Chapter 13. STATISTATICS

Class 11 Maths Chapter 13 Statistatics Exercise 13.1 Solutions :

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

1.   4, 7, 8, 9, 10, 12, 13, 17
2.  38, 70, 48, 40, 42, 55, 63, 46, 54, 44

Solution :  1. We arrange the data in ascending order :

          4, 7, 8, 9, 10, 12, 13, 17

We know that , Mean of deviations

4	|4 – 10| = 6
7	|7 – 10| = 3
8	|8 – 10| = 2
9	|9 – 10| = 1
10	|10 – 10| = 0
12	|12 – 10| = 2
13	|13 – 10| = 3
17	|17 – 10| = 7

M.D .
2. We arrange the data in ascending order :

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

We know that ,  Mean of deviations

38	|38 – 50| = 12
40	|40– 50| = 10
42	|42 – 50| = 8
44	|44 – 50| = 6
46	|46 – 50| = 4
48	|48 – 50| = 2
54	|54 – 50| = 4
55	|55 – 50| = 5
63	|63 – 50|=13
70	|70 – 50| = 20

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

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

Solution : 3. We arrange the data in ascending order :

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

Here ,  (Even)

Median (M)

	|
10 11 11 12 13 13 14 16 16 17 17 18	|10 – 13.5|= 3.5 |11– 13.5| = 2.5 |11– 13.5| = 2.5 |12– 13.5| = 1.5 |13– 13.5| = 0.5 |13– 13.5| = 0.5 |14– 13.5| = 0.5 |16– 13.5| = 2.5 |16– 13.5| = 2.5 |17– 13.5| = 3.5 |17– 13.5| = 3.5 |18– 13.5| = 4.5

 M.D. (M)

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

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

Solution :  We arrange the data in ascending order :

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

Here ,  (Even)

Median (M) 

36 42 45 46 46 49 51 53 60 72	|36 – 47.5| = 11.5 |42 – 47.5| = 5.5 |45 – 47.5| = 2.5 |46 – 47.5| = 1.5 |46 – 47.5| = 1.5 |49 – 47.5| = 1.5 |51 – 47.5| = 3.5 |53 – 47.5| = 5.5 |60 – 47.5| = 12.5 |72 – 47.5| = 24.5

M.D.(M)

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

      : 5        10      15      20      25
       : 7        4        6        3         5

Solution : 5.

5 10 15 20 25	7 4 6 3 5	35 40 90 60 125	|5 – 14| =  9 |10 – 14| = 4 |15 – 14| = 1 |20 – 14| = 6 |25 – 14| = 11	63 16 6 18 55
				158

Mean

M.D.

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

         :  10      30     50      70      90
         :   4       24     28      16        8

Solution : We have,

10 30 50 70 90	4 24 28 16 8	40 720 1400 1120 720	|10– 50| =  40 |30 – 50| = 20 |50 – 50| = 0 |70 – 50| = 20 |90 – 50| = 40	160 480 0 320 320
				1280

Mean

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

      :  5     7     9     10    12    15
       :  8     6     2       2     2       6

Solution : We have,

		C.F.
5 7 9 10 12 15	8 6 2 2 2 6	8 14 16 18 20 26	|5 – 7| =  2 |7 – 7| = 0 |9 – 7| = 2 |10 – 7| = 3 |12 – 7| = 5 |15 – 7| = 8	16 0 4 6 10 48
				84

Here ,  (even)

Median (M)

M.D. (M)

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

     :  15     21     27    30    35
     :   3      5        6      7      8

Solution : We have,

		C.F.
15 21 27 30 35	3 5 6 7 8	3 8 14 21 29	|15 – 30| =  15 |21 – 30| = 9 |27 – 30| = 3 |30 – 30| = 0 |35 – 30| = 5	45 45 18 0 40
				148

Here ,  (odd)

Median (M)

M.D. (M)
9.Find the mean deviation about the mean for the data in Exercises 9 and 10.

