Statistics – Exercise 15.1 – Class XI

  1. Find the mean deviation about the mean for the data 4, 7, 8, 9, 10, 12, 13, 17

Solution:

Given data is 4, 7, 8, 9, 10, 12, 13, 17

Mean of the data

Statistics – Exercise 15.1 – Class XI


  1. Find the mean deviation about the mean for the data 38, 70, 48, 40, 42, 55, 63, 46, 54, 44

Solution:

Given data is 38, 70, 48, 40, 42, 55, 63, 46, 54, 44

Mean of the given data

Statistics – Exercise 15.1 – Class XI


3.Find the mean deviation about the median for the data 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17

Solution:

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

Statistics – Exercise 15.1 – Class XI


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

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

Solution:

Given data is 36, 72, 46, 42, 60, 45, 53, 46, 51, 49

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

Statistics – Exercise 15.1 – Class XI


  1. Find the mean deviation about the mean for the data
xi510152025
fi74635

Solution:

xififixi|xix|fi|xix|
5735963
10440416
1569016
20360618
2551251155
 25350 158

Statistics – Exercise 15.1 – Class XI


  1. Find the mean deviation about the mean for the data
xi1030507090
fi42428168

Solution:

xififi xi|xix|fi|xix|
1044040160
302472020480
5028140000
7016112020320
90872040320
 804000 1280

Statistics – Exercise 15.1 – Class XI


  1. Find the mean deviation about the mean for the data
xi579101215
fi862226

Solution:

xific.f
588
7614
9216
10218
12220
15626

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.

7.png

|xi – M|202358
fi862226
fi|xi – M|160461048

8.png


8. Find the absolute value of the derivations from median, for the data

xi1521273035
fi35678

Solution:

xific.f
1533
2158
27614
30721
35829
   

N = 29, which is an  odd number.Statistics – Exercise 15.1 – Class XI


  1. Find the mean deviation about the mean for the data
Income per dayNumber of persons
0-1004
100-2008
200-3009
300-40010
400-5007
500-6005
600-7004
700-8003

Solution:

Income per dayNumber of personsxifixi|xix|fi|xix|
0-1004502003081232
100-200815012002081664
200-30092502250108972
300-400103503500880
400-5007450315092644
500-60055502750192960
600-700465026002921168
700-800375022503921176
 50 17900 7896

Statistics – Exercise 15.1 – Class XI


  1. Find the mean deviation about the mean for the data
Height in cmsNumber of boys
95-1059
105-11513
115-12526
125-13530
135-14512
145-15510

Solution:

Height in cmsNumber of boysxifixi|xix|fi| xix|
95-105910090025.3227.7
105-11513110143015.3198.9
115-1252612031205.3137.8
125-1353013039004.7141
135-14512140168014.7176.4
145-15510150150024.7247
   12530 1128.8

Statistics – Exercise 15.1 – Class XI


  1. Find the mean deviation about median for the following data:
MarksNumber of girls
0-106
10-208
20-3014
30-4016
40-504
50-602

Solution:

MarksNumber of girlsCumulative frequency(c.f)xi|ximedian|fi|xix|
0-1066522.85137.1
10-208141512.85102.8
20-301428252.8539.9
30-401644357.15114.4
40-504484517.1568.6
50-602505527.1554.3
 50   517.1

The class interval containing the (N/2) or 25th item is 20-30

Thus, 20-30 is the median class

12


  1. Calculate the mean deviation about median age for the age distribution of 100 persons given below:
ageNumber
16-205
21-256
26-3012
31-3514
36-4026
41-4512
46-5016
51-559

Solution:

We need to convert the given dat into continuous frequency distribution by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each interval in order to make it continuous.

ageNumberc.fxi|xix|fi|xix|
16-20551820100
21-25611231590
26-3012232810120
31-35143733570
36-4026633800
41-45127543560
46-5016914810160
51-5591005315135
 100   735

The class interval containing the N/2 th or 50th item is 35.5 – 40.5.

Therefore, 35.5 – 40.5 is the median class.

We know that,

13.png