Question

The given distribution shows the number of runs scored by some top batsmen of the world in one- day international cricket matches.
Find the mode of the data.

Answer 1

Answer:

Given data:

Modal class = 4000 – 5000,

l = 4000,

class width (h) = 1000,

fm = 18, f1 = 4 and f2 = 9

Mode Formula:

Mode = l+ [(fm-f1)/(2fm-f1-f2)]×h

Substitute the values

Mode = 4000+((18-4)/(36-4-9))×1000

Mode = 4000+(14000/23) = 4000+608.695

Mode = 4608.695

Mode = 4608.7 (approximately)

Thus, the mode of the given data is 4608.7 runs

Answer 2

SOLUTION :  

FREQUENCY DISTRIBUTION TABLE is in the attachment

For MODE :

Here the maximum frequency is 18, and the class corresponding to this frequency is 4000 – 5000. So the modal class is 4000 – 5000.

Therefore, l = 4000, h = 100,  f1= 18,  f0= 4 , f2 = 9

Mode = l + [(f1- f0) / (2f1- f0 - f2)] ×h

= 4000 + [(18 - 4)/(2 × 18 - 4 – 9) ] ×1000

= 4000 + [14 × 1000)/(36 - 13)]

= 4000 + [14000/ 23]

= 4000 + 608.7

= 4608.7

MODE = 4608.7

Hence, the mode of the data is 4608.7.

★★ Mode = l + (f1-f0/2f1-f0-f2) ×h

l = lower limit of the modal class

h = size of the class intervals

f1 = frequency of the modal class

f0 = frequency of the class preceding the modal class

f2 = frequency of the class succeed in the modal class.

HOPE THIS ANSWER WILL HELP YOU…