mean,

Now, we have.

Putting the values in the above formula, we have

   

Hence, the mean is 36.357.

Page No 687:

Question 12:

The mean of the following frequency distribution is 62.8 and the sum of all the frequencies is 50. Compute the missing frequency f1 and f2.
 

Class: 0−20 20−40 40−60 60−80 80−100 100−120
Frequency: 5 f1 10 f2 7 8
8

Answer:

It is given that mean = 62.8 and .

Let the assumed mean A = 50 and h = 20.


             .....(1)

We know that mean,

Now, we have , , .

Putting the values in the above formula, we have

            .....(2)

Putting the value of in (2), we get


                                    

Putting the value of in (1), we get

Hence, the missing frequency f1 = 8 and f2 = 12.

Page No 687:

Question 13:

In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.

Number of mangoes: 50−52 53−55 56−58 59−61 62−64
Number of boxes: 15 110 135 115 25

Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?

Answer:

The given series is an inclusive series. Firstly, make it an exclusive series.

Class Interval Mid-Value(xi) Frequency(fi) di=xi-A   =xi-57 ui=dih=di3 fiui
49.5-52.5 51 15 −6 −2 −30
52.5-55.5 54 110 −3 −1 −110
55.5-58.5 57 135 0 0 0
58.5-61.5 60 115 3 1 115
61.5-64.5 63 25 6 2 50
    fi=400     fiui=25

Let the assumed mean be A = 57 and h = 3.

We know that mean,

Now, we have.

Putting the values in the above formula, we have


   

Hence, the mean is approximately 57.19.

Page No 687:

Question 14:

To find out the concentration of SO2 in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below:
 

Concentration of SO2 (in ppm) Frequency
0.00−0.04
0.04−0.08
0.08−0.12
0.12−0.16
0.16−0.20
0.20−0.24
4
9
9
2
4
2

Find the mean of concentration of SO2 in the air.

Answer:

Let the assumed mean A = 0.1 and h = 0.04.

We know that mean,

Now, we have.

Putting the values in the above formula, we have

 
   =0.10+0.04130×-1=0.10-0.0430=0.10-0.001=0.099

Hence, the mean concentration of SO2in the air is 0.099 ppm.

Page No 687:

Question 15:

If the mean of the following frequency distribution is 18 , find the missing frequency .

Class interval :          11-13     13-15    15-17    17-19    19-21    21-23     23-25

Frequency:                  3              6            9          13        f              5            4

Answer:

It is given that mean of the data is 18.
Let the assumed mean A = 18 and h = 2.

Marks Mid-Value(xi) Frequency ui=xi-182 fiui
11–13 12 3 −3 −9
13–15 14 6 −2 −12
15–17 16 9 −1 −9
17–19 18 13 0 0
19–21 20 f 1 f
21–23 22 5 2 10
23–25 24 4 3 12
    N = 40 + f   fiui=-8+f

We know that mean,
Now, we have N=fi=40+f, h=2, A=18, fiui=-8+f
18=18+2140+f×-8+f0=-16+2f40+f-16+2f=0f=8

Page No 687:

Question 16:

The daily income of a sample of 50 employees are tabulated as follows:
Income  (in â‚¹):        1-200          201-400          401-600        601-800

No.of employees :     14                15                    14                7

Find the mean daily income of employees.

Answer:


Let the assumed mean A = 500.5 and h = 200.

Marks Mid-Value(xi) Frequency ui=xi-500.5200 fiui
1–200 100.5 14 −2 −28
201–400 300.5 15 −1 −15
401–600 500.5 14 0 0
601–800 700.5 7 1 7
    N = 50   fiui=-36

We know that mean,
Now, we have N=fi=50, h=200, A=500.5, fiui=-36
X¯=500.5+200150×-36=500.5-144=356.5



Page No 694:

Question 1:

Following are the lives in hours of 15 pieces of the components of aircraft engine. Find the median:

715, 724, 725, 710, 729, 745, 694, 699, 696, 712, 734, 728, 716, 705, 719

Answer:

First of all arranging the data in ascending order of magnitude, we have

Here,, which is an odd number

Therefore, median is the value of



Page No 695:

Question 2:

Calculate the median salary of the following data giving salaries of 280 persons:
 

Salary (in thousands) 5 – 10 10 – 15 15 – 20 20 – 25 25 – 30 30 – 35 35 – 40 40 – 45 45 – 50
No. of Persons 49 133 63 15 6 7 4 2 1

Answer:

Salary (In thousand Rs) Frequency CF
5 – 10 49 49
10 – 15 133 182
15 – 20 63 245
20 – 25 15 260
25 – 30 6 266
30 – 35 7 273
35 – 40 4 277
40 – 45 2 279
45 – 50 1 280

N2=2802=140
The cumulative frequency which is greater than and nearest to 140 is 182.
Median class = 10 – 15

We also have,

l (lower limit of median class) = 10

h (class size) = 5

n (number of observations) = 280

cf = (cumulative frequency of the class preceding the median class) = 49

f (frequency of median class) = 133


Median for grouped data is given by the formula :  

Median = l+n2-cff  ×  h

where is frequency of median class and cf is cumulative frequency of previous class.

Median = 10 + 140 - 49133×5 = Rs 13.42.
Disclaimer: The obtained answer does not match the answer in the book.

Page No 695:

Question 3:

Calculate the missing frequency from the following distribution, it being given that the median of the distribution is 24.
 

Age in years: 0−10 10−20 20−30 30−40 40−50
No. of persons: 5 25 ? 18 7

Answer:

Let the frequency of the class 20−30 be.It is given that median is 35 which lies in the class 20−30. So 20−30 is the median class.

Now, lower limit of median class

Height of the class

Frequency of median class

Cumulative frequency of preceding median class

Total frequency

Formula to be used to calculate median,

Where,

- Lower limit of median class

- Height of the class

- Frequency of median class

- Cumulative frequency of preceding median class

- Total frequency

Put the values in the above,

Hence, the required frequency is 25.

