NCERT Solutions for Class 11 Maths chapter 15-Statistics Exercise 15.2
NCERT Solutions for Class-11 Maths Chapter-15 Statistics
NCERT Solutions for Class 11 Maths
Chapter-15 Statistics Exercise 15.2 prepared by the expert of Physics Wallah score more with Physics Wallah NCERT Class 11 maths solutions. You can download and share
NCERT Solutions
for Class 11 Maths.
NCERT Solutions for Class-11 Maths Exercise 5.2
Question
1. Find the mean and variance for the data 6, 7, 10, 12, 13, 4, 8, 12
Solution :
6, 7, 10, 12, 13, 4, 8, 12
Mean,
x̅ = (6 + 7 + 10 + 12 + 13 + 4 + 8 + 12)/8
= 72/8
= 9
The following table is obtained.
|
x
i
|
(x
i
– x̅)
|
(x
i
– x̅)
2
|
|
6
|
–3
|
9
|
|
7
|
–2
|
4
|
|
10
|
–1
|
1
|
|
12
|
3
|
9
|
|
13
|
4
|
16
|
|
4
|
–5
|
25
|
|
8
|
–1
|
1
|
|
12
|
3
|
9
|
|
|
|
74
|
Question
2. Find the mean and variance for the first n natural numbers
Solution :
We know that Mean = Sum of all observations/Number of observations
∴Mean, x̅ = ((n(n + 1))2)/n
= (n + 1)/2
and also WKT Variance,
WKT, (a + b)(a – b) = a
2
– b
2
σ
2
= (n
2
– 1)/12
∴Mean = (n + 1)/2 and Variance = (n
2
– 1)/12
Question
3. Find the mean and variance for the first 10 multiples of 3
Solution :
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.
|
x
i
|
(x
i
– x̅)
|
(x
i
– x̅)
2
|
|
3
|
–13.5
|
182.25
|
|
6
|
–10.5
|
110.25
|
|
9
|
–7.5
|
56.25
|
|
12
|
–4.5
|
20.25
|
|
15
|
–1.5
|
2.25
|
|
18
|
1.5
|
2.25
|
|
21
|
4.5
|
20.25
|
|
24
|
7.5
|
56.25
|
|
27
|
10.5
|
110.25
|
|
30
|
13.5
|
182.25
|
|
|
|
742.5
|
Then, Variance
= (1/10) × 742.5
= 74.25
∴Mean = 16.5 and Variance = 74.25
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
|
Solution :
The data is obtained in tabular form as follows.
|
xi
|
f i
|
fixi
|
(x
i
– x̅)
|
(x
i
– x̅)
2
|
f
i
(x
i
– x̅)
2
|
|
6
|
2
|
12
|
–13
|
169
|
338
|
|
10
|
4
|
40
|
–9
|
81
|
324
|
|
14
|
7
|
98
|
–5
|
25
|
175
|
|
18
|
12
|
216
|
–1
|
1
|
12
|
|
24
|
8
|
192
|
5
|
25
|
200
|
|
28
|
4
|
112
|
9
|
81
|
324
|
|
30
|
3
|
90
|
11
|
121
|
363
|
|
|
40
|
760
|
|
|
1736
|
Here, N = 40,
Question
5. Find the mean and variance for the data
|
x
i
|
92
|
93
|
97
|
98
|
102
|
104
|
109
|
|
f
i
|
3
|
2
|
3
|
2
|
6
|
3
|
3
|
Solution :
The data is obtained in tabular form as follows.
|
x
i
|
f
i
|
f
i
x
i
|
(x
i
– x̅)
|
(x
i
– x̅)
2
|
f
i
(x
i
– x̅)
2
|
|
92
|
3
|
276
|
–8
|
64
|
192
|
|
93
|
2
|
186
|
–7
|
49
|
98
|
|
97
|
3
|
291
|
–3
|
9
|
27
|
|
98
|
2
|
196
|
–2
|
4
|
8
|
|
102
|
6
|
612
|
2
|
4
|
24
|
|
104
|
3
|
312
|
4
|
16
|
48
|
|
109
|
3
|
327
|
9
|
81
|
243
|
|
|
22
|
2200
|
|
|
640
|
Here, N = 22,
Question
6. Find the mean and standard deviation using short-cut method.
|
x
i
|
60
|
61
|
62
|
63
|
64
|
65
|
66
|
67
|
68
|
|
f
i
|
2
|
1
|
12
|
29
|
25
|
12
|
10
|
4
|
5
|
Solution :
The data is obtained in tabular form as follows.
