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.
Answers
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
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…
