CLASS 11 MATHS CHAPTER 15-STATISTICS

Exercise 15.1

Question1. Find the mean deviation about the mean for the data

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

Solution :
The given data is

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

Question2. Find the mean deviation about the mean for the data

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

Solution :
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

Question3. Find the mean deviation about the median for the data.

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

Solution :
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 number of observations is 12

Then,

Median = ((12/2)^th observation + ((12/2)+ 1)^th observation)/2

(12/2)^th observation = 6^th = 13

(12/2)+ 1)^th observation = 6 + 1

= 7^th = 14

Median = (13 + 14)/2

= 27/2

= 13.5

So, the absolute values of the respective deviations from the median, i.e., |x_i – 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

Mean Deviation,

= (1/12) × 28

= 2.33

Question4. Find the mean deviation about the median for the data

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

Solution :

First we have to arrange the given observations into ascending order,

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

The number of observations is 10

Then,

Median = ((10/2)^th observation + ((10/2)+ 1)^th observation)/2

(10/2)^th observation = 5^th = 46

(10/2)+ 1)^th observation = 5 + 1

= 6^th = 49

Median = (46 + 49)/2

= 95

= 47.5

So, the absolute values of the respective deviations from the median, i.e., |x_i – M| are

11.5, 5.5, 2.5, 1.5, 1.5, 1.5, 3.5, 5.5, 12.5, 24.5

Mean Deviation,

= (1/10) × 70

= 7

So, the mean deviation about the median for the given data is 7.

Question5. Find the mean deviation about the mean for the data.

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

Solution :

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

Question6. Find the mean deviation about the mean for the data

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

Solution :

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

Question7. Find the mean deviation about the median for the data.

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

Solution :
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

|xi – M|	2	0	2	3	5	8
fi	8	6	2	2	2	6
fi |xi – M|	16	0	4	6	10	48

Now, N = 26, which is even.

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

Then,

Median = (13^th observation + 14^th observation)/2

= (7 + 7)/2

= 14/2

= 7

So, the absolute values of the respective deviations from the median, i.e., |x_i – M| are shown in the table.

.
 

Question8. Find the mean deviation about the median for the data

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

Solution :
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.

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

Now, N = 29, which is odd.

So 29/2 = 14.5

The cumulative frequency greater than 14.5 is 21, for which the corresponding observation is 30.

Then,

Median = (15^th observation + 16^th observation)/2

= (30 + 30)/2

= 60/2

= 30

So, the absolute values of the respective deviations from the median, i.e., |x_i – M| are shown in the table.

Question9. 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

Solution :
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,

Question10. 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
145-155	10

Solution :
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

Question11. Find the mean deviation about median for the following data:

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

Solution :
The following table is formed.

Marks	Number of girls fi	Cumulative frequency (c.f.)	Mid-point xi	|xi – Med.|	fi |xi – Med.|
0-10	6	6	5	22.85	137.1
10-20	8	14	15	12.85	102.8
20-30	14	28	25	2.85	39.9
30-40	16	44	35	7.15	114.4
40-50	4	48	45	17.15	68.6
50-60	2	50	55	27.15	54.3
	50				517.1

The class interval containing N^th/2 or 25^th item is 20-30

So, 20-30 is the median class.

Then,

Median = l + (((N/2) – c)/f) × h

Where, l = 20, c = 14, f = 14, h = 10 and n = 50

Median = 20 + (((25 – 14))/14) × 10

= 20 + 7.85

= 27.85

Question12. 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	9

Solution :
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 N^th/2 or 50^th item is 35.5 – 40.5

So, 35.5 – 40.5 is the median class.

Then,

Median = l + (((N/2) – c)/f) × h

Where, l = 35.5, c = 37, f = 26, h = 5 and N = 100

Median = 35.5 + (((50 – 37))/26) × 5

= 35.5 + 2.5

= 38

Exercise 5.2

Question1. 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.

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

Question2. 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) = a² – b²

σ² = (n² – 1)/12

∴Mean = (n + 1)/2 and Variance = (n² – 1)/12

Question3. 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	(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

Then, Variance

= (1/10) × 742.5

= 74.25

∴Mean = 16.5 and Variance = 74.25

Question4. 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	(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

Here, N = 40,

Question5. 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

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

xi	f i	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

Here, N = 22,

Question6. 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	Yi = (xi – A)/h	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

σ² = (1²/100²) [100(286) – 0²]

= (1/10000) [28600 – 0]

= 28600/10000

= 2.86

Hence, standard deviation = σ = √2.886

= 1.691

∴ Mean = 64 and Standard Deviation = 1.691

Question7. 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	fixi	(xi – x̅)	(xi – x̅)2	fi(xi – 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

Question8. 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	Frequencyfi	Mid-point xi	fixi	(xi – x̅)	(xi – x̅)2	fi(xi – 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,

Question9. 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	Yi = (xi – A)/h	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,

Where, A = 92.5, h = 5

So, x̅ = 92.5 + ((6/60) × 5)

= 92.5 + ½

= 92.5 + 0.5

= 93

Then, Variance,

σ² = (5²/60²) [60(254) – 6²]

= (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

Question10. 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	Yi = (xi – A)/h	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,

Where, A = 42.5, h = 4

So, x̅ = 42.5 + (25/100) × 4

= 42.5 + 1

= 43.5

Then, Variance,

σ² = (4²/100²)[100(199) – 25²]

= (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.

