NCERT Solutions for Class 11 Maths Chapter 15 – Statistics Ex 15.2
Page No 371:
Question 1:
Find the mean and variance for the data 6, 7, 10, 12, 13, 4, 8, 12
Answer:
6, 7, 10, 12, 13, 4, 8, 12
Mean, 
The following table is obtained.
- xi
6–397–2410–111239134164–5258–11123974
Question 2:
Find the mean and variance for the first n natural numbers
Answer:
The mean of first n natural numbers is calculated as follows.
Question 3:
Find the mean and variance for the first 10 multiples of 3
Answer:
The first 10 multiples of 3 are
3, 6, 9, 12, 15, 18, 21, 24, 27, 30
Here, number of observations, n = 10
The following table is obtained.
- xi
3–13.5182.256–10.5110.259–7.556.2512–4.520.2515–1.52.25181.52.25214.520.25247.556.252710.5110.253013.5182.25742.5
Question 4:
Find the mean and variance for the data
| xi | 6 | 10 | 14 | 18 | 24 | 28 | 30 |
| f i | 2 | 4 | 7 | 12 | 8 | 4 | 3 |
Answer:
The data is obtained in tabular form as follows.
- xif ifixi
6212–1316933810440–98132414798–5251751812216–11122481925252002841129813243039011121363407601736
Here, N = 40, 
Question 5:
Find the mean and variance for the data
| xi | 92 | 93 | 97 | 98 | 102 | 104 | 109 |
| f i | 3 | 2 | 3 | 2 | 6 | 3 | 3 |
Answer:
The data is obtained in tabular form as follows.
- xif ifixi
923276–864192932186–74998973291–3927982196–248102661224241043312416481093327981243222200640
Here, N = 22, 
Question 6:
Find the mean and standard deviation using short-cut method.
| xi | 60 | 61 | 62 | 63 | 64 | 65 | 66 | 67 | 68 |
| fi | 2 | 1 | 12 | 29 | 25 | 12 | 10 | 4 | 5 |
Answer:
The data is obtained in tabular form as follows.
- xifi
yi2fiyifiyi2602–416–832611–39–396212–24–24486329–11–2929642500006512111212661024204067439123668541620801002200286
Mean, 
Question 7:
Find the mean and variance for the following frequency distribution.
Classes
|
0-30
|
30-60
|
60-90
|
90-120
|
120-150
|
150-180
|
180-210
|
Frequencies
|
2
|
3
|
5
|
10
|
3
|
5
|
2
|
Answer:
- ClassFrequency fiMid-point xi
yi2fiyifiyi20-30215–39–61830-60345–24–61260-90575–11–5590-120101050000120-15031351133150-1805165241020180-21021953961830276
Mean, 
Page No 372:
Question 8:
Find the mean and variance for the following frequency distribution.
| Classes | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
| Frequencies | 5 | 8 | 15 | 16 | 6 |
Answer:
Class
|
Frequency
fi
|
Mid-point xi
|  |
yi2
|
fiyi
|
fiyi2
|
0-10
|
5
|
5
|
–2
|
4
|
–10
|
20
|
10-20
|
8
|
15
|
–1
|
1
|
–8
|
8
|
20-30
|
15
|
25
|
0
|
0
|
0
|
0
|
30-40
|
16
|
35
|
1
|
1
|
16
|
16
|
40-50
|
6
|
45
|
2
|
4
|
12
|
24
|
50
|
10
|
68
|
Mean, 
Question 9:
Find the mean, variance and standard deviation using short-cut method
| Height
in cms
|
No. of children
|
70-75
|
3
|
75-80
|
4
|
80-85
|
7
|
85-90
|
7
|
90-95
|
15
|
95-100
|
9
|
100-105
|
6
|
105-110
|
6
|
110-115
|
3
|
Answer:
Class Interval
|
Frequency fi
|
Mid-point xi
|  |
yi2
|
fiyi
|
fiyi2
|
70-75
|
3
|
72.5
|
–4
|
16
|
–12
|
48
|
75-80
|
4
|
77.5
|
–3
|
9
|
–12
|
36
|
80-85
|
7
|
82.5
|
–2
|
4
|
–14
|
28
|
85-90
|
7
|
87.5
|
–1
|
1
|
–7
|
7
|
90-95
|
15
|
92.5
|
0
|
0
|
0
|
0
|
95-100
|
9
|
97.5
|
1
|
1
|
9
|
9
|
100-105
|
6
|
102.5
|
2
|
4
|
12
|
24
|
105-110
|
6
|
107.5
|
3
|
9
|
18
|
54
|
110-115
|
3
|
112.5
|
4
|
16
|
12
|
48
|
60
|
6
|
254
|
Mean, 
Question 10:
The diameters of circles (in mm) drawn in a design are given below:
| Diameters |
No. of children
|
33-36
|
15
|
37-40
|
17
|
41-44
|
21
|
45-48
|
22
|
49-52
|
25
|
Answer:
Class Interval
|
Frequency fi
|
Mid-point xi
|  |
fi2
|
fiyi
|
fiyi2
|
32.5-36.5
|
15
|
34.5
|
–2
|
4
|
–30
|
60
|
36.5-40.5
|
17
|
38.5
|
–1
|
1
|
–17
|
17
|
40.5-44.5
|
21
|
42.5
|
0
|
0
|
0
|
0
|
44.5-48.5
|
22
|
46.5
|
1
|
1
|
22
|
22
|
48.5-52.5
|
25
|
50.5
|
2
|
4
|
50
|
100
|
100
|
25
|
199
|
Here, N = 100, h = 4
Let the assumed mean, A, be 42.5.
Mean, 
Page No 375:
Question 1:
From the data given below state which group is more variable, A or B?
Marks
|
10-20
|
20-30
|
30-40
|
40-50
|
50-60
|
60-70
|
70-80
|
Group A
|
9
|
17
|
32
|
33
|
40
|
10
|
9
|
Group B
|
10
|
20
|
30
|
25
|
43
|
15
|
7
|
Answer:
Firstly, the standard deviation of group A is calculated as follows.
Marks
|
Group A fi
|
Mid-point xi
|  |
yi2
|
fiyi
|
fiyi2
|
10-20
|
9
|
15
|
–3
|
9
|
–27
|
81
|
20-30
|
17
|
25
|
–2
|
4
|
–34
|
68
|
30-40
|
32
|
35
|
–1
|
1
|
–32
|
32
|
40-50
|
33
|
45
|
0
|
0
|
0
|
0
|
50-60
|
40
|
55
|
1
|
1
|
40
|
40
|
60-70
|
10
|
65
|
2
|
4
|
20
|
40
|
70-80
|
9
|
75
|
3
|
9
|
27
|
81
|
150
|
–6
|
342
|
Here, h = 10, N = 150, A = 45
The standard deviation of group B is calculated as follows.
Marks
|
Group B
fi
|
Mid-point
xi
|  |
yi2
|
fiyi
|
fiyi2
|
10-20
|
10
|
15
|
–3
|
9
|
–30
|
90
|
20-30
|
20
|
25
|
–2
|
4
|
–40
|
80
|
30-40
|
30
|
35
|
–1
|
1
|
–30
|
30
|
40-50
|
25
|
45
|
0
|
0
|
0
|
0
|
50-60
|
43
|
55
|
1
|
1
|
43
|
43
|
60-70
|
15
|
65
|
2
|
4
|
30
|
60
|
70-80
|
7
|
75
|
3
|
9
|
21
|
63
|
150
|
–6
|
366
|
Since the mean of both the groups is same, the group with greater standard deviation will be more variable.
Thus, group B has more variability in the marks.