Page No 695:

Question 4:

The following table gives the distribution of the life time of 400 neon lamps:
 

Life time:
(in hours)
Number of lamps
1500−2000
2000−2500
2500−3000
3000−3500
3500−4000
4000−4500
4500−5000
14
56
60
86
74
62
48

Find the median life.

Answer:

We prepare the cumulative frequency table, as given below.

We have,

So,

Now, the cumulative frequency just greater than 200 is 216 and the corresponding class is .

Therefore, is the median class.

Here,

We know that

 
              
           

Hence, the median life of the lamps is approximately 3406.98 hours.

Page No 695:

Question 5:

The distribution below gives the weight of 30 students in a class. Find the median weight of students:
 

Weight (in kg): 40−45 45−50 50−55 55−60 60−65 65−70 70−75
No. of students: 2 3 8 6 6 3 2

Answer:

We prepare the cumulative frequency table, as given below.

We have,

So,

Now, the cumulative frequency just greater than 15 is 19 and the corresponding class is .

Therefore, is the median class.

Here,

We know that


             
           

Hence, the median weight of students is 56.67 kg.

Page No 695:

Question 6:

A survey regarding the height (in cm) of 51 girls of class X of a school was conducted and the following data was obtained:
 

Height in cm Number of Girls
Less than 140
Less than 145
Less than 150
Less than 155
Less than 160
Less than 165
4
11
29
40
46
51

Find the median height.

Answer:

We prepare the cumulative frequency, table as given below.

Now, we have

So,

Now, the cumulative frequency just greater than 25.5 is 40 and the corresponding class is .

Therefore, is the median class.

We know that

 
           
            = 150 − 1.59
            = 148.41

Hence, the median height is 148.41 cm.

Page No 695:

Question 7:

A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are only given to persons having age 18 years onward but less than 60 years.
 

Age in years Number of policy holders
Below 20
Below 25
Below 30
Below 35
Below 40
Below 45
Below 50
Below 55
Below 60
2
6
24
45
78
89
92
98
100

Answer:

We prepare the cumulative frequency, table as given below.

Now, we have

So,

Now, the cumulative frequency just greater than 50 is 78 and the corresponding class is .

Therefore, is the median class.

Here,

We know that

 
             
           

Hence, the median age is 35.76 years.

Page No 695:

Question 8:

The length of 40 leaves of a plant are measured correct to the nearest millimeter, and the data obtained is represented in the following table:
 

Length (in mm): 118−126 127−135 136−144 145−153 154−162 163−171 172−180
No. of leaves 3 5 9 12 5 4 2

Find the mean length of leaf.

Answer:

Calculation for mean.

Length (in mm) Mid-Values(xi) Number of Leaves(fi) fi xi
117.5–126.5 122 3 366
126.5–135.5 131 5 655
135.5–144.5 140 9 1260
144.5–153.5 149 12 1788
153.5–162.5 158 5 790
162.5–171.5 167 4 668
171.5–180.5 176 2 353
    N = 40 fixi=5880

Mean length of the leaf = 1Nfixi=140×5880=147

Calculation for median.

The given series is in inclusive form. Converting it to exclusive form and preparing the cumulative frequency table, we have

Length (in mm) Number of Leaves(fi) Cumulative Frequency (c.f.)
117.5–126.5 3 3
126.5–135.5 5 8
135.5–144.5 9 17
144.5–153.5 12 29
153.5–162.5 5 34
162.5–171.5 4 38
171.5–180.5 2 40
  N = 40  

Now, we have

So,

Now, the cumulative frequency just greater than 20 is 29 and the corresponding class is 144.5–153.5.

Therefore, 144.5–153.5 is the median class.

Here,

We know that


            =144.5+20-1712×9=144.5+2712=144.5+2.25=146.75  

Hence, the median length of leaf is 146.75 mm.

Disclaimer: If the question asks for the mean length of the leaf, then the answer is 147 mm whereas if the question asks fro the median length of the leaf, then the answer is 146.75 mm, which is same as the answer given in the book.

Page No 695:

Question 9:

An incomplete distribution is given below :
 

Variable: 10−20 20−30 30−40 40−50 50−60 60−70 70−80
Frequency: 12 30 65 25 18

You are given that the median value is 46 and the total number of items is 230.
(i) Using the median formula fill up missing frequencies.
(ii) Calculate the AM of the completed distribution.

Answer:

(i) Let the missing frequencies be x and y.

First we prepare the following cummulative table.

Here,

So,

It is given that the median is 46.

Therefore, is the median class.

We know that

 
 
 

Also,

Putting the value of x, we get

Hence, the missing frequencies are 34 and 46.

(ii)

We may prepare the table as shown.

We know that mean,

Now, we have.

Putting the values in the above formula, we have


   

Hence, the mean is 45.87.

Page No 695:

Question 10:

The median of the distribution given below is 14.4 . Find the values of x and y , if the total frequency is 20.

Class interval:       0-6      6-12         12-18           18-24         24-30

Frequency:              4          x               5                 y                  1

Answer:

The given series is in inclusive form. Converting it to exclusive form and preparing the cumulative frequency table, we have

Class interval Frequency(fi) Cumulative Frequency(c.f.)
0–6 4 4
6–12 x 4 + x
12–18 5 9 + x
18–24 y 9 + x + y
24–30 1 10 + x + y
  10 + x + y = 20  

Median = 14.4
It lies in the interval 12–18, so the median class is 12–18.
Now, we have
l=12, h=6, f=5, F=4+x, N=20
We know that

14.4=12+6×10-4-x512=36-6x6x=24x=4 
Now,
10 + x + y = 20
x+y=10y=10-4=6

Page No 695:

Question 11:

The median of the following data is 50. Find the values of p and q  , if the sum of all  the frequencies is 90 .

