NCERT Class 11 Mathematics Chapter 13 Statistics

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NCERT Class 11 Mathematics Chapter 13 Statistics

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Statistics

Chapter – 13

Exercise 13.1

Find the mean deviation about the mean for the data in Exercises 1 and 2.

1. 4, 7, 8, 9, 10, 12, 13, 17

Ans: 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 X̅, i.e. xᵢ – X̅ , are

– 6, – 3, – 2, – 1, 0, 2, 3, 7

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The absolute values of the deviations, i.e. | xᵢ – X̅ | are

6, 3, 2, 1, 0, 2, 3, 7

The required mean deviation about the mean is

2. 38, 70, 48, 40, 42, 55, 63, 46, 54, 44

Ans: 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 X̅ , i.e. xᵢ – X̅ , are

– 12,20, – 2,- 10,- 8,5,13,- 4,4,- 6

The absolute values of the deviations, i.e.| xᵢ – X̅ | are

12, 20, 2, 10, 8, 5, 13, 4, 4, 6

The required mean deviation about the mean is

Find the mean deviation about the median for the data in Exercises 3 and 4.

3. 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17.

Ans: 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. xᵢ – M, 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, |xᵢ – M|, 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

4. 36, 72, 46, 42, 60, 45, 53, 46, 51, 49.

Ans: 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. xᵢ – M, 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, |xᵢ – M|, 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

Find the mean deviation about the mean for the data in Exercises 5 and 6.

xᵢ	5	10	15	20	25
fᵢ	7	4	6	3	5

Ans:

xᵢ	fᵢ	fᵢ xᵢ	\|xᵢ – X̅\|	fᵢ\|xᵢ – X̅\|
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

xᵢ	10	30	50	70	90
fᵢ	4	24	28	16	8

Ans:

xᵢ	fᵢ	fᵢ xᵢ	\|xᵢ – X̅\|	fᵢ\|xᵢ – X̅\|
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

Find the mean deviation about the median for the data in Exercises 7 and 8.

xᵢ	5	7	9	10	12	15
fᵢ	8	6	2	2	2	6

Ans: 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. |xᵢ – M|, are

\|xᵢ â€” M\|	2	0	2	3	5	8
fᵢ	8	6	2	2	2	6
fᵢ\|xᵢ â€” M\|	16	0	4	6	10	48

xᵢ	15	21	27	30	35
fᵢ	3	5	6	7	8

Ans: 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.|xᵢ – M|, are

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

Find the mean deviation about the mean for the data in Exercises 9 and 10.

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

Ans: The following table is formed.

Income per day in	Number of persons fᵢ	Mid – point xᵢ	fᵢ xᵢ	\|xᵢ – X̅\|	fᵢ\|xᵢ – X̅\|
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

10.

Height in cms	95 – 105	105 – 115	115 – 125	125 – 135	135 – 145	145 – 155
Number of boys	9	13	26	30	12	10

Ans: The following table is formed.

Height in cms	Number of boys fᵢ	Mid – point xᵢ	fᵢ xᵢ	\|xᵢ – X̅\|	fᵢ\|xᵢ – X̅\|
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

11. Find the mean deviation about median for the following data:

Marks	0 – 10	10 – 20	20 – 30	30-40	40-50	50-60
Number of girls	6	8	14	16	4	2

Ans:

Marks	fᵢ	Cumulative frequency	xᵢ	\|xᵢ – M\|	fᵢ\|xᵢ – M\|
0 – 10	6	6	5	22.86	137.16
10 – 20	8	14	15	12.86	102.88
20 – 30	14	28	25	2.86	40.04
30 – 40	16	44	35	7.14	114.24
40 – 50	4	48	45	17.14	68.56
50 – 60	2	50	55	27.14	54.28
Total	50				517.32

Since, 25 occurs in the cumulative frequency 25.

So, the median class is 20 – 30.

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

Age ( in years)	16 – 20	21 – 25	26 – 30	31 – 35	36 – 40	41 – 45	46 – 50	51 – 55
Number	5	6	12	14	26	12	16	9

Ans: 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 Nᵗʰ/2 or 50ᵗʰ item is 35.5 – 40.5.

