Question

A purchaser of electric relays buys from two suppliers, A and B. Supplier A supplies two of every three relays used by the company. If 60 relays are selected at random from those in use by the company, find the probability that at most 38 of these relays come from supplier A. Assume that the company uses a large number of relays. (Use the normal approximation. Round your answer to four decimal places.)

Answer 1

Answer:

The probability that at most 38 of the 60 relays are from supplier A is P(X≤38)=0.3409.

Step-by-step explanation:

We can model this question with a binomial distribution random variable.

The sample size is n=60.

The probability that the relay come from supplier A is p=2/3 for any relay.

If we use a normal aproximation, we have the mean and standard deviation:

$\mu=np=60*(2/3)=40\\\\\sigma=\sqrt{npq}=\sqrt{60*(2/3)*(1/3)} =3.65$

The probability that at most 38 of the 60 relays are from supplier A is P(X≤38)=0.3409:

$P(X\leq38)=P(X<38.5)\\\\\\z=\frac{38.5-40}{3.65}= \frac{-1.5}{3.65} =-0.41\\\\\\P(X<38.5)=P(z<-0.41)=0.3409$