Chapter 14;Statistics

NCERT Solutions for Class 10th: Ch 14 Statistics Maths
Page No: 270

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 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

No. of plants
(Class interval)
No. of houses (fi)
Mid-point (xi)
fixi
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
Sum fi= 20

Sum fixi= 162

Mean = x̄ = ∑fixi/∑fi= 162/20 = 8.1
We would use  direct method because the numerical value of fiand xiare small.

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.

Answer

Here, the value of mid-point (xi) is very large, so assumed mean A = 150 and class interval is h = 20.
So, ui= (xi- A)/h = ui= (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
Sum fi= 50


Sum fiui= -12
Mean = x̄ = A + h∑fiui/∑fi=150 + (20 × -12/50) = 150 - 4.8 = 145.20
Thus, mean daily wage = Rs. 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.


Answer

Here, the value of mid-point (xi)  meanx̄= 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

Sum fixi= 752+20f

Mean = x̄ = ∑fixi/∑fi= (752+20f)/(44+f)
⇒ 18 = (752+20f)/(44+f)
⇒ 18(44+f) = (752+20f)
⇒ 792+18f = 752+20f
⇒ 792+18f = 752+20f
⇒ 792 - 752 = 20f - 18f
⇒ 40 = 2f
⇒ f = 20

Page No: 271

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


Answer

xi= (Upper limit + Lower limit)/2
Class size (h) = 3
Assumed mean (A) = 75.5

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
Sum fi= 30


Sum fiui= 4

Mean = x̄ = A + h∑fiui/∑fi= 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

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.

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

Answer

Since, the given data is not continuous so we add 0.5 to the upper limit and subtract 0.45 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
Sum fi= 400


Sum fidi= 75

Mean = x̄ = A + ∑fidi/∑fi= 57 + (75/400) = 57 + 0.1875 = 57.19

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

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

Answer

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
Sum fi= 25


Sum fidi= -350

Mean = x̄ = A + ∑fidi/∑fi= 225 + (-350/25) = 225 - 14 = 211
The mean daily expenditure on food is 211

7. To find out the concentration of SO2in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below:

Find the mean concentration of SO2in the air.

Answer

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
Sum fi= 30

Sum (fixi) = 2.96

Mean = x̄ = ∑fixi/∑fi
= 2.96/30 = 0.099 ppm

Page No. 272

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

Answer

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
Sumfi= 40

Sumfixi= 499

Mean = x̄ = ∑fixi/∑fi
= 499/40 = 12.48 days

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

Answer

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
Sumfi= 35



Sumfiui= -2

Mean = x̄ = a + (∑fiui/∑fi) х h
= 70 + (-2/35) х 10 = 69.42

Page No. 275
Exercise 14.2

1. The following table shows the ages of the patients admitted in a hospital during a year:

Age (in years)
5-15
15-25
25-35
35-45
45-55
55-65
Number of patients
6
11
21
23
14
5
Find the mode and the mean of the data given above. Compare and interpret the two
measures of central tendency.

Answer

Modal class = 35 – 45, l = 35, class width (h) = 10, fm= 23, f1= 21 and f2= 14


Calculation of Mean:

Class Interval
Frequency (fi)
Mid-point (xi)
fixi
5-15
6
10
60
15-25
11
20
220
25-35
21
30
630
35-45
23
40
920
45-55
14
50
700
55-65
5
60
300
Sumfi= 80

Sumfixi= 2830

Mean = x̄ = ∑fixi/∑fi
= 2830/80 = 35.37 yr

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

Lifetime (in hours)
0-20
20-40
40-60
60-80
80-100
100-120
Frequency
10
35
52
61
38
29
Determine the modal lifetimes of the components.

Answer

Modal class of the given data is 60–80.
Modal class = 60-80, l = 60, fm= 61, f1= 52, f2= 38 and h = 20


3. The following data gives the distribution of total monthly household expenditure of 200
families of a village. Find the modal monthly expenditure of the families. Also, find the
mean monthly expenditure :

Expenditure
Number of families
1000-1500
24
1500-2000
40
2000-2500
33
2500-3000
28
3000-3500
30
3500-4000
22
4000-4500
16
4500-5000
7

Answer

Modal class = 1500-2000, l = 1500, fm= 40, f1= 24, f2= 33 and h = 500


Calculation for mean:

Class Interval
fi
xi
di = xi - a
ui = di/h
fiui
1000-1500
24
1250
-1500
-3
-72
1500-2000
40
1750
-1000
-2
-80
2000-2500
33
2250
-500
-1
-33
2500-3000
28
2750
0
0
0
3000-3500
30
3250
500
1
30
3500-4000
22
3750
1000
2
44
4000-4500
16
4250
1500
3
48
4500-5000
7
4750
2000
4
28
fi = 200



fiui = -35

Mean = x̄ = a + (∑fiui/∑fi) х h
= 2750 + (35/200) х 500
= 2750 - 87.50 = 2662.50