Income per day (in Rs)	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 : We have ,

Income per day	Mid-points ()	No. of Students ()
0 – 100 100 – 200 200 – 300 300 – 400 400 – 500 500 – 600 600 – 700 700 – 800	50 150 250 350 450 550 650 750	4 8 9 10 7 5 4 3	200 1200 2250 3500 3150 2750 2600 2250	|50 - 358| = 308 |150 – 358| = 208 |250 – 358|=108 |350 – 358| = 8 |450 – 358|=92 |550 – 358|=192 |650 – 358|= 292 |750 – 358|= 392	1232 1664 972 80 644 960 1168 1176
		50	17900		7896

 

M.D.

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

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 : We have,

Class interval
95 – 105 105 – 115 115 – 125 125 – 135 135 – 145 145 – 155	9 13 26 30 12 10	100 110 120 130 140 150	– 3 – 2  – 1 0 1 2	– 27 –26 –26 0 12 20	|100 – 125.3| = 25.3 |110 – 125.3| = 15.3 |120 – 125.3| = 5.3 |130 – 125.3| = 4.7 |140 – 125.3| = 14.7 |150 – 125.3| = 24.7	227.7 198.9 137.8 141.0 176.4 247.0
	100			– 47		1128.8

Here,

 So, 

M.D.

11. 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 : We have

Marks			C.F.
0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60	5 15 25 35 45 55	6 8 14 16 4 2	6 14 28 44 48 50	|5 – 27.68| = 22.86 |15 – 27.68|= 12.68 |25 – 27.68| = 2.68 |35 – 27.68| = 7.14 |45 – 27.68| = 17.14 |55 – 27.68| = 27.14	137.16 102.88 40.04 114.24 68.56 54.28
		50			517.16

Here, , , , C.F. = 14

Median

M.D.(M)

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

Age (in years)	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

[Hint Convert the given data into continuous frequency distribution by subtracting 0.5
from the lower limit and adding 0.5 to the upper limit of each class interval]

Solution : We have,

Age			CF
15.5 – 20.5 20.5 – 25.5 25.5 – 30.5 30.5 – 35.5 35.5 – 40.5 40.5 – 45.5 45.5 – 50.5 50.5 – 55.5	18 23 28 33 38 43 48 53	5 6 12 14 26 12 16 9	5 11 23 37 63 75 91 100	|18 – 38|= 20 |23 – 38| = 15 |28 – 38| = 10 |33 – 38| = 5 |38 – 38| = 0 |43 – 38| = 5 |48 – 38| = 10 |53 – 38| = 15	100 90 120 70 0 60 160 135
		100			735

Here,   ,, CF = 37

Median

M.D.(M)

Class 11 Maths Chapter 13 Statistatics Exercise 13.2 Solutions :

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

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

Solution : We have ,

6 7 10 12 13 4 8 12	36 49 100 144 169 16 64 144

Mean

Variance

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

First natural numbers

Solution : We have ,

1 2 3	1² 2² 3²

Mean

Variance

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

First 10 multiples of 3

Solution :  List of  10 multiples of 3 :

   3 , 6 , 9 , 12 , 15 , 18 , 21 , 24 , 27 , 30

3 6 9 12 15 18 21 24 27 30	9 36 81 144 225 324 441 576 729 900

Mean

Variance

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

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

Solution : We have ,

6 10 14 18 24 28 30	2 4 7 12 8 4 3	36 100 196 324 576 784 900	12 40 98 216 192 112 90	72 400 1372 3888 4608 3136 2700
130	n = 40		760	16176

Mean

Variance

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

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

Solution : We have,

92 93 97 98 102 104 109	3 2 3 2 6 3 3	8464 8649 9409 9604 10404 10816 11881	276 186 291 196 612 312 327	25392 17298 28227 19208 62424 32448 35643

Mean

Variance

6. Find the mean and standard deviation using short-cut method.

	60	61	62	63	64	65	66	67	68
	2	1	12	29	25	12	10	4	5

Solution : We have,

60 61 62 63 64(A) 65 66 67 68	2 1 12 29 25 12 10 4 5	– 4  – 3  – 2  – 1 0 1 2 3 4	16 9 4 1 0 1 4 9 16	– 8  – 3  – 24 – 29 0 12 20 12 20	32 9 48 29 0 12 40 36 80
		0

