# 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

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

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

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

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 |xi–x| fi|xi–x| 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

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 |xi – x| fi|xi – x| 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

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.

 |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. 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.

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 |xi –x| fi|xi –x| 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

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 |xi – x| fi| xi – x| 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

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 |xi – median| fi|xi –x| 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

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 |xi – x| fi|xi–x| 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,