Therefore, 35.5 – 40.5 is the median class.

It is known that,

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

Thus, mean deviation about the median is given by,

Exercise 13.2

Find the mean and variance for each of the data in Exercise 1 to 5.

1. 6, 7, 10, 12, 13, 4, 8, 12.

Ans: 6, 7, 10, 12, 13, 4, 8, 12

Mean,

The following table is obtained.

xᵢ	(xᵢ – X̅)	(xᵢ – 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

2. First n natural numbers.

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

3. First 10 multiples of 3.

Ans: 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	(xᵢ – X̅)	(xᵢ – 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

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

Ans: The data is obtained in tabular form as follows.

xᵢ	fᵢ	fᵢxᵢ	xᵢ – X̅	(xᵢ – X̅)²	fᵢ(xᵢ – 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

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

Ans: The data is obtained in tabular form as follows.

xᵢ	fᵢ	fᵢxᵢ	xᵢ – X̅	(xᵢ – X̅)²	fᵢ(xᵢ – 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

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

Ans: 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

Find the mean and variance for the following frequency distributions in Exercises 7 and 8.

Classes	0 – 30	30 – 60	60 – 90	90 – 120	120 – 150	150 – 180	180 – 210
Frequencies	2	3	5	10	3	5	2

Ans:

Class	Frequencyfᵢ	Mid – point xᵢ		yᵢ²	fyᵢ	fyᵢ²
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

Classes	0 – 10	10 – 20	20 – 30	30 – 40	40 – 50
Frequencies	5	8	15	16	6

Ans:

Class	Frequencyfᵢ	Mid – point xᵢ		yᵢ²	fyᵢ	fyᵢ²
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

Heightin cms	70-75	75-80	80-85	85-90	90-95	95-100	100-105	105-110	110-115
No. ofchildren	3	4	7	7	15	9	6	6	3

Ans:

10. The diameters of circles (in mm) drawn in a design are given below:

Diameters	33 – 36	37 – 40	41 – 44	45 – 48	49 – 52
No. of circles	15	17	21	22	25

Ans:

Class Interval	Frequencyfᵢ	Mid – point xᵢ		fᵢ²	fyᵢ	fyᵢ²
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.

Miscellaneous Exercise on Chapter 12

1. 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.

Ans: Let the remaining two observations be x and y.

Therefore, the observations are 6, 7, 10, 12, 12, 13, х, у.

⇒ 60 + x + y = 72

⇒ x + y = 12 …(1)

⇒ x² + y² = 80 …(2)

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= A ± 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.

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

Ans: Let the remaining two observations be x and y.

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

⇒ 56 = 42 + x + y

⇒x + y = 14 ….(1)

⇒ x² + y² = 112 – 12 = 100

x² + y² = 100 …..(2)

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= A ± 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.

3. 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.

Ans: 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 y, then

= 3 × 8 ….[Using(1)]

= 24

Standard deviation,

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

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

Therefore, variance of new observations =

Hence, the standard deviation of new observations is √144 = 12

4. Given that X̅ is the mean and Ïf² is the variance of n observations x₁, x₂ -| xₙ.Prove that the mean and variance of the observations ax₁, ax₂, ax₃ -axₙ are aX̅ and a² Ïf², respectively, ( a ≠ 0).

Ans: The given n observations are x₁ ,x₂ -| xₙ.

Mean = X̅

Variance = Ïf²

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

Therefore, mean of the observations, ax₁, ax₂ -| axₙ, is aX̅.

Substituting the values of xand X̅ in (1), we obtain

Thus, the variance of the observations, ax₁,ax₂ -| axₙ, is a² Ïf² .

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 the wrong item is omitted.

Ans: 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) If it is replaced by 12.

Ans: When 8 is replaced by 12,

Incorrect sum of observations = 200

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

6. 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.

Ans: Number of observations (n) = 100

Incorrect mean ( X̅ )=20

Incorrect standard deviation (σ) = 3