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```Chapter 32: Statistics – Exercise 32.3

Statistics – Exercise – 32.3 – Q.1

We have to calculate mean deviation from the median. So, first we calculate the median.

CI
x
F
cf
d = (x – med)
fd

0 – 10
5
5
5
20
100

10 – 20
15
10
15
10
100

20 – 30
25
20
35
0
0

30 – 40
35
5
91
10
50

40 – 50
45
10
101
20
200

50

450

Statistics – Exercise – 32.3 – Q.2(i)

CI
x
f
xf
d = (x-mean)
fd

0 – 100
50
4
200
308
1232

100 – 200
150
8
1200
208
1664

200 – 300
250
9
2250
108
972

300 – 400
350
10
3500
8
80

400 – 500
450
7
3150
92
644

500 – 600
550
5
2750
192
960

600 – 700
650
4
2600
292
1168

700 – 800
750
3
2250
392
1176

50
17900

7896

Statistics – Exercise – 32.3 – Q.2(ii)

Classes
fi
xi
di
fidi

95 – 105
9
100
-3
-27
28.58
257.22

105 – 115
13
110
-2
-26
18.58
241.54

115 – 125
16
120
-1
-16
8.58
137.28

125 – 135
26
130
0
0
1.42
36.92

135 – 145
30
140
1
30
11.42
342.6

145 – 155
12
150
2
24
21.42
257.04

N = 106

Total = -15

Total = 1272.60

Statistics – Exercise – 32.3 – Q.2(iii)

CI
x
f
xf
d = (x - mean)
fd

0 – 10
5
6
30
22
132

10 – 20
15
8
120
12
96

20 – 30
25
14
350
2
28

30 – 40
35
16
560
8
128

40 – 50
45
4
180
18
72

50 – 60
55
2
110
28
56

50
1350

512

Mean
27

Deviation
10.24

Statistics – Exercise – 32.3 – Q.3

Find the mean deviation from the mean for the data:

Classes
0 -10
10 – 20
20 – 30
30 – 40
40 – 50
50 – 60

Frequencies
6
8
14
16
4
2

Statistics – Exercise – 32.3 – Q.4

We have to calculate mean deviation from the median. So, first we calculate the median.

CI
x
f
cf
D = (x – med)
fd

17 – 19.5
18.25
5
5
20
100

20 – 25.5
22.75
16
21
15.5
248

26 – 35.5
30.75
12
33
7.5
90

36 – 40.5
38.25
26
59
0
0

41 – 50.5
45.75
14
73
7.5
105

51 – 55.5
53.25
12
85
15
180

56 – 60.5
58.25
6
91
20
120

61 – 70.5
65.75
5
96
27.5
137.5

96

980.5

We have N = 96 ⟹ N/2 = 48

The cumulative frequency just greater than N/2 is 59 and the corresponding value of x is 38.25.

Hence, median = 38.25

Statistics – Exercise – 32.3 – Q.5

M.D from Median

Marks
Students
xi
Cum. Freq

fidi

0 – 10
5
5
5
55/3
275/3

10 – 20
8
15
13
25/3
200/3

20 – 30
15
25
28
3-May
75/3

30 – 40
16
35
44
35/3
560/3

40 – 50
6
45
50
65/3
390/3

N = 50

Total = 500

M.D from mean

Marks
Students
xi

fidi
|xi - 27|
fi|xi - 27|

0 – 10
5
5
-3
-15
22
110

10 – 20
8
15
-2
-16
12
96

20 – 30
15
25
-1
-15
2
30

30 – 40
16
35
0
0
8
128

40 – 50
6
45
1
6
18
108

N = 50

Total = 40

Total = 472

Statistics – Exercise – 32.3 – Q.6

Converting the given data into continuous frequency distribution by sub tracing 0.5 from the lower limit and adding 0.5 to the upper limit of each class interval.

Age
x:
f:
Cumulative frequency
|d:|=|x:-38|
f:|d:|

15.5 – 20.5
18
5
5
20
100

20.5 – 25.5
23
6
11
15
90

25.5 – 30.5
28
12
23
10
120

30.5 – 35.5
33
14
37
5
70

35.5 – 40.5
38
26
63
0
0

40.5 – 45.5
43
12
75
5
60

45.5 – 50.5
48
16
91
10
160

50.5 – 55.5
53
9
100
15
135

N = Σf: = 100

Σf:|d:| = 735

Clearly, N = 100 ⟹ N/2 = 50.

Cumulative frequency is just greater than N/2 is 63 and the corresponding class is 35.5 - 40.5.

I = 35.5, f = 26, h = 5, F = 37

Statistics – Exercise – 32.3 – Q.7

Classes
fi
xi
fixi
|xi - 9.2|
fi|xi - 9.2|

0 – 4
4
2
8
7.2
28.8

4 – 8
6
6
35
3.2
19.2

8 – 12
8
10
80
0.8
6.4

12 – 16
5
14
70
4.8
24

16 – 20
2
18
36
8.8
17.6

N = 25

Total = 230

Total = 96.0

Statistics – Exercise – 32.3 – Q.8

Classes
fi
xi
fixi
|xi - 14.1|
fi|xi - 14.1|

0 – 6
4
3
12
11.1
44.4

6 – 12
5
9
45
5.1
25.5

12 – 18
3
15
45
0.9
2.7

18 – 24
6
21
126
6.9
41.4

24 – 30
2
27
54
12.9
25.8

N = 20

Total = 282

Total = 139.8

```

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