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

Solution:

xi fi fixi |xix| fi|xix|
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

Statistics – Exercise 15.1 – Class XI


  1. 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 |xix| fi|xix|
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

Statistics – Exercise 15.1 – Class XI


  1. Find the mean deviation about the mean for the data
xi 5 7 9 10 12 15
fi 8 6 2 2 2 6

Solution:

xi fi c.f
5 8 8
7 6 14
9 2 16
10 2 18
12 2 20
15 6 26

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

8.png


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

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

Solution:

xi fi c.f
15 3 3
21 5 8
27 6 14
30 7 21
35 8 29
     

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 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:

Income per day Number of persons xi fixi |xix| fi|xix|
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

Statistics – Exercise 15.1 – Class XI


  1. 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:

Height in cms Number of boys xi fixi |xix| fi| xix|
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
      12530   1128.8

Statistics – Exercise 15.1 – Class XI


  1. 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:

Marks Number of girls Cumulative frequency(c.f) xi |ximedian| fi|xix|
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 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:
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:

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.

age Number c.f xi |xix| fi|xix|
16-20 5 5 18 20 100
21-25 6 11 23 15 90
26-30 12 23 28 10 120
31-35 14 37 33 5 70
36-40 26 63 38 0 0
41-45 12 75 43 5 60
46-50 16 91 48 10 160
51-55 9 100 53 15 135
  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


 

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