Class  10 
Chapter  Statistics 
Subject  Math 
Category  Important Question Answer 
Class 10 Math Chapter 13 Important Question Answer
Q1. 100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows:
Number of letters  14  47  710  1013  1316  1619 
Number of surnames  6  30  40  16  4  4 
Find the median numbers of letters in the surnames.
Ans.
Number of letters  Number of surnames  Cumulative Frequency 
14
47
710
1013
1316
1619 
6
30
40
16
4
4 
6
36
76
92
96
100 
Now, n=100, . This observation lies in the class 710.
Then, l (lower limit) = 7,
cf (the cumulative frequency of the class preceding 710) = 36
f (the frequency of the median class 710) = 40,
h (the class size) = 3.
Using the formula, Median = l + × h
Median =7 + =7 + 1.05 = 9.05
Q2. The median of the following data is 28.5. Find the values of x and y, if the total frequency is 60.
Class interval  010  1020  2030  3040  4050  5060 
Frequency  5  x  20  15  y  5 
Ans.
ClassInterval  Frequency  Cumulative Frequency 
010
1020
2030
3040
4050
5060

5
x
20
15
y
5 
5
5 + x
25 + x
40 + x
40 + x + y
45 + x + y 
It is given that n = 60
So, 45 + x + y = 60 i.e., x + y = 15
The Median is 28.5, which lies in the class 2030
So, l = 20, f = 20, cf = 5 + x, h = 10
Using the formula, Median = l + × h
28.5 = 20 +
28.5 – 20 =
17 = 25 – x
x = 8
putting vlaue of x in x + y = 15,
we get y = 7
Q3. Consider the distribution of daily wages of 50 workers of a factory :
Daily Wages (in Rs.)  100120  120140  140160  160180  180200 
Number of workers  12  14  8  6  10 
Find the mean daily wages of the workers of the factory by using an appropriate method.
Ans.
As the given data is large, we will use step deviation method to find the mean.
Here a = 150, h = 20
Class Interval  Frequency
(f_{i}) 
Class Marks
(u_{i}) 
f_{i}u_{i}  
100120
120140 140160 160180 180200 
12
14 8 6 10 
110
130 150 (a) 170 190 
2
1 0 1 2

24
14 0 6 20 
Step Deviation Method : = a + × h
= 150 + = 150 –
= 145.20
Hence, the mean daily expenditure on food is Rs.145.20
Q4. The following distribution shows the daily pocket allowance of children of a locality: The mean pocket allowance is Rs. 18. Find the missing frequency f.
Daily pocket (in Rs.)  1113  1315  1517  1719  1921  2123  2325 
Number of chapter  7  6  9  13  f  5  4 
Ans.
Daily pocket allowances ( in Rs.)  Class Mark
(x_{i}) 
Number of children
(f_{i}) 
D_{i }= x_{i }– 18  f_{i}d_{i} 
1113
1315 1517 1719 1921 2123 2325 
12
14 16 18 ( let a = 18) 20 22 24 
7
6 9 13 f 5 4 
6
4 2 0 2 4 6 
42
24 18 0 2f 20 24 
Total 
We have,
Assumed Mean Method : = a +
18 = 18 +
0 =
2f – 40 = 0
f = 20
Therefore, required frequency is 20.
Q5. The following data gives the information on the lifetime (in hours) of 75 electrical instruments:
Lifetime (in hours)  020  2040  4060  6080  80100  100120 
Frequency  10  15  12  21  8  9 
Determine the modal lifetime of the components.
Ans.
Here the maximum frequency is 21. So the modal class is 6080.
Now,
Modal class = 6080, lower limit (l) of modal class = 60, class size (h) = 20
Frequency (f_{1}) of modal class = 21
Frequency (f_{0}) of class preceding the modal class = 12,
Frequency (f_{2}) of class succeeding the modal class = 8
Now, Using Mode = l + × h
Mode = 60 + = 60 + = 68.18
Q6. The median of the following data is 525. Find the value of x and y, if the total frequency is 100:
ClassInterval  Frequency 
0100
100200 200300 300400 400500 500600 600700 700800 800900 9001000 
2
5 x 12 17 20 y 9 7 4 
Ans.
ClassInterval  Frequency  Cumulative Frequency 
0100
100200
200300
300400
400500
500600
600700
700800
800900
9001000 
2
5
x
12
17
20
y
9
7
4 
2
7
7+x
19 + x
36 + x
56 + x
56 + x + y
65 + x + y
72 + x + y
76 + x + y