Marks:        20-30        30-40       40-50        50-60        60-70         70-80       80-90

Frequency:    p               15            25            20               q                 8              10

Answer:

The given series is in inclusive form. Converting it to exclusive form and preparing the cumulative frequency table, we have

Class interval Frequency(fi) Cumulative Frequency (c.f.)
20–30   p p
30–40 15 p + 15
40–50 25 p + 40
50–60 20 p + 60
60–70 q p + q + 60
70–80 8 p + q + 68
80–90 10 p + q + 78
  78 + p + q = 90  

Median = 50
It lies in the interval 50–60, so the median class is 50–60.
Now, we have
l=50, h=10, f=20, F=p+40, N=90
We know that

50=50+45-p+4020×100=5-p2p=5And, p+q+78=90 p+q=12q=12-5=7



Page No 703:

Question 1:

Find the mode of the following data:

(i) 3, 5, 7, 4, 5, 3, 5, 6, 8, 9, 5, 3, 5, 3, 6, 9, 7,4
(ii) 3, 3, 7, 4, 5, 3, 5, 6, 8, 9, 5, 3, 5, 3, 6, 9, 7, 4
(iii) 15, 8, 26, 25, 24, 15, 18, 20, 24, 15, 19, 15

Answer:

(i) 3, 5, 7, 4, 5, 3, 5, 6, 8, 9, 5, 3, 5, 3, 6, 9, 7, 4

The frequency table for the given data

Value x 3 4 5 6 7 8 9
Frequency f 4 2 5 2 2 1 2


We observe that the value 5 has the maximum frequency.

Hence, the mode of data is 5.

 

(ii) 3, 3, 7, 4, 5, 3, 5, 6, 8, 9, 5, 3, 5, 3, 6, 9, 7, 4

The frequency table for the given data

Value x 3 4 5 6 7 8 9
Frequency f 5 2 4 2 2 1 2

We observe that the value 3 has the maximum frequency.

Hence, the mode of data is 3.


(iii) 15, 8, 26, 25, 24, 15, 18, 20, 24, 15, 19, 15

The frequency table for the given data

Value x 8 15 18 19 20 24 25 26
Frequency f 1 4 1 1 1 2 1 1

We observe that the value 15 has the maximum frequency.

Hence, the mode of data is 15.

Page No 703:

Question 2:

The shirt sizes worn by a group of 200 persons, who bought the shirt from a store, are as follows:

Shirt Size: 37 38 39 40 41 42 43 44
Number of persons: 15 25 39 41 36 17 15 12

Find the modal shirt size worn by the group.

Answer:

 

Shirt Size 37 38 39 40 41 42 43 44
Number of persons 15 25 39 41 36 17 15 12


Here, shirt size 40 has the maximum number of persons.

Hence, the mode shirt size is 40.

Page No 703:

Question 3:

Find the mode of the following distribution:

(i)

Class-interval: 25−30 30−35 35−40 40−45 45−50 50−55
Frequency: 25 34 50 42 38 14

(ii) â€‹
Class: 0−10 10−20 20−30 30−40 40−50 50−60 60–70
Frequency: 8 10 10 16 12 6 7

Answer:

(i) Here, maximum frequency is 50 so the modal class is 35−40.

Therefore,



(ii)
Here, maximum frequency is 16 so the modal class is 30−40.

Therefore,

Lower limit of the modal class, l = 30

Width of the modal class, h = 10

Frequency of the modal class, f = 16

Frequency of the class preceding the modal class,  f1 = 10

Frequency of the class following the modal class,  f2 = 12

∴ Mode of the distribution

=l+f-f12f-f1-f2×h=30+16-102×16-10-12×10=30+610×10=36

Hence, the mode of the distribution is 36.

Page No 703:

Question 4:

Compare the modal ages of two groups of students appearing for an entrance test:
 

Age (in years): 16−18 18−20 20−22 22−24 24−26
Group A: 50 78 46 28 23
Group B: 54 89 40 25 17

Answer:

Age (in years) Group ‘A’ Group ‘B’
16–18  50 54
18–20  78 89
20–22  46 40
22–24 28 25
24–25 23 17

For group “A”

The maximum frequency is 78 so the modal class is 18–20.

Therefore,

For group “B”

The maximum frequency 89 so modal class 18–20.

Therefore,

Thus, the modal age of group A is 18.93 years whereas the modal age of group B is 18.83 years.

Page No 703:

Question 5:

The following data is gives the information on the observed lifetimes (in hours) of 225 electrical components:

Lifetimes (in hours) 0−20 20−40 40−60 60−80 80−100 100−120
No. of components 10 35 52 61 38 29

Determine the modal lifetimes of the components.

Answer:

Lifetimes (in hours) 0−20 20−40 40−60 60−80 80−100 100−120
No. of components 10 35 52 61 38 29

Here, the maximum frequency of electrical components is 61 so the modal class is 60−80.

Therefore,



Thus, the modal lifetimes of the components is 65.625 hours.

Page No 703:

Question 6:

The following table gives the daily income of 50 workers of  a factory :

Daily income (in â‚¹)          100-120        120-140       140-160        160-180       180-200

Number of workers :             12              14                  8                    6                  10

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

Answer:

Let the assumed mean a be 150.

The table for the given data can be drawn as

Class Interval

Number of Workers(fi)

Classmark(xi)

di = xi − 150

fidi

Cumulative Frequency(c.f.)

100−120

12

110

−40

−480

12

120−140

14

130

−20

−280

26

140−160

8

150

0

0

34

160−180

6

170

20

120

40

180−200

10

190

40

400

50

Total

50

   

−240

 

Mean is given by .

Thus, the mean of the given data is 145.2.

It can be seen in the data table that the maximum frequency is 14. The class corresponding to this frequency is 120−140.

∴ Modal class = 120 − 140

Lower limit of modal class (l) = 120

Class size (h) = 140 − 120 = 20

Frequency of modal class (f1) = 14

Frequency of class preceding the modal class (f1) = 12

Frequency of class succeeding the modal class (f1) = 8

Mode is given by

Thus, the mode of the given data is 125.

Here, number of observations (n) = 50

This observation lies in class interval 120−140.

