Amanufacturer produces a large number of microwaves. from past experience, the manufacturer knows that approximately 1.5% of all their microwaves are defective. consumer reports randomly selects 20 of these microwaves for testing. we want to determine the probability that no more than one of these microwaves is defective?

a) what is the probability that exactly one of the microwaves is defective?

b) what is the probability that at most two of the microwaves are defective?

c) find the mean and standard deviation.

This problem represents "binomial probability," because any given "experiment" can have only two possible outcomes:  defective or not defective.

Here the "population" probability that a given microwave unit is defective is 0.015.  We arbitrarily define "defective microwave" as a "success" and "non-defective microwave" as a "failure."

We can obtain binomial probability values from a calculator such as the TI-83, from a table or from the binomial probabilty formula.  In all of these cases the number of samples is given and is n=20; the probability of "success" is also given and is p=0.015.

Case 1:  What is the probability that exactly 1 microwave unit is defective?
Using the TI-83:     binompdf(20,0.015,1) = 0.225, or 9/40.  

Case 2:  What is the p. that at most 2 are defective?  Add together the binomial probabilities binompdf(20,0.015,0),binompdf(20,0.015,1),binompdf(20,0.015,2).

Result:  0.7391 + 0.2250 + 0.0326 = 0.9967.

Mean:  The mean of a binomial probability such as this one is simply np, which in his case is 20(0.015)=0.30.  Find and apply the formula for the standard deviation.