Question

Give answer of this question ​

Answer 1

A. Solution:

Mean = A + (∑fd/n) • h

= 30 + (24/73) • 10

= 30 + 0.3288 • 10

= 30 + 3.2877

= 33.2877

B. Mode

The maximum frequency is 20.

The mode class is 40-50.

L = lower boundary point of mode class = 40

f1 = frequency of the mode class = 20

f0 = frequency of the preceding class = 15

f2 = frequency of the succedding class = 11

c = class length of mode class = 10

Z = L + (f1 - f0)/(2 • f1 - f0 - f2) • c

= 40 + (20-15)/(2 • 20 - 15 - 11)• 10

= 40 + (5/14) 10

= 40 + 3.5714

= 43.5714

C. Median

Median Class = value of (n/2)th observation

= value of (100/2)th observation

= value of 50th observation

From the column of cumulative frequency cf, we find that the 50th observation lies in the class 45-50.

The median class is 45-50.

Now

L = lower boundary point of median class = 45

n = Total frequency = 100

cf = Cumulative frequency of the class preceding the median class = 48

f = Frequency of the median class = 23

c = class length of median class = 5

Median M = L + [(n/2) - cf]/f • c

= 45 + (50 - 48)/23 • 5

= 45 + (2/23) • 5

= 45 + 0.4348

= 45.4348

Hope this would help.