Question

The following data gives the information on the (in hour) of 225 electrical compounds​

Answer 1

Answer:

Step-by-step explanation:

SOLUTION :  

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.

FREQUENCY DISTRIBUTION TABLE is in the attachment  

Here the maximum frequency is 61, and the class corresponding to this frequency is 60 – 80. So the modal class is 60 - 80.

Therefore,l = 60 , h = 80 – 60 = 20,  f1 = 61, , f0 = 52, f2 = 38

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

= 60 + [(61 - 52)/(2 × 61 - 52 – 38) ]×20

= 60 +[(9 × 20)/(122 - 90)]

= 60 + [180/32]

= 60 + 5.625

= 65.625 hours

Hence, the modal lifetimes of the components is 65.625 hours  

HOPE THIS ANSWER WILL HELP YOU…