Vendor A supplies parts, each of which has probability 0.03 of being defective. Vendor B also supplies parts, each of which has probability 0.05 of being defective. You receive a shipment of 100 parts from each vendor.

a. Let X be the number of defective parts in the shipment from vendor A and let Y be the number of defective parts in the shipment from vendor B. What are the distributions of X and Y?

b. Generate simulated samples of size 1000 from the distributions of X and Y.

c. Use the samples to estimate the probability that the total number of defective parts is less than 10.

d. Use the samples to estimate the probability that the shipment from vendor A contains more defective parts than the shipment from vendor B.

e. Construct a normal probability plot for the total number of defective parts. Is this quantity approximately normally distributed?

Step 1 of 4:

Vendor A suppliers parts which has probability 0.03 of being defective.

Vendor B suppliers parts which have probability 0.05 of being defective.

There is a shipment of 100 parts from each vendor.

Let X be the number of defective parts in the shipment from the vendor and Y be the number of defective parts in the shipment from vendor B.

Step 2 of 4:

We have to find the distribution of X and Y.

According to the definition of binomial distribution:

It a total of n trials conducted,then

The trials must be independent. Each trials have same success probability P . Each trial have only two possibilities.Since from the given data it is clear that each parts can be either defective or nondefective,With a given probabilities of being defective.

P(X) = 0.03, P(Y)= 0.05

And each trials are independent. We assume that the number of defective as the number of success .

X ~ B(100, 0.03) and Y~B(100,0.05)

That means both X and Y follows a binomial distribution.

(b) We have to generate a random sample of 100 from the distribution of X and Y.

From the given data it is clear that the variables X and Y follows binomial distribution.

By using software like minitab we can generate the random sample for any distribution as:

Calc- random data- distribution - parameters (with sample size)- ok( store the data in column 1 and column 2).

The generated sample can be different.