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NCERT Solutions for Class 11 Maths Chapter 15 – Statistics Ex 15.1

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

Answer:

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

Answer:

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

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.
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

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.
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
∴ 

Page No 361:

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

Answer:

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

Answer:

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

Answer:

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 theor 25th item is 20 – 30.
Therefore, 20 – 30 is the median class.
It is known that,
Here, l = 20, C = 14, f = 14, h = 10, and N = 50
∴ 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

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.
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 theor 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, f = 26, h = 5, and N = 100
 
Thus, mean deviation about the median is given by,

Courtesy : CBSE
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