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NCERT Solutions for Class 11 Maths Chapter 15

Statistics Class 11

Chapter 15 Statistics Exercise 15.1, 15.2, 15.3, miscellaneous Solutions

�Exercise 15.1 : Solutions of Questions on Page Number : 360��

Q1 :

Find the mean deviation about the mean for the data 4, 7, 8, 9, 10, 12, 13, 17

Answer :

The given data is

4, 7, 8, 9, 10, 12, 13, 17

Mean of the data,

The deviations of the respective observations from the mean are –6, – 3, –2, –1, 0, 2, 3, 7

The absolute values of the deviations, i.e. , are 6, 3, 2, 1, 0, 2, 3, 7

The required mean deviation about the mean is

Q2 :

Find the mean deviation about the mean for the data 38, 70, 48, 40, 42, 55, 63, 46, 54, 44

Answer :

The given data is

38, 70, 48, 40, 42, 55, 63, 46, 54, 44

Mean of the given data,

The deviations of the respective observations from the mean�� are –12, 20, –2, –10, –8, 5, 13, –4, 4, –6

The absolute values of the deviations, i.e. , are 12, 20, 2, 10, 8, 5, 13, 4, 4, 6

The required mean deviation about the mean is

Q3 :

Find the mean deviation about the median for the data. 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17

Answer :

The given data is

13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17

Here, the numbers of observations are 12, which is even. Arranging the data in ascending order, we obtain

10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18

The deviations of the respective observations from the median, i.e. are

–3.5, –2.5, –2.5, –1.5, –0.5, –0.5, 0.5, 2.5, 2.5, 3.5, 3.5, 4.5

The absolute values of the deviations, , are 3.5, 2.5, 2.5, 1.5, 0.5, 0.5, 0.5, 2.5, 2.5, 3.5, 3.5, 4.5

The required mean deviation about the median is

Q4 :

Find the mean deviation about the median for the data 36, 72, 46, 42, 60, 45, 53, 46, 51, 49

Answer :

The given data is

36, 72, 46, 42, 60, 45, 53, 46, 51, 49

Here, the number of observations is 10, which is even. Arranging the data in ascending order, we obtain

36, 42, 45, 46, 46, 49, 51, 53, 60, 72

The deviations of the respective observations from the median, i.e. are –11.5, –5.5, –2.5, –1.5, –1.5, 1.5, 3.5, 5.5, 12.5, 24.5

The absolute values of the deviations, , are 11.5, 5.5, 2.5, 1.5, 1.5, 1.5, 3.5, 5.5, 12.5, 24.5

Thus, the required mean deviation about the median is

Q5 :

Find the mean deviation about the mean for the data.

xi	5	10	15	20	25
fi	7	4	6	3	5

Answer :

xi	fi	fi� �xi
5	7	35	9	63
10	4	40	4	16
15	6	90	1	6
20	3	60	6	18
25	5	125	11	55
	25	350		158

Q6 :

Find the mean deviation about the mean for the data

xi	10	30	50	70	90
fi	4	24	28	16	8

Answer :

xi	fi	fi� �xi
10	4	40	40	160
30	24	720	20	480
50	28	1400	0	0
70	16	1120	20	320
90	8	720	40	320
	80	4000		1280

Q7 :

Find the mean deviation about the median for the data.

xi	5	7	9	10	12	15
fi	8	6	2	2	2	6

Answer :

The given observations are already in ascending order.

Adding a column corresponding to cumulative frequencies of the given data, we obtain the following table.

xi	fi	c.f.
5	8	8
7	6	14

9	2	16
10	2	18
12	2	20
15	6	26

Here, N = 26, which is even.

Median is the mean of 13th and 14th observations. Both of these observations lie in the cumulative frequency 14, for which the corresponding observation is 7.

The absolute values of the deviations from median, i.e. are

|xi� �– M|

fi� �|xi� �– M|

and

Q8 :

Find the mean deviation about the median for the data

xi	15	21	27	30	35
fi	3	5	6	7	8

Answer :

The given observations are already in ascending order.

