Question

Suppose that 1% of all computers of a certain type experience CPU failure during the warranty period. If a simple random sample of 500 such computers is taken,
(a) What is the expected value and the standard deviation of the number of computers in sample that experienced CPU failure during the warranty period?
(b) What is approximately the probability that none of the computers in the sample experienced CPU failure during the warranty period? (Remark: Can you compute the exact value of this probability?)
(c) What is approximately the probability that more than 4 computers in the above sample experience CPU failure during the warranty period?

Answer 1

Answer:

A)

μ = 5

σ = 2.2

B)

P(0) = 0.0066

C)

P(x ≥ 4) = 0.74

Step-by-step explanation:

A) The expected value is going to be the same as the mean μ.

We can use the binomial probability distribution because there are a fixed number of trials (500), each trial is independent and there are only two possible outcomes (CPU failure or not) and the probability of failure has the fixed value 0.01.

For binomial distributions:

μ = np, where p is the probability (in this case of CPU failure) and n is the number of trials (500)

μ = (500)(0.01) = 5

σ = $\sqrt{npq}$, where q = 1 - p

σ = $\sqrt{(500)(0.01)(0.99)}$ = 2.2

B) We can use the binomial probability formula

$P(x)=(\frac{n!}{(n-x)!x!})$ × pˣ × qⁿ⁻ˣ

$P(0)=(\frac{500!}{(500-0)!0!})$ × 0.01⁰ × 0.99⁵⁰⁰

C) In this case wee need to use a binomial distribution calculator. Setting n as 500, π = 0.01 and finding the probability above 3.