Question

a manufacturer of pencils has 60 dozen pencils randomly chosen from each day's productionand checked for defects. a defect rate of 10% is considered acceptable. assuming that 10%of all the pencils are actually defective, what is the probability of finding 80 or more defectivepencils in the sample? if a normal approximation is justifiable, use it to find this probability.

Answer 1

The question involves finding the probability of 80 or more defective pencils in a sample of 720 pencils with a 10% defect rate using the binomial distribution and its normal approximation. The normal approximation involves calculating the mean and standard deviation and then determining the area to the right of the calculated z-score.

Step-by-step explanation:

The student asked what the probability is of finding 80 or more defective pencils in a sample of 60 dozen (720) pencils if the defect rate is 10%. This is a question of applying the binomial distribution, and possibly its normal approximation, to find the probability of a certain number of successes (defects) given a certain number of trials (pencils tested).

To solve the problem using the normal approximation to the binomial distribution, we calculate the mean (μ) and standard deviation (σ) of the binomial distribution. For a binomial distribution with n trials and a probability of success p, μ = np and σ = √(np(1-p)). In this case, n = 720 and p = 0.10, so μ = 72 and σ = √(64.8) or approximately 8.05. We then use the standard normal distribution to find the probability that X, the number of defective pencils, is greater than or equal to 80.

To find P(X ≥ 80), we compute the z-score for X = 80: z = (X - μ) / σ and use the standard normal distribution table (or a calculator) to find the corresponding probability. Please note that because we're interested in the probability of finding 80 or more defects, we need to find the area to the right of our z-score.