Adding a column corresponding to cumulative frequencies of the given data, we obtain the following table.

xi	fi	c.f.
15	3	3
21	5	8

27	6	14
30	7	21
35	8	29

Here, N = 29, which is odd.

observation = 15th observation

This observation lies in the cumulative frequency 21, for which the corresponding observation is 30.

∴ Median = 30

The absolute values of the deviations from median, i.e. are

\|xi� �– M\|	15	9	3	0	5
fi	3	5	6	7	8
fi� �\|xi� �– M\|	45	45	18	0	40

∴

Q9 :

Find the mean deviation about the mean for the data.

Income per day	Number of persons
0-100	4
100-200	8
200-300	9
300-400	10
400-500	7
500-600	5
600-700	4
700-800	3

Answer :

The following table is formed.

Income per day	Number of persons fi	Mid-point xi	fi� �xi
0 – 100	4	50	200	308	1232
100 – 200	8	150	1200	208	1664
200 – 300	9	250	2250	108	972
300 – 400	10	350	3500	8	80
400 – 500	7	450	3150	92	644
500 – 600	5	550	2750	192	960
600 – 700	4	650	2600	292	1168
700 – 800	3	750	2250	392	1176
	50		17900		7896

Here,

Q10 :

Find the mean deviation about the mean for the data

Height in cms	Number of boys
95-105	9
105-115	13
115-125	26
125-135	30
135-145	12

Answer :

The following table is formed.

Height in cms	Number of boys fi	Mid-point xi	fi� �xi
95-105	9	100	900	25.3	227.7
105-115	13	110	1430	15.3	198.9
115-125	26	120	3120	5.3	137.8
125-135	30	130	3900	4.7	141
135-145	12	140	1680	14.7	176.4
145-155	10	150	1500	24.7	247

Here,

Q11 :

Calculate the mean deviation about median age for the age distribution of 100 persons given below:

Age	Number
16-20	5
21-25	6
26-30	12
31-35	14
36-40	26
41-45	12
46-50	16

51-55

Answer :

The given data is not continuous. Therefore, it has to be converted into continuous frequency distribution by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each class interval.

The table is formed as follows.

Age	Number fi	Cumulative frequency (c.f.)	Mid-point xi	\|xi� �– Med.\|	fi\|xi� �– Med.\|
15.5-20.5	5	5	18	20	100
20.5-25.5	6	11	23	15	90
25.5-30.5	12	23	28	10	120
30.5-35.5	14	37	33	5	70
35.5-40.5	26	63	38	0	0
40.5-45.5	12	75	43	5	60
45.5-50.5	16	91	48	10	160
50.5-55.5	9	100	53	15	135
	100				735

The class interval containing the or 50th item is 35.5 – 40.5. Therefore, 35.5 – 40.5 is the median class.

It is known that,

Here, l = 35.5, C = 37, f = 26, h = 5, and N = 100

�

Thus, mean deviation about the median is given by,

�Exercise 15.2 : Solutions of Questions on Page Number : 371��

Q1 :

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	–3	9
7	–2	4
10	–1	1
12	3	9
13	4	16
4	–5	25
8	–1	1
12	3	9
		74

Q2 :

Find the mean and variance for the first n natural numbers

Answer :

The mean of first n natural numbers is calculated as follows.

Q3 :

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	–13.5	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

Q4 :

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.

xi	*f i*	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

Here, N = 40,

Q5 :

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.

xi	*f i*	fixi
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,

Q6 :

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.

xi	fi		yi2	fiyi	fiyi2
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

Mean,

Q7 :

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 :

Class	Frequency fi	Mid-point xi		yi2	fiyi	fiyi2
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

Mean,

Q8 :

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,

Q9 :

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,

Q10 :

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,

�Exercise 15.3 : Solutions of Questions on Page Number : 375��

Q1 :

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.

Q2 :

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
Σ	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
Σ	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.

