CBSE Board NCERT Solutions for Class 10th Mathematics Chapter14 : Statistics
CBSE NCERT Solutions for Class Ten Mathematics Chapter14 – Statistics
CBSE NCERT Solutions For Class 10th Maths Chapter 14 


NCERT Solutions for Class X Maths Chapter 14 Statistics – Mathematics CBSE
Exercise 14.1


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 plants0-2 2-44-66-88-1010-1212-14
Number of houses1215623

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

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

Mean can be calculated as follows:

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

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

Daily wages (in Rs) 100-120120-140140-160160-180  180-200
Number of workers12148610

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 Interval fixi di = xi – afidi
100-12012110-40-480
120-14014130-20-280
140-160815000
160-180617020120
180-2001019040400
 Sum fi = 50   Sum fidi = -240

Now, mean of deviations can be calculated as follows:

Mean can be calculated as follows:

x = d + a = -4.8 + 150 = 145.20

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)11-13 13-15 15-1717-19  19-2121-2323-25
Number of children76913f56

Solution:

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

We have;



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 min65-6868-7171-7474-7777-8080-8383-86
Number of women2438741

Solution:

Class Intervalfi xi di = xi – a  fidi
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
 Sum fi = 30    Sum fidi = 12

 Now, mean can be calculated as follows:


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 mangoes50-5253-5556-5889-6162-64
Number of boxes1511013511525

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

Solution:

Class Interval fi xi di = x – a  fidi
50-521551-690
53-5511054-3-330
56-581355700
59-61115603345
62-6425636150
 Sum fi = 400   Sum fidi = 75

Mean can be calculated as follows:


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

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

Daily expenditure (in Rs) 100-150150-200200-250250-300300-350
Number of households451222

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

Solution:

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

Mean can be calculated as follows:


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
0.16-0.20 4
0.20-0.242

Find the mean concentration of SO2 in the air.
Solution:

Class Interval fi xi fixi
0.00-0.0440.020.08
0.04-0.0890.060.54
0.08-0.12 90.100.90
0.12-0.1620.140.28
0.16-0.2040.180.72
0.20-0.2420.220.44
 Sum fi = 30 Sum fixi = 2.96

Mean can be calculated as follows:

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 days0-6 6-1010-1414-2020-2828-3838-40
Number of students 111074431

Solution:

Class Interval fixi fixi
0-611333
6-1010880
10-1471284
14-204768
20-2842496
28-3833399
38-4013939
 Sum fi = 40 Sum fixi = 499

Mean can be calculated as follows:


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

Literacy rate (in %)45-5555-6565-75 75-85 85-98
Number of cities3101183

Solution:

Class Interval  fi xidi = xi – a ui = di/hfiui
45-55350-20-2-6
55-651060-10-1-10
65-751170000
75-858801018
85-953902026
 Sum fi = 35    Sum fiui = -2

 Mean can be calculated as follows: