Class 11 NCERT Solutions- Chapter 15 Statistics – Exercise 15.1

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

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

Solution:

Given observations are = 4, 7, 8, 9, 10, 12, 13, 17

Therefore, Total Number of observations = n = 8

Step 1: Calculating mean(a) for given data about which we have to find the mean deviation.

Mean(a) = ∑(all data points) / Total number of data points             

= (4 + 7 + 8 + 9 + 10 + 12 + 13 + 17) / 8

= 80 / 8

a = 10

Step 2: Finding deviation of each data point from mean(x_i – a)

xi	4	7	8	9	10	12	13	17
xi – a	4 – 10 = -6	7 – 10 = -3	8 – 10 = -2	9 – 10 = -1	10 – 10 = 0	12 – 10 = 2	13 – 10 = 3	17 – 10 = 7

Step 3: Taking absolute value of deviations, we get 

|x_i – a| = 6, 3, 2, 1, 0, 2, 3, 7

Step 4: Required mean deviation about the mean is

M.D (a) = ∑ (|x_i – a|) / n

= (6 + 3 + 2 + 1 + 0 + 2 + 3 + 7) / 8

= 24 / 8 

= 3

So, mean deviation for given observations is 3

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

Solution:

Given observations are = 38, 70, 48, 40, 42, 55, 63, 46, 54, 44 

Therefore, Total Number of observations = n = 10

Step 1: Calculating mean(a) for given data about which we have to find the mean deviation.

Mean(a) = ∑(all data points)/ Total number of data points

= (38 + 70 + 48 + 40 + 42 + 55 + 63 + 46 + 54 + 44)/ 10

= 500 / 10

a = 50

Step2 : Finding deviation of each data point from mean(x_i – a)

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

Step 3: Taking absolute value of deviations, we get 

|x_i – a| = 12, 20, 2, 10, 8, 5, 13, 4, 4, 6

Step 4: Required mean deviation about the mean is

M.D(a) = ∑(|x_i – a|) / n

= (12 + 20 + 2 + 10 + 8 + 5 + 13 + 4 + 4 + 6) / 10

= 84 / 10

= 8.4

So, mean deviation for given observations is 8.4

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

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

Solution:

Given observations are = 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17 

Therefore, Total Number of observations = n = 12 

Step 1: Calculating median(M) for given data about which we have to find the mean deviation.

To calculate median for given data we have to arrange observations in ascending order

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

As number of observations are even median is given by following formula

Median = [(n/2)^th observation+ ((n/2) + 1)^th observation] / 2

(n/2)^th observation = 12 / 2 = 6^th observation= 13

((n/2)+1)^th observation = (12/2) + 1= 7^thobservation = 14 

Therefore, Median = (13 + 10) / 2 = 13.5

Step 2: Finding deviation of each data point from median(x_i – M)       

xi	10	11	11	12	13	13	14	16	16	17	17	18
xi -M	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

Step 3: Taking absolute values of deviations we get

|x_i – M| = 3.5, 2.5, 2.5, 1.5, 0.5, 0.5, 0.5, 2.5, 2.5, 3.5, 3.5, 4.5

Step 4: Required mean deviation about median is given by

M.D (M) = ∑(|x_i – M|) / n

= (3.5 + 2.5 + 2.5 + 1.5 + 0.5 + 0.5 + 0.5 + 2.5 + 2.5 + 3.5 + 3.5 + 4.5) / 12

= 28 / 12

= 2.33

So, mean deviation about median for given observations is 2.33

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

Solution:

Given observations are = 36, 72, 46, 42, 60, 45, 53, 46, 51, 49 

Therefore, Total Number of observations = n = 10

Step 1: Calculating median(M) for given data about which we have to find the mean deviation.

To calculate median for given data we have to arrange observations in ascending order

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

As number of observations are even median is given by following formula

Median = [(n/2)^th observation+ ((n/2) + 1)^th observation] / 2

(n/2)^th observation = 10 / 2 = 5^th observation= 46

((n/2) + 1)^th observation = (10/2) + 1 = 6^th observation = 49

Therefore, Median = (46 + 49) / 2 = 47.5

Step 2: Finding deviation of each data point from median(x_i -M) 

xi	36	42	45	46	46	49	51	53	60	72
xi – M	-11.5	-5.5	-2.5	-1.5	-1.5	1.5	3.5	5.5	12.5	24.5

Step 3: Taking absolute values of deviations we get

|x_i – M| = 11.5, 5.5, 2.5, 1.5, 1.5, 1.5, 3.5, 5.5, 12.5, 24.5

Step 4: Required mean deviation about median is given by

M.D (M) = ∑(|x_i – M|) / n

= (11.5 + 5.5 + 2.5 + 1.5 + 1.5 + 1.5 + 3.5 + 5.5 + 12.5 + 24.5) / 10

= 70 / 10

= 7

So, mean deviation about median for given observations is 7

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

Question 5. 