Mean

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

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

Solution : We have ,

Classes
0 – 30 30 – 60 60 – 90 90 – 120 120 – 150 150 – 180 180 – 210	15 45 75 105 (A) 135 165 195	2 3 5 10 3 5 2	– 3  – 2 – 1 0 1 2    3	9 4 1 0 1 4 9	– 6 – 8 – 1 0 1 8 27	18 12 5 0 3 20 18
		30			2	76

Here,  ,  ,  ,    

Mean

Variance

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

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

Solution : We have,

Classes
0 – 10 10 – 20 20 – 30 30 – 40 40 – 50	5 15 25 35 45	5 8 15 16 6	– 2 – 1 0 1 2	4 1 0 1 4	– 10  – 8 0 16 12	20 8 0 16 24
		50			10	68

Here,   ,  ,  ,    

Mean

Variance

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 : We have,

Classes
70 – 75	72.5	3	– 4	16	– 12	48
75 – 80	77.5	4	– 3	9	– 12	36
80 – 85	82.5	7	– 2	4	– 14	28
85 – 90	87.5	7	– 1	1	– 7	7
90 – 95	92.5 (A)	15	0	0	0	0
95 – 100	97.5	9	1	1	9	9
100 – 105	102.5	6	2	4	24	24
105 – 110	107.5	6	3	9	18	54
110 – 115	112.5	3	4	16	12	48
		60			6	254

Here,  ,  ,  ,   

Mean

Variance

Standard deviation

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

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

Calculate the standard deviation and mean diameter of the circles.
[ Hint First make the data continuous by making the classes as 32.5-36.5 , 36.5-40.5 , 40.5-44.5 , 44.5 - 48.5 , 48.5 - 52.5 and then proceed.]

Solution : We have,

Classes
32.5 – 36.5 36.5 – 40.5 40.5 – 44.5 44.5 – 48.5 48.5 – 52.5	34.5 38.5 42.5 (A) 46.5 50.5	15 17 21 22 25	– 2  – 1 0 1 2	4 1 0 1 4	– 30 – 17 0 22 50	60 17 0 22 100
Total		100			25	199

Here,  ,  ,  ,   

Mean

Variance

Class 11 Maths Chapter 13 Statistatics Miscellaneous Exercise Solutions :

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.

Solution :  Let the other two observations be  and   .

Here , , and

So, the series is  6 , 7 , 10 , 12 , 12 , 13 , ,   .

Mean

 

Variance

 [From (i) ]

  or 

   , 

  

Putting  in (i) , we get 

Putting   in (i) , we get

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.

Solution :  Let the other two observations be  and  .

Here ,  , and

So, the series is  2 , 4 , 10 ,12 , 14 , ,

Mean

Variance

 [From (i) ]

 or

   ,

  

Putting   in (i) , we get 

Putting   in (i) , we get

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.

Solution : Let the observations be   and be their mean

Here ,  , and

We have,  Variance

 

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

From (i) , (ii) and (iii) , we get   

 

Thus the variance of new observations

4. Given that  is the mean and  is the variance of  observations  Prove that the mean and variance of the observations  are   and , respectively, (

Solution : Mean of the observations :   .

Mean

Again , mean of the observations  :

Mean

Hence, the mean of    is .

Variance of the observations : 

Variance

Again , variance of the observations  :   .

Variance

Hence, the variance of  is   .

5. 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.

 Mean

 Standard deviation

Number of observations

   Mean,

 

     [8 is omit ]

Correct   

Number of correct observations = 20 – 1 = 19 [one observation omit .]

New mean

We have,

 

 

Correct

The standard deviation

(ii) If it is replaced by 12.

    

   [ 8 is replaced by 12 ]

Correct   

New mean

We have,

 

 

Correct

The standard deviation

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.

Solution : Given,  Mean

 Standard deviation

Number of observations

          

  

The incorrect observations are: 21, 21, 18.

 

Correct  

Number of correct observations = 100 – 3 = 97

New mean

We have,

 

The incorrect observations are: 21, 21, 18 .

 Correct

The standard deviation