# NCERT solution class 11 chapter 15 Statistics exercise 15.1 mathematics

## EXERCISE 15.1

#### Question 1:

Find the mean deviation about the mean for the data

4, 7, 8, 9, 10, 12, 13, 17

The given data is

4, 7, 8, 9, 10, 12, 13, 17

Mean of the data, The deviations of the respective observations from the mean are

–6, – 3, –2, –1, 0, 2, 3, 7

The absolute values of the deviations, i.e. , are

6, 3, 2, 1, 0, 2, 3, 7

The required mean deviation about the mean is #### Question 2:

Find the mean deviation about the mean for the data

38, 70, 48, 40, 42, 55, 63, 46, 54, 44

The given data is

38, 70, 48, 40, 42, 55, 63, 46, 54, 44

Mean of the given data, The deviations of the respective observations from the mean are

–12, 20, –2, –10, –8, 5, 13, –4, 4, –6

The absolute values of the deviations, i.e. , are

12, 20, 2, 10, 8, 5, 13, 4, 4, 6

The required mean deviation about the mean is #### Question 3:

Find the mean deviation about the median for the data.

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

The 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 The deviations of the respective observations from the median, i.e. are

–3.5, –2.5, –2.5, –1.5, –0.5, –0.5, 0.5, 2.5, 2.5, 3.5, 3.5, 4.5

The absolute values of the deviations, , are

3.5, 2.5, 2.5, 1.5, 0.5, 0.5, 0.5, 2.5, 2.5, 3.5, 3.5, 4.5

The required mean deviation about the median is #### Question 4:

Find the mean deviation about the median for the data

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

The given data is

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

Here, 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 The deviations of the respective observations from the median, i.e. are

–11.5, –5.5, –2.5, –1.5, –1.5, 1.5, 3.5, 5.5, 12.5, 24.5

The absolute values of the deviations, , are

11.5, 5.5, 2.5, 1.5, 1.5, 1.5, 3.5, 5.5, 12.5, 24.5

Thus, the required mean deviation about the median is #### Question 5:

Find the mean deviation about the mean for the data.

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

 xi fi fi xi    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  #### Question 6:

Find the mean deviation about the mean for the data

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

 xi fi fi xi    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 #### Question 7:

Find the mean deviation about the median for the data.

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

The given observations are already in ascending order.

Adding a column corresponding to cumulative frequencies of the given data, we obtain the following table.

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

Here, 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. The absolute values of the deviations from median, i.e. are

 |xi – M| 2 0 2 3 5 8 fi 8 6 2 2 2 6 fi |xi – M| 16 0 4 6 10 48 and  #### Question 8:

Find the mean deviation about the median for the data

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

The given observations are already in ascending order.

Adding a column corresponding to cumulative frequencies of the given data, we obtain the following table.

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

Here, N = 29, which is odd. observation = 15th observation

This observation lies in the cumulative frequency 21, for which the corresponding observation is 30.

∴ Median = 30

The absolute values of the deviations from median, i.e. are

 |xi – M| 15 9 3 0 5 fi 3 5 6 7 8 fi |xi – M| 45 45 18 0 40 ∴  #### Question 9:

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

The following table is formed.

 Income per day Number of persons fi Mid-point xi fi xi    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

Here,   #### Question 10:

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

The following table is formed.

 Height in cms Number of boys fi Mid-point xi fi xi    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

Here,   #### Question 11:

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

The following table is formed.

 Marks Number of girls fi Cumulative frequency (c.f.) Mid-point xi |xi â€“ Med.| fi |xi â€“ Med.| 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 or 25th item is 20 â€“ 30.

Therefore, 20 â€“ 30 is the median class.

It is known that, Here, l = 20, C = 14, f = 14, h = 10, and N = 50

âˆ´ Median =  Thus, mean deviation about the median is given by, #### Question 12:

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

The given data is not continuous. Therefore, it has to be converted into continuous frequency distribution by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each class interval.

The table is formed as follows.

 Age Number fi Cumulative frequency (c.f.) Mid-point xi |xi – Med.| fi |xi – Med.| 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 75 43 5 60 45.5-50.5 16 91 48 10 160 50.5-55.5 9 100 53 15 135 100 735

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

Therefore, 35.5 – 40.5 is the median class.

It is known that, Here, l = 35.5, C = 37, = 26, h = 5, and N = 100  Thus, mean deviation about the median is given by, error: Content is protected !! 