NCERT Solutions for Class 11 Maths Chapter 15 – Statistics Ex 15.1
Page No 360:
Question 1:
Find the mean deviation about the mean for the data
4, 7, 8, 9, 10, 12, 13, 17
Answer:
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
Answer:
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
Answer:
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
Answer:
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.
x_{i}

5

10

15

20

25

f_{i}

7

4

6

3

5

Answer:
 x_{i}f_{i}f_{i} x_{i}573596310440416156901620360618255125115525350158
Question 6:
Find the mean deviation about the mean for the data
x_{i}

10

30

50

70

90

f_{i}

4

24

28

16

8

Answer:
 x_{i}f_{i}f_{i} x_{i}104404016030247202048050281400007016112020320908720403208040001280
Question 7:
Find the mean deviation about the median for the data.
x_{i}

5

7

9

10

12

15

f_{i}

8

6

2

2

2

6

Answer:
The given observations are already in ascending order.
Adding a column corresponding to cumulative frequencies of the given data, we obtain the following table.
 x_{i}f_{i}c.f.58876149216102181222015626
Here, N = 26, which is even.
Median is the mean of 13^{th} and 14^{th} 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
 x_{i} – M202358f_{i}862226
f_{i} x_{i} – M160461048
and
Question 8:
Find the mean deviation about the median for the data
x_{i}

15

21

27

30

35

f_{i}

3

5

6

7

8

Answer:
The given observations are already in ascending order.
Adding a column corresponding to cumulative frequencies of the given data, we obtain the following table.
 x_{i}f_{i}c.f.15332158276143072135829
Here, N = 29, which is odd.
observation = 15^{th} 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
x_{i} – M 159305f_{i}35678f_{i} x_{i} – M 454518040
∴
Page No 361:
Question 9:
Find the mean deviation about the mean for the data.
Income per day

Number of persons

0100

4

100200

8

200300

9

300400

10

400500

7

500600

5

600700

4

700800

3

Answer:
The following table is formed.
 Income per dayNumber of persons f_{i}Midpoint x_{i}f_{i} x_{i}0 – 1004502003081232100 – 200815012002081664200 – 30092502250108972300 – 400103503500880400 – 5007450315092644500 – 60055502750192960600 – 700465026002921168700 – 80037502250392117650179007896
Here,
Question 10:
Find the mean deviation about the mean for the data
Height in cms

Number of boys

95105

9

105115

13

115125

26

125135

30

135145

12

145155

10

Answer:
The following table is formed.
 Height in cmsNumber of boys f_{i}Midpoint x_{i}f_{i} x_{i}95105910090025.3227.710511513110143015.3198.91151252612031205.3137.81251353013039004.714113514512140168014.7176.414515510150150024.7247
Here,
Question 11:
Find the mean deviation about median for the following data:
Marks

Number of girls

010

6

1020

8

2030

14

3040

16

4050

4

5060

2

Answer:
The following table is formed.
 MarksNumber of girls f_{i}Cumulative frequency (c.f.)Midpoint x_{i}x_{i} â€“ Med.f_{i} x_{i} â€“ Med.01066522.85137.110208141512.85102.820301428252.8539.930401644357.15114.440504484517.1568.650602505527.1554.350517.1
The class interval containing theor 25^{th} 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

1620

5

2125

6

2630

12

3135

14

3640

26

4145

12

4650

16

5155

9

Answer:
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.
 AgeNumber f_{i}Cumulative frequency (c.f.)Midpoint x_{i}x_{i} – Med.f_{i} x_{i} – Med.15.520.555182010020.525.561123159025.530.51223281012030.535.514373357035.540.52663380040.545.512754356045.550.51691481016050.555.591005315135100735
The class interval containing theor 50^{th} 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, f = 26, h = 5, and N = 100
Thus, mean deviation about the median is given by,