Therefore, the median class is 120−140.

Lower limit of median class (l) = 120

Cumulative frequency of class preceding the median class(c.f.) = 12

Frequency of median class(f) = 14

Median of the data is given by

Thus, the median of the given data is 138.57.

Page No 703:

Question 7:

The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret, the two measures:
 

Number of students per Teacher Number of States/U.T Number of students per Teacher Number of State/ U.T
15−20
20−25
25−30
30−35
3
8
9
10
35−40
40−45
45−50
50−55
3
0
0
2

Answer:

Number of
students per teacher
Number of
states/U.T. (f
i)

xi

f i xi
15−20 3 17.5 52.5
20−25 8 22.5 180.0
25−30 9 27.5 247.5
30−35 10 32.5 325.0
35−40 3 37.5 112.5
40−45 0 42.5 0
45−50 0 47.5 0
50−55 2 52.5 105.0
   

Here, the maximum frequency is 10 so the modal class is 30−35.

Therefore,



Thus, the mode of the data is 30.6.

Now,
Mean of the data = fixifi=1022.535=29.2
Thus, the mean of the data is 29.2.

The mode of the number of students per teacher in the states is more than the mean of the number of students per teacher in the states.

Page No 703:

Question 8:

Find the mean, median and mode of the following data:
 

Classes: 0−50 50−100 100−150 150−200 200−250 250−300 300−350
Frequency: 2 3 5 6 5 3 1

Answer:

Classes Frequency (fi) xi fi xi C.f.
0−50 2 25 50 2
50−100 3 75 225 5
100−150 5 125 625 10
150−200 6 175 1050 16
200−250 5 225 1125 21
250−300 3 275 825 24
300−350 1 325 325 25
    fixi=4225  

Here, the maximum frequency is 6 so the modal class 150−200.

Therefore,

  

Thus, the mean of the data is 169.



Thus, the median of the data is 170.83.

      
           = 175

Thus, the mode of the data is 175.

Page No 703:

Question 9:

A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised in the table given below. Find the mode of the data:
 

Number of cars: 0−10 10−20 20−30 30−40 40−50 50−60 60−70 70−80
Frequency: 7 14 13 12 20 11 15 8

Answer:

The given data is shown below.

Number of cars Frequency
0−10 7
10−20 14
20−30 13
30−40 12
40−50 20
50−60 11
60−70 15
70−80 8

Here, the maximum frequency is 20 so the modal class is 40−50.

Therefore,



Mode=l+f-f12f-f1-f2×h=40+20-122×20-12-11×10=40+817×10=40+4.7=44.7

Thus, the mode of the data is 44.7.



Page No 704:

Question 10:

Find the mean, median and mode of the following data:
 

Classes: 0−20 20−40 40−60 60−80 80−100 100−120 120−140
Frequency: 6 8 10 12 6 5 3

Answer:

Consider the following data.

Class Frequency (fi) xi fi xi C.f.
0−20 6 10 60 6
20−40 8 30 240 14
40−60 10 50 500 24
60−80 12 70 840 36
80−100 6 90 540 42
100−120 5 110 550 47
120−140 3 130 390 50
     

Here, the maximum frequency is 12 so the modal class is 60−80.

Therefore,



Thus, the median of the data is 61.66.



Thus, the mean of the data is 62.4.

Mode


Thus, the mode of the data is 65.

Page No 704:

Question 11:

The frequency distribution table of agriculture holdings in a village is given below:

Area of land (in hectares):       1-3      3-5     5-7     7-9     9-11     11-13

Number of families:                 20       45      80      55        40         12

Find the modal agricultural holdings of the village .

Answer:

The maximum class frequency is 80. The class corresponding to this frequency is 5–7.
So, the modal class is 5–7.
l (the lower limit of modal class) = 5
f1 (frequency of the modal class) = 80
f0 (frequency of the class preceding the modal class) = 45
f2 (frequency of the class succeeding the modal class) = 55
h (class size) = 2
Mode=l+f1-f02f1-f0-f2×h=5+80-452×80-45-55×2=5+3560×2=5+3530=6.2

Hence, the modal agricultural holdings of the village is 6.2 hectares.

Page No 704:

Question 12:

The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.
 

Runs scored Number of batsman Runs scored Number of Batsman
3000−4000
4000−5000
5000−6000
6000−7000
4
18
9
7
7000−8000
8000−9000
9000−10000
10000−11000
6
3
1
1

Find the mode of data.

Answer:

The given data is shown below.

Runs scored Number of batsmen
3000−4000 4
4000−5000 18
5000−6000 9
6000−7000 7
7000−8000 6
8000−9000 3
9000−10000 1
10000−11000 1

Here, the maximum frequency is 18 so the modal class is 4000-5000.

Therefore,



Mode=l+f-f12f-f1-f2×h=4000+18-42×18-4-9×1000=4000+1423×1000=4000+608.7=4608.7


Thus, the mode of the data (or runs scored) is 4608.7.

Page No 704:

Question 13:

 The monthly income of 100 families are given as below :
    Income in ( in  â‚¹)                                     Number of families

       0-5000                                                           8
5000-10000                                                        26
10000-15000                                                      41
15000-20000                                                      16
20000-25000                                                      3
25000-30000                                                      3
30000-35000                                                      2
35000-40000                                                      1
Calculate the modal income.

Answer:

The maximum class frequency is 41. The class corresponding to this frequency is 10000–15000.
So, the modal class is 10000–15000.
l (the lower limit of modal class) = 10000
f1 (frequency of the modal class) = 41
f0 (frequency of the class preceding the modal class) = 26
f2 (frequency of the class succeeding the modal class) = 16
h (class size) = 5000
Mode=l+f1-f02f1-f0-f2×h=10000+41-262×41-26-16×5000=10000+1540×5000=10000+1875=11875
Thus, the modal income is ₹11875.



