NCERT Solutions For Class 11 Maths Chapter 15 Statistics Exercise 15.2

Class 11 Maths NCERT Solutions for Chapter 15 Statistics Exercise 15.2

Statistics Exercise 15.2 Solutions

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

The following table is obtained. 

xi
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

2. Find the mean and variance for the first n natural numbers 

Solution

The mean of first n natural numbers is calculated as follows. 
Mean = (Sum of all observations)/(Number of observations) 

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. 

xi
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

4. Find the mean and variance for the data

xi	6	10	14	18	24	28	30
fi	2	4	7	12	8	4	3

Solution

The data is obtained in tabular form as follows. 

xi	fi	fixi
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

5. Find the mean and variance for the data

xi	92	93	97	98	102	104	109
fi	3	2	3	2	6	3	3

Solution

The data is obtained in tabular form as follows.

xi	fi	fi xi
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

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

Solution

The data is obtained in tabular form as follows. 

xi	fi		yi2	fi yi	fi yi2
60	2	-4	15	-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

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 fi	Mid – point xi		yi2	fi yi	fi yi2
0 – 30	2	15	-3	9	-6	18
30 – 60	3	45	-2	4	-6	12
60 – 90	5	75	-1	1	-5	5
90 – 120	10	105	0	0	0	0
120 – 150	3	135	1	1	3	3
150 – 180	5	165	2	4	10	20
180 – 210	2	195	3	9	6	18
	30				2	76

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 xi		yi2	fi yi	fi yi2
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

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 fi	Mid – point xi		yi2	fi yi	fi yi2
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

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 fi	Mid – point xi		fi2	fi yi	fi yi2
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.