Q3 :

From the prices of shares X and Y below, find out which is more stable in value:

X	35	54	52	53	56	58	52	50	51	49
Y	108	107	105	105	106	107	104	103	104	101

Answer :

The prices of the shares X are

35, 54, 52, 53, 56, 58, 52, 50, 51, 49

Here, the number of observations, N = 10

The following table is obtained corresponding to shares X.

xi
35	–16	256
54	3	9
52	1	1
53	2	4
56	5	25

58	7	49
52	1	1
50	–1	1
51	0	0
49	–2	4
		350

The prices of share Y are

108, 107, 105, 105, 106, 107, 104, 103, 104, 101

The following table is obtained corresponding to shares Y.

yi
108	3	9
107	2	4
105	0	0
105	0	0
106	1	1
107	2	4
104	–1	1
103	–2	4
104	–1	1
101	–4	16
		40

C.V. of prices of shares X is greater than the C.V. of prices of shares Y. Thus, the prices of shares Y are more stable than the prices of shares X.

Q4 :

An analysis of monthly wages paid to workers in two firms A and B, belonging to the same industry, gives the following results:

	Firm A	Firm B
No. of wage earners	586	648
Mean of monthly wages	Rs 5253	Rs 5253
Variance of the distribution of wages	100	121

(i) Which firm A or B pays larger amount as monthly wages?

(ii) Which firm, A or B, shows greater variability in individual wages?

Answer :

(i) Monthly wages of firm A = Rs 5253 Number of wage earners in firm A = 586

∴Total amount paid = Rs 5253 x 586

Monthly wages of firm B = Rs 5253 Number of wage earners in firm B = 648

∴Total amount paid = Rs 5253 x 648

Thus, firm B pays the larger amount as monthly wages as the number of wage earners in firm B are more than the number of wage earners in firm A.

(ii) Variance of the distribution of wages in firm A�� = 100

∴ Standard deviation of the distribution of wages in firm A ((σ1) =

Variance of the distribution of wages in firm�� = 121

∴ Standard deviation of the distribution of wages in firm

The mean of monthly wages of both the firms is same i.e., 5253. Therefore, the firm with greater standard deviation will have more variability.

Thus, firm B has greater variability in the individual wages.

Q5 :

The following is the record of goals scored by team A in a football session:

No. of goals scored	0	1	2	3	4
No. of matches	1	9	7	5	3

For the team B, mean number of goals scored per match was 2 with a standard deviation 1.25 goals. Find which team may be considered more consistent?

Answer :

The mean and the standard deviation of goals scored by team A are calculated as follows.

No. of goals scored	No. of matches	fixi	xi2	fixi2
0	1	0	0	0
1	9	9	1	9
2	7	14	4	28
3	5	15	9	45
4	3	12	16	48
	25	50		130

Thus, the mean of both the teams is same.

The standard deviation of team B is 1.25 goals.

The average number of goals scored by both the teams is same i.e., 2. Therefore, the team with lower standard deviation will be more consistent.

Thus, team A is more consistent than team B.

Q6 :

The sum and sum of squares corresponding to length x (in cm) and weight y

(in gm) of 50 plant products are given below:

Which is more varying, the length or weight?

Answer :

Here, N = 50

∴ Mean,

Mean,

Thus, C.V. of weights is greater than the C.V. of lengths. Therefore, weights vary more than the lengths.

�Exercise Miscellaneous : Solutions of Questions on Page Number : 380��

Q1 :

The mean and variance of eight observations are 9 and 9.25, respectively. If six of the observations are 6, 7, 10, 12, 12 and 13, find the remaining two observations.

Answer :

Let the remaining two observations be x and y. Therefore, the observations are 6, 7, 10, 12, 12, 13, x, y.