Page No 715:

Question 1:

Draw an ogive to represent the following frequency distribution:

Class-interval: 0−4 5−9 10−14 15−19 20−24
Frequency: 2 6 10 5 3

Answer:

Firstly, prepare the cumulative frequency table.

Class Interval No. of students Less than Cumulative frequency Suitable points
0-4 2 4 2 (4, 2)
5-9 6 9 8 (9, 8)
10-14 10 14 18 (14, 18)
15-19 5 19 23 (19, 23)
20-24 3 24 26 (24, 26)

Now, plot the less than ogive using the suitable points.

Page No 715:

Question 2:

The monthly profits (in Rs.) of 100 shops are distributed as follows:
 

Profits per shop: 0−50 50−100 100−150 150−200 200−250 250−300
No. of shops: 12 18 27 20 17 6

Draw the frequency polygon for it.

Answer:

Firstly, we make a cumulative frequency table.

Profit per shop No. of shop More than profit Cumulative frequency Suitable points
0-50 12 0 100 (0, 100)
50-100 18 50 88 (50, 88)
100-150 27 100 70 (100, 70)
150-200 20 150 43 (150, 43)
200-250 17 200 23 (200, 23)
250-300 6 250 6 (250, 6)

Now, plot the frequency polygon (or more than ogive) using suitable points.

Page No 715:

Question 3:

The following distribution gives the daily income of 50 workers of a factory:

Daily income in (Rs): 100−120 120−140 140−160 160−180 180−200
Number of workers: 12 14 8 6 10

Convert the above distribution to a less than type cumulative frequency distribution and draw its ogive.

Answer:

Daily income
(in Rs.)
No. of workers (f) Daily income (less than) Cumulative frequency Suitable points
100−120 12 120 12 (120, 12)
120−140 14 140 26 (140, 26)
140−160 8 160 34 (160, 34)
160−180 6 180 40 (180, 40)
180−200 10 200 50 (200, 50)


Now, plot the less than ogive with suitable points.
 

Page No 715:

Question 4:

The following table gives production yield per hectare of wheat of 100 farms of a village:
 

Production yield in kg per hectare: 50−55 55−60 60−65 65−70 70−75 75−80  
Number of farms: 2 8 12 24 38 16  

Draw 'less than' ogive and 'more than' ogive.

Answer:

Prepare a table for less than type.
 


Production
yield

No. of
farms
Production
yield
(less than)

 

Cumulative
frequency

Suitable
points
50−55 2 55 2 (55, 2)
55−60 8 60 10 (60, 10)
60−65 12 65 22 (65, 22)
65−70 24 70 46 (70, 46)
70−75 38 75 84 (75, 84)
75−80 16 80 100 (80, 100)

Now, plot the less than ogive using suitable points.

Again, prepare a table for more than type.
 

Production
yield
No. of
farms
Production
yield
(more than)
Cumulative
frequency
Suitable
points
50-55 2 50 100 (50, 100)
55-60 8 55 98 (55, 98)
60-65 12 60 90 (60, 90)
65-70 24 65 78 (65, 78)
70-75 38 70 54 (70, 54)
75-80 16 75 16 (75, 16)


Now, plot the more than ogive with suitable points.


Page No 715:

Question 5:

During the medical check up of 35 students of a class, their weights were recorded as follows:
 

Weight (in kg) Number of students
Less than 38
Less than 40
Less than 42
Less than 44
Less than 46
Less than 48
Less than 50
Less than 52
0
3
5
9
14
28
32
35

Draw a less than type ogive for the given data. Hence, obtain the median weight from the graph and verify the result by using the formula.

Answer:

Prepare the table for cumulative frequency for less than type.

Weight
(in kg)
No. of
students
Weight
(less than)
Cumulative
frequency
Suitable
points
36–38 0 38 0 (38, 0)
38–40 3 40 3 (40, 3)
40–42 2 42 5 (42, 5)
42–44 4 44 9 (44, 9)
44–46 5 46 14 (46, 14)
46–48 14 48 28 (48, 28)
48–50 4 50 32 (50, 32)
50–52 3 52 35 (52, 35)

Now, draw the less than ogive using suitable points.



Here, =

From (0, 17.5) draw a line parallel to horizontal axis, which intersects the graph at the point M having x-coordinate as 46.5.

Therefore, the median is 46.5 kg.

Calculation for median using formula.

Now,

So, 46–48 is the median class.

Here,

Hence, both the methods give same result.

Page No 715:

Question 6:

The annual rainfall record of a city for 66 days is given in the following table :
Rainfall (in cm ):        0-10      10-20    20-30      30-40      40-50     50-60

Number of days :        22            10        8              15          5              6

Calculate the median rainfall using ogives of more than type and less than type.

Answer:

Prepare a table for less than type.
 

Rainfall
(in cm)
Number of
Days
Rainfall
(Less Than)
Cumulative
Frequency
Suitable
Points
0−10 22 10 22 (10, 22)
10−20 10 20 32 (20, 32)
20−30 8 30 40 (30, 40)
30−40 15 40 55 (40, 55)
40−50 5 50 60 (50, 60)
50−60 6 60 66 (60, 66)

Now, plot the less than ogive using suitable points.


Here, N = 66.
N2=33
In order to find the median rainfall, we first locate the point corresponding to 33rd day on the y-axis. Let the point be P. From this point draw a line parallel to the x-axis cutting the curve at Q. From this point Q, draw a line parallel to the y-axis and meeting the x-axis at the point R. The x-coordinate of R is 21.25.

Thus, the median rainfall is 21.25 cm.

Let us now prepare a table for more than type.

Rainfall (in cm) Number of
Days
Rainfall
(More Than)
Cumulative
Frequency
Suitable
Points
0−10 22 0 66 (0, 66)
10−20 10 10 44 (10, 44)
20−30 8 20 34 (20, 34)
30−40 15 30 26 (30, 26)
40−50 5 40 11 (40, 11)
50−60 6 50 6 (50, 6)

Now, plot the more than ogive with suitable points.


