In a factory it is assumed that the probability any individu

In a factory, it is assumed that the probability any individual item will be defective is 0.03. Each day, we select 50 items, and count the number of defective items in the group of 50. Suppose we examine the results of a random sample of 75 days. What is the probability that the average number of defective items per day over the 75 days is 1 or less?


We are using the program R to calculate. And I believe we use the central limit theorem.

Solution

Here no of defective is binomial with mean = np = 50(0.03) = 1.5

For 75 days, mean follows a normal with mean = 15 and std error = rt of npq//75

= 0.44054

P(def <=1) = P(Z<=-0.5/0.44054)

=P(Z<-1.1352)

= 0.5-0.4192

= 0.0808

In a factory, it is assumed that the probability any individual item will be defective is 0.03. Each day, we select 50 items, and count the number of defective

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