|
xi
|
fi
|
Y
i
= (x
i
– A)/h
|
yi
2
|
fiyi
|
fiyi
2
|
|
60
|
2
|
–4
|
16
|
–8
|
32
|
|
61
|
1
|
–3
|
9
|
–3
|
9
|
|
62
|
12
|
–2
|
4
|
–24
|
48
|
|
63
|
29
|
–1
|
1
|
–29
|
29
|
|
64
|
25
|
0
|
0
|
0
|
0
|
|
65
|
12
|
1
|
1
|
12
|
12
|
|
66
|
10
|
2
|
4
|
20
|
40
|
|
67
|
4
|
3
|
9
|
12
|
36
|
|
68
|
5
|
4
|
16
|
20
|
80
|
|
|
100
|
220
|
|
0
|
286
|
σ
2
= (1
2
/100
2
) [100(286) – 0
2
]
= (1/10000) [28600 – 0]
= 28600/10000
= 2.86
Hence, standard deviation = σ = √2.886
= 1.691
∴ Mean = 64 and Standard Deviation = 1.691
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
|
Solution :
|
Class
|
Frequency f
i
|
Mid-point x
i
|
f
i
x
i
|
(x
i
– x̅)
|
(x
i
– x̅)
2
|
f
i
(x
i
– x̅)
2
|
|
0-30
|
2
|
15
|
30
|
-92
|
8464
|
16928
|
|
30-60
|
3
|
45
|
135
|
-62
|
3844
|
11532
|
|
60-90
|
5
|
75
|
375
|
-32
|
1024
|
5120
|
|
90-120
|
10
|
105
|
1050
|
-2
|
4
|
40
|
|
120-150
|
3
|
135
|
405
|
28
|
784
|
2352
|
|
150-180
|
5
|
165
|
825
|
58
|
3364
|
16820
|
|
180-210
|
2
|
195
|
390
|
88
|
7744
|
15488
|
|
|
30
|
|
3210
|
|
2
|
68280
|
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
|
Solution :
|
Class
|
Frequency
fi
|
Mid-point x
i
|
f
i
x
i
|
(x
i
– x̅)
|
(x
i
– x̅)
2
|
f
i
(x
i
– x̅)
2
|
|
0-10
|
5
|
5
|
25
|
-22
|
484
|
2420
|
|
10-20
|
8
|
15
|
120
|
-12
|
144
|
1152
|
|
20-30
|
15
|
25
|
375
|
-2
|
4
|
60
|
|
30-40
|
16
|
35
|
560
|
8
|
64
|
1024
|
|
40-50
|
6
|
45
|
270
|
18
|
324
|
1944
|
|
|
50
|
|
1350
|
|
|
6600
|
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
|
Solution :
|
Class Interval
|
Frequency f
i
|
Mid-point x
i
|
Y
i
= (x
i
– A)/h
|
y
i
2
|
f
i
y
i
|
f
i
y
i
2
|
|
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,
Where, A = 92.5, h = 5
So, x̅ = 92.5 + ((6/60) × 5)
= 92.5 + ½
= 92.5 + 0.5
= 93
Then, Variance,
σ
2
= (5
2
/60
2
) [60(254) – 6
2
]
= (1/144) [15240 – 36]
= 15204/144
= 1267/12
= 105.583
Hence, standard deviation = σ = √105.583
= 10.275
∴ Mean = 93, variance = 105.583 and Standard Deviation = 10.275
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
|
Solution :
|
Class Interval
|
Frequency f
i
|
Mid-point x
i
|
Y
i
= (x
i
– A)/h
|
f
i
2
|
f
i
y
i
|
f
i
y
i
2
|
|
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,
Where, A = 42.5, h = 4
So, x̅ = 42.5 + (25/100) × 4
= 42.5 + 1
= 43.5
Then, Variance,
σ
2
= (4
2
/100
2
)[100(199) – 25
2
]
= (1/625) [19900 – 625]
= 19275/625
= 771/25
= 30.84
Hence, standard deviation = σ = √30.84
= 5.553
∴ Mean = 43.5, variance = 30.84 and Standard Deviation = 5.553.