Here, N = 66.
N2=33
In order to find the median rainfall, we first locate the point corresponding to 33rd day on the y-axis. Let the point be P. From this point draw a line parallel to the x-axis cutting the curve at Q. From this point Q, draw a line parallel to the y-axis and meeting the x-axis at the point R. The x-coordinate of R is 21.25.



Thus, the median rainfall is 21.25 cm.

Page No 715:

Question 7:

Change the following distribution to a 'More than type' distribution. Hence, draw the more than type' ogive for this distribution.
 

Class – interval: 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 70 – 80 80 – 90
Frequency: 10 8 12 24 6 25 15

Answer:


The cumulative frequency table of 'More than type' is given below.
 

Class Interval Cumulative Frequency
More than 20 100
More than 30 90
More than 40 82
More than 50 70
More than 60 46
More than 70 40
More than 80 15

Mark the lower class class limits along the x-axis and the cumulative frequencies along y-axis.

Plot the points (20, 100), (30, 90), (40, 82), (50, 70), (60, 46), (70, 40) and (80, 15). Join the points to obtain the 'More than type' ogive.

Page No 715:

Question 8:

The following distribution gives daily income of 50 workers of a factory:
 

Daily income(in ₹): 200 – 220 220 – 240 240 – 260 260 – 280 280 – 300
Number of workers: 12 14 8 6 10

Convert the distribution above to a 'less than type' cumulative frequency distribution and draw its ogive.

Answer:


The cumulative frequency table of 'Less than type' is given below.
 

Daily income(in ₹) Cumulative Frequency
Less than 220 12
Less than 240 26
Less than 260 34
Less than 280 40
Less than 300 50


Mark the upper class class limits along the x-axis and the cumulative frequencies along y-axis.

Plot the points (220, 12), (240, 26), (260, 34), (280, 40) and (300, 50). Join the points to obtain the 'Less than type' ogive.

Page No 715:

Question 9:

The following table gives the height of trees:
 

Height No. of trees
Less than 7
Less than 14
Less than 21
Less than 28
Less than 35
Less than 42
Less than 49
Less than 56
26
57
92
134
216
287
341
360

Draw 'less than' ogive and 'more than' ogive.

Answer:

Consider the following table.
 

Height (less than) Height-class No. of trees Cumulative frequency Suitable points
7 0-7 26 26 (7, 26)
14 7-14 31 57 (14, 57)
21 14-21 35 92 (21, 92)
28 21-28 42 134 (28, 134)
35 28-35 82 216 (35, 216)
42 35-42 71 287 (42, 287)
49 42-49 54 341 (49, 341)
56 49-56 19 360 (56, 360)

Now, draw the less than ogive using suitable points.

Now, prepare the cumulative frequency table for more than series.

Height No. of trees Height (more than) Cumulative frequency Suitable points
0-7 26 0 360 (0, 360)
7-14 31 7 334 (7, 334)
14-21 35 14 303 (14, 303)
21-28 42 21 268 (21, 268)
28-35 82 28 226 (28, 226)
35-42 71 35 144 (35, 144)
42-49 54 42 73 (42, 73)
49-56 19 49 54 (49, 54)

Now, draw the more than ogive using suitable points.



Page No 716:

Question 10:

The annual profits earned by 30 shops of a shopping complex in a locality give rise to the following distribution:
 

Profit (in lakhs in Rs) Number of shops (frequency)
More than or equal to 5
More than or equal to 10
More than or equal to 15
More than or equal to 20
More than or equal to 25
More than or equal to 30
More than or equal to 35
30
28
16
14
10
7
3

Draw both ogives for the above data and hence obtain the median.

Answer:

Firstly, we prepare the cumulative frequency table for less than type.
 


Profit
(In lakh in Rs.)

No. of shops

(frequency)


Profit (less than)

Cumulative
frequency

Suitable
points
5-10 2 10 2 (10, 2)
10-15 12 15 14 (15, 14)
15-20 2 20 16 (20, 16)
20-25 4 25 20 (25, 20)
25-30 3 30 23 (30, 23)
30-35 4 35 27 (35, 27)
35-40 3 40 30 (40, 30)


Again, prepare the cumulative frequency table for more than type.
 


Profit
(In lakh in Rs.)

No. of shops

(frequency)



Profit (more than)

Cumulative
frequency

Suitable
points

5-10 2 5 30 (5, 30)
10-15 12 10 28 (10, 28)
15-20 2 15 16 (15, 16)
20-25 4 20 14 (20, 14)
25-30 3 25 10 (25, 10)
30-35 4 30 7 (30, 7)
35-40 3 35 3 (35, 3)


Now, “more than ogive” and “less than ogive” can be drawn as follows:
 


The x-coordinate of the point of intersection of the “more-than ogive” and “less-than ogive” gives the median of the given distribution..


So, the corresponding median is Rs 17.5 lakh.

Page No 716:

Question 1:

Define mean.

Answer:

Mean

The mean of a set of observation is equal to sum of observations divided by the number of observations.

(Where = sum of the observation and n = number of observations)

Page No 716:

Question 2:

What is the algebraic sum of deviation of a frequency distribution about its mean?

Answer:

The algebraic sum of deviation of a frequency distribution about its mean is zero.

 

 

Page No 716:

Question 3:

Which measure of central tendency is given by the x-coordinate of the point of intersection of the 'more than' ogive and 'less than' ogive?

Answer:

Median

 

Page No 716:

Question 4:

Write the empirical relation between mean, mode and median.

Answer:

The empirical relation between mean, median and mode is

Mode = 3 Median − 2 Mean

 

Page No 716:

Question 5:

Which measure of central tendency can be determine graphically?

Answer:

Median can be determined graphically.

Page No 716:

Question 6:

What is the value of the median of the data using the graph in the following figure of less than ogive and more than ogive?
 

Answer:

We know that the abscissa of the point of intersection of two ogives gives the median.

From the given figure, it can be seen that both the ogives intersect at the point (4, 15).

