Question

An important part of the customer service responsibilities of a cable company relates to the speed with which trouble in service can be repaired. Historically, the data show that the likelihood is 0.75 that troubles in a residential service can be repaired on the same day. For the first five troubles reported on a given day, what is the probability that at least 3 troubles will be repaired on the same day?

Answer 1

BIONOMIAL DISTRIBUTION

pmf of B.D is = f ( k ) = ( n k ) p^k * ( 1- p) ^ n-k

where    

k = number of successes in trials      

n = is the number of independent trials      

p = probability of success on each trial

I.

mean = np

where

n = total number of repetitions experiment is excueted

p = success probability

mean = 5 * 0.75

= 3.75

II.

variance = npq

where

n = total number of repetitions experiment is excueted

p = success probability

q = failure probability

variance = 5 * 0.75 * 0.25

= 0.9375

III.

standard deviation = sqrt( variance ) = sqrt(0.9375)

=0.968246

P( X < 3)  = P(X=2) + P(X=1) + P(X=0)     

=  ( 5  2 ) * 0.75^2 * ( 1- 0.75 ) ^3 + ( 5  1 ) * 0.75^1 * ( 1- 0.75 ) ^4 + ( 5  0 ) * 0.75^0 * ( 1- 0.75 ) ^5    

= 0.103516

P( X > = 3 ) = 1 - P( X < 3) = 0.896484