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Class 11 NCERT Solutions- Chapter 15 Statistics – Exercise 15.1

  • Last Updated : 11 Feb, 2021

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

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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(xi – a)

xi478910121317
xi – a4 – 10 = -67 – 10 = -38 – 10 = -29 – 10 = -110 – 10 = 012 – 10 = 213 – 10 = 317 – 10 = 7

Step 3: Taking absolute value of deviations, we get 

|xi – a| = 6, 3, 2, 1, 0, 2, 3, 7



Step 4: Required mean deviation about the mean is

M.D (a) = ∑ (|xi – 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(xi – a)

xi38704840425563465444
xi  – a38 – 50 = -1270 – 50 = 2048 – 50 = -240 – 50 = -1042 – 50 = -855 – 50 = 563 – 50 = 1346 – 50 = -454 – 50 = 444 – 50 = -6

Step 3: Taking absolute value of deviations, we get 

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

Step 4: Required mean deviation about the mean is

M.D(a) = ∑(|xi – 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 = 6th observation= 13

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

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

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

xi101111121313141616171718
xi -M10 – 13.5 = -3.511 – 13.5 = -2.511 – 13.5 = -2.512 – 13.5 = -1.513 – 13.5 = -0.513 – 13.5 = 0.514 – 13.5 = 0.516 – 13.5 = 2.516 – 13.5 = 2.517 – 13.5 = 3.517 – 13.5 = 3.518 – 13.5 = 4.5

Step 3: Taking absolute values of deviations we get

|xi – 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) = ∑(|xi – 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 = 5th observation= 46

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

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

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

xi36424546464951536072
xi – M-11.5-5.5-2.5-1.5-1.51.53.55.512.524.5

Step 3: Taking absolute values of deviations we get

|xi – 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) = ∑(|xi – 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. 

xi510152025
fi74635

Solution: 



Given table data is of discrete frequency distribution as 

we have n = 5 distinct values(xi) along with their frequencies(fi).

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

xifixi * fi|xi – a|fi * |xi  – a|
5735963
10440416
1569016
20360618
2551251155
Total   25350 158

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

N = ∑ fi = (7 + 4 + 6 + 3 + 5) = 25

∑ xi * fi = (35 + 40 + 90 + 60 + 125) = 350

Step 3: Find the mean using following formula

Mean (a) = ∑(xi * fi)/ N = 350 / 25 = 14

Step 4: Using above mean find absolute values of deviations 

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

 Now,

∑ fi *|xi – a| = (63 + 16 + 6 + 18 + 55) = 158

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

= 158 / 25 

= 6.32

So, mean deviation for given observations is 6.32

Question 6. 

xi1030507090
fi42428168

Solution: 

Given table data is of discrete frequency distribution as 

we have n = 5 distinct values(xi) along with their frequencies(fi).

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

xifixi * fi|xi – a|fi * |xi – a|
1044040160
302472020480
5028140000
7016112020320
90872040320
Total804000 1280

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



N = ∑(fi) = 80 & ∑(xi * fi) = 4000

Step 3: Find the mean using following formula

Mean (a) = ∑ (xi * fi)/ N = 4000 / 80 = 50

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

of fi * |xi -a| column.

Now, from above table,

∑ fi *|xi – a| = (160 + 480 + 0 + 320 + 320) = 1280

Using formula, M.D (a) = ∑ fi *|xi -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. 

xi579101215
fi862226

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.

xifiC.F|xi – M|fi * |xi – M|
588216
761400
921624
1021836
12220510
15626848
Total26  84

Step 2: Identifying the observation whose cumulative frequency is equal 

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

N = ∑fi = 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

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



= (7 + 7) / 2 = 7

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

i.e., |xi – M| and fi * |xi – M| as shown in above table.

∑ fi * |xi – M| = (16 + 4 + 6 + 10 + 48) = 84

Using formula, M.D (M) = ∑ (fi * |xi – M|)/ N

= 84 / 26

= 3.23

So, mean deviation about median for given observations is 3.23

Question 8. 

xi1521273035
fi35678

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.

xifiC.F|xi – M|fi * |xi – M|
15331545
2158945
27614318
3072100
35829540
Total29  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 = ∑fi = 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

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

= (30 + 30)/2 = 30

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

i.e., |xi – M| and fi * |xi – M| as shown in above table .

 ∑fi * |xi – M| = 148

Using formula, M.D (M) = ∑(fi * |xi  – 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.

0-100100-200200-300300-400400-500500-600600-700700-800

Number Of 



Persons  

489107543

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|
0-1004502003081232
100-200815012002081664
200-30092502250108972
300-400103503500880
400-5007450315092644
500-60055502750192960
600-700465026002921160
700-800375022503921176
Total50 17900 7896

 Step 2: Finding the sum of frequencies fi‘s and sum of fixi‘s

N = ∑fi = 50

∑fixi = 17900

Then mean of given data is given by

a = ∑ (fixi)/ N 

= 17900 / 50

a = 358

Step 3: Computing sum of column fi * |xi – a|

∑fi * |xi – a| = 7896

Thus mean deviation about mean is given by 

M.D (a) = ∑fi * |xi – a| / N

= 7896 / 50 

= 157.92

Therefore, mean deviation about mean for given data is 157.92

Question 10. 

  Height in 

     cms

95 – 105105 – 115115 – 125 125 – 135 135 – 145 145 – 155 

Number of

    boys

91326301210

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|
95-105910090025.3227.7
105-11513110143015.3198.9
115-1252612031205.3137.8
 125-1353013039004.7141
135-14512140168014.7176.4
145-15510150150024.7247
Total100 12530 1128.8

 Step 2: Finding the sum of frequencies fi‘s and the sum of fixi‘s



N = ∑fi = 100

∑ fixi = 17900

Then mean of given data is given by

a = ∑(fixi) / N

= 12530 / 100

a = 125.3

Step 3: Computing sum of column fi * |xi – a|

∑fi * |xi – a| = 1128.8

Thus mean deviation about mean is given by 

M.D(a) = ∑(fi * |xi – 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.

Marks0 – 1010 – 2020 – 3030 – 4040 – 5050 – 60

Number of

     Girls

68141642

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|
0-1066522.85137.1
10-208141512.85102.8
20-301428252.8539.9
30-401644357.15114.4
40-504484517.1568.6
50-602505527.1554.3
Total50   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 fi * |xi – M|

∑fi * |xi – M| = 517.1

Thus, mean deviation about median is given by 

M.D(M) = ∑fi * |xi – 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 – 3536 – 4041 – 4546 – 5051 – 55
Number5612142612169

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|
15.5-20.5551820100
20.5-25.5611231590
25.5-30.512232810120
30.5-35.5143733570
35.5-40.526633800
40.5-45.5127143560
45.5-50.516954810160
50.5-55.591005315135
Total100   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 fi * |xi – M|

∑fi * |xi – M| = 735

Thus, mean deviation about median is given by

M.D(M) = ∑fi * |xi – M| / N

= 735 / 100

= 7.35 

So, mean deviation about median for given observations is 7.35




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