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

Solution: 

Given table data is of discrete frequency distribution as 

we have n = 5 distinct values(x_i) along with their frequencies(f_i).

Step 1: Let us make a table of given data and append other columns after calculation

xi	fi	xi * fi	|xi – a|	fi * |xi  – a|
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
Total	25	350		158

Step2 : Now to find the mean we have to first calculate the sum of given data

N = ∑ f_i = (7 + 4 + 6 + 3 + 5) = 25

∑ x_i * f_i = (35 + 40 + 90 + 60 + 125) = 350

Step 3: Find the mean using following formula

Mean (a) = ∑(x_i * f_i)/ N = 350 / 25 = 14

Step 4: Using above mean find absolute values of deviations 

from mean i.e |x_i – a| and also find values of f_i *|xi -a| column.

 Now,

∑ f_i *|x_i – a| = (63 + 16 + 6 + 18 + 55) = 158

Using formula, M.D (a) = ∑(f_i * |x_i – a|)/ N

= 158 / 25 

= 6.32

So, mean deviation for given observations is 6.32

Question 6. 

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

Solution:

Given table data is of discrete frequency distribution as 

we have n = 5 distinct values(x_i) along with their frequencies(f_i).

Step 1: Let us make a table of given data and append other columns after calculation

xi	fi	xi * fi	|xi – a|	fi * |xi – a|
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
Total	80	4000		1280

Step 2: Now to find the mean we have to first calculate the sum of given data. From the above table,

N = ∑(f_i) = 80 & ∑(x_i * f_i) = 4000

Step 3: Find the mean using following formula

Mean (a) = ∑ (x_i * f_i)/ N = 4000 / 80 = 50

Step 4: Using above mean find absolute values of deviations from mean i.e |x_i – a| and also find values  

of f_i * |x_i -a| column.

Now, from above table,

∑ f_i *|x_i – a| = (160 + 480 + 0 + 320 + 320) = 1280

Using formula, M.D (a) = ∑ f_i *|x_i -a| / N

= 1280 / 80

= 16

So, mean deviation for given observations is 16

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

Question 7. 

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

Solution: 

For the given discrete frequency distribution we have to find a mean deviation about the median.

Step 1: Let us make a table of given data and append another column of cumulative frequencies. 

The given observations are already in ascending order.

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

Step 2: Identifying the observation whose cumulative frequency is equal 

to or just greater than N / 2 and then Finding the median.

N = ∑f_i = 26 is even. We divide N by 2. Thus, 26/2 = 13

The cumulative frequency for greater than 13 is 14, for which corresponding observation is 7

Median = [(N/2)^th observation + ((N/2) + 1)^th observation] / 2

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

= (7 + 7) / 2 = 7

Step 3: Now, find absolute values of the deviations from median,

i.e., |x_i – M| and f_i * |x_i – M| as shown in above table.

∑ f_i * |x_i – M| = (16 + 4 + 6 + 10 + 48) = 84

Using formula, M.D (M) = ∑ (f_i * |x_i – M|)/ N

= 84 / 26

= 3.23

So, mean deviation about median for given observations is 3.23

Question 8. 

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

Solution: 

For the given discrete frequency distribution we have to find a mean deviation about the median.

Step 1: Let us make a table of given data and append another column of cumulative frequencies. 

The given observations are already in ascending order.

xi	fi	C.F	|xi – M|	fi * |xi – M|
15	3	3	15	45
21	5	8	9	45
27	6	14	3	18
30	7	21	0	0
35	8	29	5	40
Total	29			148

Step 2: Identifying the observation whose cumulative frequency is equal to or

just greater than N / 2 and then Finding the median.

Here, N = ∑f_i = 29 is odd we divide N by 2. Thus, 29 / 2 = 14.5

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

Therefore, Median = [(N/2)^th observation+ ((N/2) + 1)^th observation] / 2

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

= (30 + 30)/2 = 30

Step 3: Now, find absolute values of the deviations from median,

i.e., |x_i – M| and f_i * |x_i – M| as shown in above table .

 ∑f_i * |x_i – M| = 148

Using formula, M.D (M) = ∑(f_i * |x_i  – M|) / N           

= 148 / 29

= 5.1

So, mean deviation about median for given observations is 5.1

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

Question 9. 

