Class 10 Maths

# Statistics

## NCERT Exercise 14.1

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 plants in 20 houses in a locality. Find the mean number of plants per house.

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

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

Solution:

Class Intervalfixifixi
0-2111
2-4236
4-6155
6-85735
8-106954
10-1221122
12-1431339
Σ fi = 20Σ fixi = 162

Mean can be calculated as follows:

x=(Σf_ix_1)/(Σf_1)=(162)/(20)=8.1

In this case, the values of fi and xi are small hence direct method has been used.

Question 2. Consider the following distribution of daily wages of 50 workers of a factory.

 Daily wages (in Rs) Number of workers 100-120 120-140 140-160 160-180 180-200 12 14 8 6 10

Find the mean daily wages of the workers of the factory by using an appropriate method.

Solution: In this case, value of xi is quite large and hence we should select the assumed mean method.

Let us take assumed mean a = 150

Class Intervalfixidi = xi - afidi
100-12012110-40-480
120-14014130-20-280
140-160815000
160-180617020120
180-2001019040400
Σ fi = 50Σ fidi = -240

Now, mean of deviations can be calculated as follows:

d=(Σf_i\d_i)/(Σf_i)=(-240)/(50)=-4.8

Mean can be calculated as follows:

x = d + a = -4.8 + 150 = 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 Rs) Number of children 11-13 13-15 15-17 17-19 19-21 21-23 23-25 7 6 9 13 f 5 4

Solution:

Class Intervalfixifixi
11-1371284
13-1561484
15-17916144
17-191318234
19-21f2020f
21-23522110
23-2542496
Σ fi = 44 + fΣ fixi = 752 + 20f

We have;

x=(Σf_i\x_i)/(Σf_i)=18

Or, 18=(752+20f)/(44+f)

Or, 18(44+f)=752+20f
Or, 2f=40
Or, 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 min Number of women 65-68 68-71 71-74 74-77 77-80 80-83 83-86 2 4 3 8 7 4 2

Solution:

Class Intervalfixidi = xi - afidi
65-68266.5-9-18
68-71469.5-6-24
71-74372.5-3-9
74-77875.500
77-80778.5321
80-83481.5624
83-86284.5918
Σ fi = 30Σ fidi = 12

Now, mean can be calculated as follows:

d=(Σf_i\d_i)/(Σf_1)=12/30=0.4

x=d+a=0.4+75.5=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 according to the number of boxes.

 Number of mangoes Number of boxes 50-52 53-55 56-58 89-61 62-64 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:

Class Intervalfixidi = x - afidi
50-521551-690
53-5511054-3-330
56-581355700
59-61115603345
62-6425636150
Σ fi = 400Σ fidi = 75

Mean can be calculated as follows:

d=(Σf_i\d_i)/(Σf_1)=75/400=0.1875

x=d+a=0.1875+57=57.1875

In this case, there are wide variations in fi and hence assumed mean method is used.

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

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

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

Solution:

Class Intervalfixidi = xi - aui = di/hfiui
100-1504125-100-2-8
150-2005175-50-1-5
200-25012225000
250-30022755012
300-350232510024
Σ fi = 25Σ fiui = -7

Mean can be calculated as follows:

x=a+(Σf_i\u_i)/(Σf_i)xxh

=225+(-7)/(25)xx50=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
0.00-0.044
0.04-0.089
0.08-0.129
0.12-0.162
0.16-0.204
0.20-0.242

Find the mean concentration of SO2 in the air.

Solution:

Class Intervalfixifixi
0.00-0.0440.020.08
0.04-0.0890.060.54
0.08-0.1290.100.90
0.12-0.1620.140.28
0.16-0.2040.180.72
0.20-0.2420.220.44
Σ fi = 30Σ fixi = 2.96

Mean can be calculated as follows:

x=(Σf_i\x_i)/(Σf_i)=(2.96)/(30)=0.099 pp\m

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 Number of students 0-6 6-10 10-14 14-20 20-28 28-38 38-40 11 10 7 4 4 3 1

Solution:

Class Intervalfixifixi
0-611333
6-1010880
10-1471284
14-2041768
20-2842496
28-3833399
38-4013939
Σ fi = 40Σ fixi = 499

Mean can be calculated as follows:

x=(Σf_i\x_i)/(Σf_i)=(499)/(40)=12.4 (approx)

Question 9. The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.

 Literacy rate (in %) Number of cities 45-55 55-65 65-75 75-85 85-98 3 10 11 8 3

Solution:

Class Intervalfixidi = xi - aui = di/hfiui
45-55350-20-2-6
55-651060-10-1-10
65-751170000
75-858801018
85-953902026
Σ fi = 35Σ fiui = -2

Mean can be calculated as follows:

x=a+(Σf_i\u_i)/(Σf_i)xxh

=70+(-2)/(35)xx10=69.42