Guest

Chapter 32: Statistics – Exercise 32.2

Statistics – Exercise – 32.2 – Q.1

xi fi Cum.Freq |di| = |xi - 61| fi|di|
58 15 15 3 45
59 20 35 2 40
60 32 67 1 32
61 35 102 0 0
62 35 137 1 35
63 22 159 2 44
64 20 179 3 60
65 10 189 4 40
66 8 197 5 40
  N = 197     Total = 336

Corresponding value for median is 61

 

Statistics – Exercise – 32.2 – Q.2

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

x f cf d = (x – med) fd
0 14 14 4 56
1 21 35 3 63
2 25 60 2 50
3 43 103 1 43
4 51 154 0 0
5 40 194 1 40
6 39 233 2 78
7 12 245 3 36
245 366

we have N = 245 ⟹ N/2 = 122.5

The cumulative frequency just greater than N/2 is 154 and the corresponding value of x is 4.

Hence, Median = 4

 

Statistics – Exercise – 32.2 – Q.3

xi fi Cum fre |di| = |xi - 13| fi|di|
5 2 2 8 16
7 4 6 6 24
9 6 12 4 24
11 8 20 2 16
13 10 30 0 0
15 12 42 2 24
17 8 50 4 32
  N = 50     Total = 136

Value corresponding to 25 is median = 13

 

Statistics – Exercise – 32.2 – Q.4(i)

xi fi fixi |di| = |xi - 9| fi|di|
5 8 40 4 32
7 6 42 2 12
9 2 18 0 0
10 2 20 1 2
12 2 24 3 6
15 6 90 6 36
  26 Total = 234   Total = 88

Mean = 9

 

Statistics – Exercise – 32.2 – Q.4(ii)

x f xf d = (x - mean) fd
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

 

Statistics – Exercise – 32.2 – Q.4(iii)

x f xf d = (x – mean) fd
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 180

 

Statistics – Exercise – 32.2 – Q.4(iv)

xi fi fixi |di| = |xi - 21.65| fi|di|
20 6 120 1.65 9.9
21 4 84 0.65 2.6
22 5 110 0.35 1.75
23 1 23 1.35 1.35
24 4 96 2.35 9.4
  20 Total = 433   Total = 25

Mean = 21.65

 

Statistics – Exercise – 32.2 – Q.5

xi fi Cum. Freq |di| = |xi - 30| fi|di|
15 3 3 15 45
21 5 8 9 45
27 6 14 3 18
30 7 21 0 0
35 8 29 5 40
  29     Total = 148

Median = 30

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

x f cf d = (x-med) fd
35 4 4 39 456
42 2 6 32 64
54 4 10 20 80
74 20 30 0 0
89 12 42 15 180
91 5 47 17 85
94 3 50 20 60
50 625

We have N = 50 ⟹ N/2 = 25

The cumulative frequency just greater than N/2 is 30 and the corresponding value of x is 74.

Hence, median = 74

xi fi Cum. Freq |di| = |xi - 12| fi|di|
10 2 2 2 4
11 3 5 1 3
12 8 13 0 0
14 3 16 2 6
15 4 20 3 12
  20     Total = 25

Median = 12


TOP Your EXAMS!

Upto 50% Scholarship on Live Classes

Course Features

  • Video Lectures
  • Revision Notes
  • Previous Year Papers
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Test paper with Video Solution

r