From (1), we obtain

x2 + y2 + 2xy = 144 …(3) From (2) and (3), we obtain

2xy = 64 … (4)

Subtracting (4) from (2), we obtain

x2 + y2 – 2xy = 80 – 64 = 16

⇒ x – y = ± 4 … (5)

Therefore, from (1) and (5), we obtain

x = 8 and y = 4, when x – y = 4

x = 4 and y = 8, when x – y = –4

Thus, the remaining observations are 4 and 8.

Q2 :

The mean and variance of 7 observations are 8 and 16, respectively. If five of the observations are 2, 4, 10, 12 and 14. Find the remaining two observations.

Answer :

Let the remaining two observations be x and y.

The observations are 2, 4, 10, 12, 14, x, y.

From (1), we obtain

x2 + y2 + 2xy = 196 … (3) From (2) and (3), we obtain

2xy = 196 – 100

⇒ 2xy = 96 … (4)

Subtracting (4) from (2), we obtain

x2 + y2 – 2xy = 100 – 96

⇒ (x – y)2 = 4

⇒ x – y = ± 2 … (5)

Therefore, from (1) and (5), we obtain

x = 8 and y = 6 when x – y = 2

x = 6 and y = 8 when x – y = – 2

Thus, the remaining observations are 6 and 8.

Q3 :

The mean and standard deviation of six observations are 8 and 4, respectively. If each observation is multiplied by 3, find the new mean and new standard deviation of the resulting observations.

Answer :

Let the observations be x1, x2, x3, x4, x5, and x6.

It is given that mean is 8 and standard deviation is 4.

If each observation is multiplied by 3 and the resulting observations are yi, then

From (1) and (2), it can be observed that,

Substituting the values of xi and ��in (2), we obtain

Therefore, variance of new observations = �

�Hence, the standard deviation of new observations is

Q4 :

Given that is the mean and σ2 is the variance of n observations x1, x2 … xn. Prove that the mean and variance of the observations ax1, ax2, ax3 …axn are ��and a2 σ2, respectively (a ≠ 0).

Answer :

The given n observations are x1, x2 … xn. Mean =

Variance = σ2

If each observation is multiplied by a and the new observations are yi, then

Therefore, mean of the observations, ax1, ax2 … axn, is

. Substituting the values of xiand

�in (1), we obtain

Thus, the variance of the observations, ax1, ax2 … axn, is a2 σ2.

Q5 :

The mean and standard deviation of 20 observations are found to be 10 and 2, respectively. On rechecking, it was found that an observation 8 was incorrect. Calculate the correct mean and standard deviation in each of the following cases:

(i) If wrong item is omitted.

(ii) If it is replaced by 12.

Answer :

(i) Number of observations (n) = 20 Incorrect mean = 10

Incorrect standard deviation = 2

That is, incorrect sum of observations = 200 Correct sum of observations = 200 – 8 = 192

∴ Correct mean

��

(ii) When 8 is replaced by 12, Incorrect sum of observations = 200

∴ Correct sum of observations = 200 – 8 + 12 = 204

Q6 :

The mean and standard deviation of marks obtained by 50 students of a class in three subjects, Mathematics, Physics and Chemistry are given below:

Subject	Mathematics	Physics	Chemistry
Mean	42	32	40.9
Standard deviation	12	15	20

Which of the three subjects shows the highest variability in marks and which shows the lowest?

Answer :

Standard deviation of Mathematics = 12 Standard deviation of Physics = 15

Standard deviation of Chemistry = 20

The coefficient of variation (C.V.) is given by�� .

The subject with greater C.V. is more variable than others.

Therefore, the highest variability in marks is in Chemistry and the lowest variability in marks is in Mathematics.

Q7 :

The mean and standard deviation of a group of 100 observations were found to be 20 and 3, respectively. Later on it was found that three observations were incorrect, which were recorded as 21, 21 and 18. Find the mean and standard deviation if the incorrect observations are omitted.

Answer :

Number of observations (n) = 100

Incorrect mean

Incorrect standard deviation

∴ Incorrect sum of observations = 2000

⇒ Correct sum of observations = 2000 – 21 – 21 – 18 = 2000 – 60 = 1940