# NCERT Solutions for Class 10 Maths:Chapter 14 Statistics Exercise 14.1

In this page we have NCERT Solutions for Class 10 Maths:Chapter 14 Statistics for Exercise 14.1 . Hope you like them and do not forget to like , social_share and comment at the end of the page.
Question 1
A survey was conducted by a group of students as a part of their environment awareness programmes, in which they collected the following data regarding the number of 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?
Solution
We will use direct method because the numerical value of fi and xi are small
 No. of plants  (Class interval) No. of houses (fi) Mid-point (xi) fi xi 0-2 1 1 1 2-4 2 3 6 4-6 1 5 5 6-8 5 7 35 8-10 6 9 54 10-12 2 11 22 12-14 3 13 39 ∑fi = 20 ∑fi xi = 162

Mean(M) = 162/20 = 8.1

Question 2
Consider the following distribution of daily wages of 50 workers of a factory.
 Daily wages (in Rs.) 100-120 120-140 140-160 160-180 180-200 Number of workers 12 14 8 6 10
Find the mean daily wages of the workers of the factory by using an appropriate method.
Solution
Here The value of mid-point (xi) is very large. So, assumed mean or step deviation method would be simple to calculate
Here the class interval is h = 20.
We can see that step deviation method would be best here
Now assumed mean A = 150 and class interval is h = 20.
So, u= (xi - A)/h
u= (xi - 150)/20
 Daily wages    (Class interval) Number of workers frequency (fi) Mid-point (xi) ui = (xi - 150)/20 fiui 100-120 12 110 -2 -24 120-140 14 130 -1 -14 140-160 8 150 0 0 160-180 6 170 1 6 180-200 10 190 2 20 Total ∑ fi = 50 ∑ fiui = -12

=150 + (20 × -12/50) = 150 - 4.8 = 145.20
Thus, mean daily wage = Rs. 145.20
Question 3
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 Allowance (in c) 11-13 13-15 15-17 17-19 19-21 21-23 23-25 Number of Children 7 6 9 13 f 5 4

Solution
Here, Mean= 18
 Class interval Number of children (fi) Mid-point (xi) fixi 11-13 7 12 84 13-15 6 14 84 15-17 9 16 144 17-19 13 18 = A 234 19-21 f 20 20f 21-23 5 22 110 23-25 4 24 96 Total ∑fi = 44+f ∑ fixi = 752+20f

18= (752+20f)/(44+f)
18(44+f) = (752+20f)
40 = 2f
f = 20

Question 4
Thirty women were examined in a hospital by a doctor and the number of heart beats per minute were recorded and summarized as follows. Find the mean heart beats per minute for these women, choosing a suitable method.
 Number of heart beats per minute 65-68 68-71 71-74 74-77 77-80 80-83 83-86 Number of Women 2 4 3 8 7 4 3

Solution
x= (Upper limit + Lower limit)/2
Here The value of mid-point (xi) is very large. So, assumed mean or step deviation method would be simple to calculate
Now Class size (h) = 3
We can see that step deviation method would be best here
Now assumed mean A = 75.5 and class interval is h = 3
So, u= (xi - A)/h
u= (xi – 75.5)/3

 Class Interval Number of women (fi) Mid-point (xi) ui = (xi - 75.5)/h fiui 65-68 2 66.5 -3 -6 68-71 4 69.5 -2 -8 71-74 3 72.5 -1 -3 74-77 8 75.5 0 0 77-80 7 78.5 1 7 80-83 4 81.5 3 8 83-86 2 84.5 3 6 ∑fi= 30 ∑ fiui = 4

= 75.5 + 3×(4/30) = 75.5 + 4/10 = 75.5 + 0.4 = 75.9
The mean heart beats per minute for these women is 75.9

Question 5
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 per 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?

Solution
Since, the given data is not continuous so we add 0.5 to the upper limit and subtract 0.5 from the lower limit.
Here, assumed mean (A) = 57
Class size (h) = 3
 Class Interval Number of boxes (fi) Mid-point (xi) di = xi - A fidi 49.5-52.5 15 51 -6 90 52.5-55.5 110 54 -3 -330 55.5-58.5 135 57 = A 0 0 58.5-61.5 115 60 3 345 61.5-64.5 25 63 6 150 ∑ fi = 400 ∑ fidi = 75

= 57 + (75/400) = 57 + 0.1875 = 57.19

Question 6
The table below shows the daily expenditure on food of 25 households in a locality.
 Daily expenditure (in c) 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.

Solution
Here, assumed mean (A) = 225
 Class Interval Number of households (fi) Mid-point (xi) di = xi - A fidi 100-150 4 125 -100 -400 150-200 5 175 -50 -250 200-250 12 225 0 0 250-300 2 275 50 100 300-350 2 325 100 200 ∑ fi = 25 ∑ fidi = -350

= 225 + (-350/25) = 225 - 14 = 211
The mean daily expenditure on food is 211

Question 7
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 4 0.04-0.08 9 0.08-0.12 9 0.12-0.16 2 0.16-0.20 4 0.20-0.24 2

Find the mean concentration of SO2 in the air.
Solution
 Concentration of SO2 (in ppm) Frequency (fi) Mid-point (xi) fixi 0.00-0.04 4 0.02 0.08 0.04-0.08 9 0.06 0.54 0.08-0.12 9 0.10 0.90 0.12-0.16 2 0.14 0.28 0.16-0.20 4 0.18 0.72 0.20-0.24 2 0.20 0.40 Total ∑ fi = 30 ∑ (fixi) = 2.96

= 2.96/30 = 0.099 ppm

Question 8
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.
 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

Solution
 Class interval Frequency (fi) Mid-point (xi) fixi 0-6 11 3 33 6-10 10 8 80 10-14 7 12 84 14-20 4 17 68 20-28 4 24 96 28-38 3 33 99 38-40 1 39 39 ∑ fi = 40 ∑ fixi = 499

= 499/40 = 12.48 days

Question 9.
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 85-98 Number of cities 3 10 11 8 3

Solution
We will use here the step deviation method
Assumed mean =70
H=10

 Class Interval Frequency (fi) (xi) di = xi - a ui = di/h fiui 45-55 3 50 -20 -2 -6 55-65 10 60 -10 -1 -10 65-75 11 70 0 0 0 75-85 8 80 10 1 8 85-95 3 90 20 2 6 ∑ fi  = 35 ∑ fiui  = -2

= 70 + (-2/35) ? 10 = 69.42