Income per  day in Rs.	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

Solution: 

Given data is in continuous intervals along with their frequencies so it is in the 

continuous frequency distribution. In this case, we assume that the frequency 

in each class is centered at its mid-point.

Step 1: So, we find a midpoint for each interval.

Then we append other columns similar to discrete frequency distribution.

Income per day in Rs.	Number Of Person fi	Midpoints       xi	fixi	|xi – a|	fi * |xi – a|
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	1160
700-800	3	750	2250	392	1176
Total	50		17900		7896

 Step 2: Finding the sum of frequencies f_i‘s and sum of f_ix_i‘s

N = ∑f_i = 50

∑f_ix_i = 17900

Then mean of given data is given by

a = ∑ (f_ix_i)/ N 

= 17900 / 50

a = 358

Step 3: Computing sum of column f_i * |x_i – a|

∑f_i * |x_i – a| = 7896

Thus mean deviation about mean is given by 

M.D (a) = ∑f_i * |x_i – a| / N

= 7896 / 50 

= 157.92

Therefore, mean deviation about mean for given data is 157.92

Question 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

Solution:

Given data is in the continuous frequency distribution.

Step 1: We find a midpoint for each interval and then append other columns.

Height in     cms	Number of     boys	Midpoints       xi	fixi	|xi – a|	fi * |xi – a|
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
Total	100		12530		1128.8

Step 2: Finding the sum of frequencies f_i‘s and the sum of f_ix_i‘s

N = ∑f_i = 100

∑ f_ix_i = 17900

Then mean of given data is given by

a = ∑(f_ix_i) / N

= 12530 / 100

a = 125.3

Step 3: Computing sum of column f_i * |x_i – a|

∑f_i * |x_i – a| = 1128.8

Thus mean deviation about mean is given by 

M.D(a) = ∑(f_i * |x_i – a|)/ N

= 1128.8 / 100

= 11.288

Therefore, the mean deviation about mean for given data is 11.288

Find the mean deviation about the median for the following data in questions 11 and 12.

Question 11.

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

Solution:

Given data is in the continuous frequency distribution &

here the only difference is that we have to calculate the median.

Step 1: First we have to compute cumulative frequencies, then we find a midpoint for each interval and then append other columns.

The data is already arranged in ascending order

Marks	Number of      Girls fi	C.F	Midpoints      xi	|xi – M|	fi * |xi – M|
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
Total	50				517.1

Step 2: First identifying the interval in which median lies and then applying the formula to compute median

The class interval whose cumulative frequency is greater than equal to N/2 = 25 is 20 – 30.

So, 20 – 30 is the median class.

Then applying the formula

Median (M) = l + {[(N / 2) – C] / f} * h

where, l = lower limit of the median class

h = width of median class

N = sum of frequencies

C = cumulative frequency of the class just preceding the median class

Therefore,

M = 20 + {[(25 – 14) / 14] * 10}

M = 27.85

Step 3: Finding absolute values of the deviations from median as shown in table

computing sum of column f_i * |x_i – M|

∑f_i * |x_i – M| = 517.1

Thus, mean deviation about median is given by 

M.D(M) = ∑f_i * |x_i – M| / N

= 517.1 / 50

= 10.34

So, mean deviation about median for given observations is 10.34

Question12. 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

Solution:

Converting 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. 

Check if data is arranged in ascending order.

Step 1: Finding midpoints and C.Fs and then appending other columns

Age (in years)	Number     fi	C.F	Midpoints         xi	|xi – M|	fi * |xi – M|
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	71	43	5	60
45.5-50.5	16	95	48	10	160
50.5-55.5	9	100	53	15	135
Total	100				735

Step 2: First identifying the interval in which median lies and then applying the formula to compute median

The class interval whose cumulative frequency is greater than equal to N/2 = 50 is 35.5-40.5.

So, 35.5-40.5 is the median class.

Then applying the formula

Median (M) = l + {[(N / 2) – C] / f} * h

where l = lower limit of the median class = 35.5

h = width of median class = 5

N = sum of frequencies = 100

C = cumulative frequency of the class just preceding the median class = 37

f = frequency = 26

Therefore,

M = 35.5 + {[(50 – 37) / 26] * 5}

M = 38

Step 3: Finding absolute values of the deviations from median as shown in table

computing sum of column f_i * |x_i – M|

∑f_i * |x_i – M| = 735

Thus, mean deviation about median is given by

M.D(M) = ∑f_i * |x_i – M| / N

= 735 / 100

= 7.35 

So, mean deviation about median for given observations is 7.35