It is given that n = 100
So, 76 + x + y = 100 i.e., x + y = 24
The Median is 525, which lies in the class 500600
So, l = 500, f = 20, cf = 36 + x, h = 100
Using the formula, Median = l + × h
525 = 500 +
525 – 500 = (14 – x)×5
25 = 70 – 5x
5x = 70 – 25 = 45
x = 9
putting value of x in x + y = 24,
we get 9 + y = 24
y = 15
Q7. A survey regarding the heights (in cm) of 51 girls of class X of a school was conducted and the following data was :
Obtained height (in cm)  Less than 140  Less than 145  Less than 150  Less than 155  Less than 160  Less than 165 
Number of girls  4  11  29  40  46  51 
Find the median height.
Ans.
The given distribution being of the less than type. Therefore, the classes should be below 140, 140145, 145150, ….., 160165.
The frequency of class interval below 140 is 4, frequency of class interval 140145 is 114=7. Similarly the frequency of 145150 is 2911=18 and so on.
So, our frequency distribution table with the given cumulative frequencies becomes :
Class Interval  Frequency  Cumulative Frequency (cf) 
Below 140  4  4 
140145  7  11 
145150  18  29 
150155  11  40 
155160  6  46 
160165  5  51 
Now, n=51, . This observation lies in the class 145150.
Then, l (lower limit) = 145,
cf (the cumulative frequency of the class preceding 145150) = 11
f (the frequency of the median class 145150) = 18,
h (the class size) = 5.
Using the formula, Median = l + × h
Median = 145 +
= 145 + = 149.03
Q8. The table below shows daily expenditure on food of 25 households in a locality :
Daily Expenditure (in Rs.)  100150  150200  200250  250300  300350 
No. of households.  4  5  12  2  2 
Find the mean daily expenditure on food by suitable method.
Ans.
Here, a = 225 and h = 50
Class Interval  Frequency  Class Marks  f_{i}u_{i}  
100150
150200 200250 250300 300350 
4
5 12 2 2 
125
175 225 (a) 275 325 
2
1 0 1 2

8
5 0 2 4 
Step Deviation Method : = a + × h
= 225 + = 225 – 14
= 211.
Hence, the mean daily expenditure on food is Rs.211.
Q9. The following distribution shows the daily pocket money of children of a school:
Daily Pocket Money (Rs.)  1113  1315  1517  1719  1921  2123  2325 
Number of Children  7  6  9  13  20  5  4 
Find the average daily pocket money of children.
Ans.
Daily pocket allowances ( in Rs.)  Class Mark
(x_{i}) 
Number of children
(f_{i}) 
D_{i} = x_{i} – 18  f_{i}d_{i} 
1113
1315 1517 1719 1921 2123 2325 
12
14 16 18 ( let a = 18) 20 22 24 
7
6 9 13 20 5 4 
6
4 2 0 2 4 6 
42
24 18 0 40 20 24 
Total 
We have,
Assumed Mean Method : = a +
= 18 + = 18
Therefore, the average daily pocket money of children is Rs. 18.
Q10. The following distribution gives the monthly consumption of consumers of a locality. Find the median of the distribution.
Monthly consumption (in units)  6585  85105  105125  125145  145165  165185 
Number of consumers  4  8  13  20  14  4 
Ans.
Monthly consumption
(in units) 
Number of consumers
(f) 
Cumulative Frequency
(cf) 
6585  4  4 
85105  8  12 
105125  13  25 
125145  20  45 
145165  14  59 
165185  4  63 
Now, n = 63, . This observation lies in the class 125145 .
Then, l (lower limit) = 125,
cf (the cumulative frequency of the class preceding 125145) = 25,
f (the frequency of the median class 125145) = 20,
h (the class size) = 20.
Using the Formula, Median = l + × h
Median = 125 + = 125 + 6.5 = 126.5
Q11. The wickets taken by a bowler in 10 cricket matches are as follows:
3 5 2 1 2 0 5 1 2 4
Find the mode of the data.
Ans. Arranging data in ascending order : 0, 1, 2, 2, 2, 3, 4, 5, 5
From above data we can clearly conclude that 3 is the mode of data as it is used in max times.
Therefore,
Mode = 3
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