∴ Median of the data = 4

Page No 716:

Question 7:

A student draws a cumulative frequency curve for the marks obtained by 40 students of a class as shown below. Find the median marks obtained by the students of the class.
 

Answer:

Here, N = 40

So,

Draw a line parallel to x-axis from the point (0, 20), intersecting the graph at point P.



Now, draw PM from P on the x-axis. The x-coordinate of M gives us the median.

∴ Median = 50

Page No 716:

Question 8:

Write the modal class for the following frequency distribution:
 

Class-interval: 10−15 15−20 20−25 25−30 30−35 35−40
Frequency: 30 35 75 40 30 15

Answer:

Class Interval 10−15 15−20 20−25 25−30 30−35 35−40
Frequency 30 35 75 40 30 15

Here, the maximum frequency is 75 and the corresponding class-interval is 20−25.

Therefore, 20−25 is the modal class.

Page No 716:

Question 9:

Write the median class for the following frequency distribution:
 

Class-interval: 0−10 10−20 20−30 30−40 40−50 50−60 60−70 70−80
Frequency: 5 8 7 12 28 20 10 10

Answer:

We are given the following table.

Class Interval Frequency Cumulative Frequency
0−10 5 5
10−20 8 13
20−30 7 20
30−40 12 32
40−50 28 60
50−60 20 80
60−70 10 90
70−80 10 100
  N = 100  

Here, N = 100

N2=50

The cumulative frequency just greater than 50 is 60.

So, the median class is 40−50.

Page No 716:

Question 10:

In the graphical representation of a frequency distribution, if the distance between mode and mean is k times the distance between median and mean, then write the value of k.

Answer:

We have,

Mode = 3 Median − 2 Mean

⇒ Mode − Mean = 3 Median − 3 Mean

⇒ Mode − Mean = 3(Median − Mean)             .....(1)

 It is given that,

 Mode − Mean = k (Median − Mean)               .....(2)

From (1) and (2), we get

k = 3

Page No 716:

Question 11:

Find the class marks of classes 10−25 and 35−55.

Answer:

Class mark of 10−25

Class mark of 35−55



Page No 717:

Question 12:

Write the median class of the following distribution:
 

Class-interval: 0−10 10−20 20−30 30−40 40−50 50−60 60−70
Frequency: 4 4 8 10 12 8 4

Answer:

Consider the following distribution table.

Class Frequency C.F.
0−10 4 4
10−20 4 8
20−30 8 16
30−40 10 26
40−50 12 38
50−60 8 46
60−70 4 50
  N = 50  

Here,

The cumulative frequency just greater than 25 is 26.

So, the median class is 30−40.

Page No 717:

Question 1:

If the mean of x, y, z is y, then mean of x and z is ________.

Answer:


It is given that the mean of x, y, z is y.

x+y+z3=y                            Mean=Sum of observationsTotal number of observationsx+y+z=3yx+z=3y-y=2y            .....1

Mean of x and z =x+z2=2y2=y         [Using (1)]

If the mean of x, y, z is y, then mean of x and z is ____y____.

Page No 717:

Question 2:

The mode of the data 2, x, 3, 4, 5, 2, 4, 6 where x > 2, is ________.

Answer:


Disclaimer: The question has been modified to match the answer given in the book as given below. The answer provided for the question is 4.

The mode of the data 2, x, 3, 4, 5, 2, 4, 6, 4 where x > 2, is ________.

Solution:

Mode is the value which occur most frequently in a set of observations.

Here, x is a variable which can take any value greater than 2. It can be 3 or 4 or 5 or .... .

However, in this set of observations, the value 4 occurs most frequently i.e. 3 times.

So, the mode of the given data is 4.

The mode of the data 2, x, 3, 4, 5, 2, 4, 6, 4 where x > 2, is ____4____.



Solution for question given in the book.

Mode is the value which occur most frequently in a set of observations.

Here, x is a variable which can take any value greater than 2. It can be 3 or 4 or 5 or .... .

However, in this set of observations, the value 2 and 4 occurs most frequently i.e. 2 times.

So, the mode of the given data is 2 and 4.


When x = 3, then the data set is 2, 3, 3, 4, 5, 2, 4, 6. So, the mode is 2, 3 and 4.

When x = 4, then the data set is 2, 4, 3, 4, 5, 2, 4, 6. So, the mode is 4.

When x = 5, then the data set is 2, 5, 3, 4, 5, 2, 4, 6. So, the mode is 2, 4 and 5.

When x = 6, then the data set is 2, 6, 3, 4, 5, 2, 4, 6. So, the mode is 2, 4 and 6.

For x = 7, 8, 9,...

The mode is 2 and 4.

Page No 717:

Question 3:

The mean of first 673 natural numbers is _________.

Answer:


The first 673 natural numbers are 1, 2, 3, ..., 673.

These are in AP with a = 1, l = 673 and n = 673.

∴ Sum of first 673 natural numbers

=1+2+3+...+673=67321+673                 Sn=n2a+l=6732×674=673×337

Now,

Mean of first 673 natural numbers =Sum of first 673 natural numbers673=673×337673=337

The mean of first 673 natural numbers is ___337___.

Page No 717:

Question 4:

The mean of the observations 425, 430, 435, ..., 495 is ___________.

Answer:


The observations 425, 430, 435, ..., 495 are in AP.

Here, a = 425 and d = 5.

Let the number of observation be n.

495=425+n-1×5              an=a+n-1d5n-1=495-425=70n-1=14n=15

Sum of the given observations =152425+495=152×920=15×460                   Sn=n2a+l

∴ Mean of the given observations

=Sum of the given observationsNumber of observations=15×46015=460

The mean of the observations 425, 430, 435, ..., 495 is _____460______.

Page No 717:

Question 5:

If the mean and median of a unimodal data are 34.5 and 32.5 respectively, then mode of the data is __________.