Exercise 15.3

Question1. 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

Solution :
Group A

Where A = 45,

and y_i = (x_i – A)/h

Here h = class size = 20 – 10

h = 10

So, x̅ = 45 + ((-6/150) × 10)

= 45 – 0.4

= 44.6

σ² = (10²/150²) [150(342) – (-6)²]

= (100/22500) [51,300 – 36]

= (100/22500) × 51264

= 227.84

Hence, standard deviation = σ = √227.84

= 15.09

∴ C.V for group A = (σ/ x̅) × 100

= (15.09/44.6) × 100

= 33.83

Group B

Where A = 45,

h = 10

So, x̅ = 45 + ((-6/150) × 10)

= 45 – 0.4

= 44.6

σ² = (10²/150²) [150(366) – (-6)²]

= (100/22500) [54,900 – 36]

= (100/22500) × 54,864

= 243.84

Hence, standard deviation = σ = √243.84

= 15.61

∴ C.V for group B = (σ/ x̅) × 100

= (15.61/44.6) × 100

= 35

By comparing C.V. of group A and group B.

C.V of Group B > C.V. of Group A

So, Group B is more variable.

Question2. 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

Solution :

We have to calculate Mean for x,

Mean x̅ = ∑x_i/n

Where, n = number of terms

= 510/10

= 51

= (1/10²)[(10 × 26360) – 510²]

= (1/100) (263600 – 260100)

= 3500/100

= 35

WKT Standard deviation = √variance

= √35

= 5.91

So, co-efficient of variation = (σ/ x̅) × 100

= (5.91/51) × 100

= 11.58

Now, we have to calculate Mean for y,

Mean ȳ = ∑y_i/n

Where, n = number of terms

= 1050/10

= 105

= (1/10²)[(10 × 110290) – 1050²]

= (1/100) (1102900 – 1102500)

= 400/100

= 4

WKT Standard deviation = √variance

= √4

= 2

So, co-efficient of variation = (σ/ x̅) × 100

= (2/105) × 100

= 1.904

By comparing C.V. of X and Y.

C.V of X > C.V. of Y

So, Y is more stable than X.

Question3. 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?

Solution :
(i) Monthly wages of firm A = Rs 5253

Number of wage earners in firm A = 586

∴Total amount paid = Rs 5253 × 586

Monthly wages of firm B = Rs 5253

Number of wage earners in firm B = 648

∴Total amount paid = Rs 5253 × 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 firm A = 100

We know that, standard deviation (σ)= √100

=10

Variance of firm B = 121

Then,

Standard deviation (σ)=√(121 )

=11

Hence the standard deviation is more in case of Firm B that means in firm B there is greater variability in individual wages.

Question4. 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?

Solution :
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

Question5. The sum and sum of squares corresponding to lengths x(in cm) and weight y(in gm) of 5- plant products are given below:

Which is more varying, the length or weight?

Solution :
Given:

Miscellaneous Exercise

Question1. 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.

Solution :
Let two required observations be x and y Then,

Question 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.

Solution :
Let two required observations be x and y Then,

so, from eqn (i) ,we have

Question3. 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.

Solution :
 

Question4. Given that x̅ is the mean and σ² is the variance of n .observations x₁, x₂, …,x_n ..Prove that the mean and variance of the observations ax₁, ax₂, ax₃, …., ax_n are ax̅ and a²σ² respectively (a ≠ 0).

Solution :

From the question it is given that, n observations are x₁, x₂,…..x_n

Mean of the n observation = x̅

Variance of the n observation = σ²

As we know that,

Question5. 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.

Solution :

(i) If wrong item is omitted,

From the question it is given that,

The number of observations i.e. n = 20

The incorrect mean = 20

The incorrect standard deviation = 2

(ii) If it is replaced by 12,

From the question it is given that,

The number of incorrect sum observations i.e. n = 200

The correct sum of observations n = 200 – 8 + 12

n = 204

Then, correct mean = correct sum/20

= 204/20

= 10.2

Question6. The mean and standard deviation of marks obtained by 50 students of a class in three subjects, Mathematics, Physics and Chemistry are given below:

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

Solution :

From the question it is given that,

Mean of Mathematics = 42

Standard deviation of Mathematics = 12

Mean of Physics = 32

Standard deviation of physics = 15

Mean of Chemistry = 40.9

Standard deviation of chemistry = 20

As we know that,

Therefore, Chemistry with highest C.V. shows highest variability and Mathematics with lowest C.V. shows lowest variability.

Question7. 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 12, 21 and 18. Find the mean and standard deviation if the incorrect observations are omitted.

Solution :

From the question it is given that,

The total number of observations (n) = 100

Incorrect mean, (x̅) = 20

And, Incorrect standard deviation (σ) = 3