Answer:


Mean of the data = 34.5

Median of the data = 32.5

Now,

Mode = 3 × Median − 2 × Mean
         =3×32.5-2×34.5=97.5-69=28.5

If the mean and median of a unimodal data are 34.5 and 32.5 respectively, then mode of the data is ___28.5___.

Page No 717:

Question 6:

The mean of first n odd natural numbers is _________.

Answer:


The first n odd natural numbers are 1, 3, 5, ..., 2n − 1. 

This forms an AP with a = 1 and l = 2n − 1.

Sum of first n odd natural numbers =n21+2n-1=n2                 Sn=n2a+l

∴ Mean of first n odd natural numbers=Sum of first n odd natural numbersn=n2n=n

The mean of first n odd natural numbers is ____n____.

Page No 717:

Question 7:

The mean of the observations 1, 3, 5, 7, 9, ..., 99 is _________.

Answer:


The observations 1, 3, 5, 7, 9, ..., 99 are first 50 odd natural numbers.

We know that the mean of first n odd natural numbers is n.

Here, n = 50

So, the mean of the observations 1, 3, 5, 7, 9, ..., 99 is 50.

The mean of the observations 1, 3, 5, 7, 9, ..., 99 is ____50____.

Page No 717:

Question 8:

If the mean of first n natural numbers is 20, then n = __________.

Answer:


The first n natural numbers are 1, 2, 3, ..., n.

This forms an AP with a = 1 and d = 1.

Sum of first n natural numbers = 1 + 2 + 3 + ... + n = n21+n=nn+12                 Sn=n2a+l

∴ Mean of first n natural numbers = Sum of first n natural numbersn=nn+12n=n+12

Now,

n+12=20             Givenn+1=40n=40-1=39

If the mean of first n natural numbers is 20, then n = _____39_____.

Page No 717:

Question 9:

If a mode exceeds a mean by 12, then the mode exceeds median by ___________.

Answer:


Given: Mode − Mean = 12        .....(1)

We know that,

Mode = 3Median − 2Mean

∴ Mode − Mean = 3(Median − Mean)

⇒ 3(Median − Mean) = 12

⇒ Median − Mean = 4           .....(2)

Subtracting (2) from (1), we get

(Mode − Mean) − (Median − Mean) = 12 − 4

⇒ Mode − Median = 8

So, the mode exceeds median by 8.

If a mode exceeds a mean by 12, then the mode exceeds median by ____8____.

Page No 717:

Question 10:

If the median of the observations x1, x2, x3, x4, x5, x6, x7, x8 is m, then the median of the observations x3, x4, x5, x6 (where x1 < x2 < x3 < x4 < x< x< x< x8) is _________.

Answer:


If the number of observations i.e. n is even, then

Median = Mean of n2th observation and n2+1th observation

Now,

The median of the observations x1, x2, x3, x4, x5, x6, x7, x8 is m. Here, n = 8.

4th observation+5th observation2=mx4+x52=mx4+x5=2m         .....1


Consider the observations x3, x4, x5, x6. Here, n = 4.

∴ Median of the observations  x3, x4, x5, x6 = 2nd observation+3rd observation2=x4+x52=2m2=m    [Using (1)]

If the median of the observations x1, x2, x3, x4, x5, x6, x7, x8 is m, then the median of the observations x3, x4, x5, x6 (where x1 < x2 < x3 < x4 < x< x< x< x8) is ____m____.

Page No 717:

Question 11:

If mode – median = 2, then median – mean = ___________.

Answer:


We know that,

Mode = 3Median − 2Mean

⇒ Mode − Median = 2(Median − Mean)

∴ 2(Median − Mean) = 2                   (Mode − Median = 2)

⇒ Median − Mean = 1

If mode – median = 2, then median – mean = ____1____.

Page No 717:

Question 12:

If the average of a, b,c,d is the average of b and c, then the value of a – b – c + d is ___________.

Answer:


Average of a, b, c, d = a+b+c+d4

Average of b and c = b+c2

It is given that,

Average of a, b, c, d = Average of b and c

a+b+c+d4=b+c2a+b+c+d=2b+2ca+b+c+d-2b-2c=0a-b-c+d=0

If the average of a, b, c, d is the average of b and c, then the value of a – b – c + d is _____0_____.

Page No 717:

Question 13:

Given that a, b, c, d are non-zero integers such that a < b < c < d. If the mean and median of a, b, c, d are equal to zero, then a = ________ and b = ________.

Answer:


Mean of a, b, c, d =a+b+c+d4
a+b+c+d4=0               Givena+b+c+d=0               .....1

Now,

Median of a, b, c, d = Mean of 2nd observation and 3rd observation =b+c2
b+c2=0                 Givenb+c=0                .....2

From (1) and (2), we get

a+d=0a=-d

Also, b+c=0b=-c

Given that a, b, c, d are non-zero integers such that a < b < c < d. If the mean and median of a, b, c, d are equal to zero, then a = ___−d___ and b = ___−c___.

Page No 717:

Question 14:

If a < b < 2a and mean and median of a, b and 2a are 15 and 12 respectively, then a = _________.

Answer:


The given observations are a, b and 2a.

Mean of the given observations = 15
a+b+2a3=153a+b=45            .....1

If the number of observations i.e. n is odd, then

Median = n+12th observation

Here, n = 3.

Median of the given observations = 2nd observation = b = 12         .....(2)

From (1) and (2), we get

3a+12=453a=45-12=33a=11

If a < b < 2a and mean and median of a, b and 2a are 15 and 12 respectively, then a = ____11____.

Page No 717:

Question 15:

If the mean of 26, 19, 15, 24 and x is x, then median of the data is __________.

Answer:


The given data is 26, 19, 15, 24 and x.

Mean of the given data = x

26+19+15+24+x5=x84+x=5x4x=84x=21

The given data is 26, 19, 15, 24 and 21.

Arrange the data in increasing order.

15, 19, 21, 24, 26

Here, n = 5

∴ Median of the data = 5+12th observation = 3rd observation = 21

If the mean of 26, 19, 15, 24 and x is x, then median of the data is _____21_____.



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