MATHEMATICS HIGH SCHOOL

A manufacturing process is designed to produce bolts with a 0.75-inch diameter. Once each day, a random sample of 36 bolts is selected and the bolt diameters recorded. If the resulting sample mean is less than 0.730 inches or greater than 0.770 inches, the process is shut down for adjustment. The standard deviation for diameter is 0.04 inches. What is the probability that the manufacturing line will be shut down unnecessarily? (Hint: Find the probability of observing an x in the shutdown range when the true process mean really is 0.75 inches. Round your answer to four decimal places.)

Answers

Answer 1
Answer:

Answer:

0.0026 = 0.26% probability that the manufacturing line will be shut down unnecessarily

Step-by-step explanation:

We need to understand the normal probability distribution and the central limit theorem to solve this question.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = (X - \mu)/(\sigma)

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = (\sigma)/(√(n)).

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

\mu = 0.75, \sigma = 0.04, n = 36, s = (0.04)/(√(36)) = 0.0067

What is the probability that the manufacturing line will be shut down unnecessarily?

Less than 0.73 or more than 0.77.

Less than 0.73

pvalue of Z when X = 0.73

Z = (X - \mu)/(\sigma)

By the Central Limit Theorem

Z = (X - \mu)/(s)

Z = (0.73 - 0.75)/(0.0067)

Z = -3

Z = -3 has a pvalue of 0.0013

More than 0.77

Z = (X - \mu)/(s)

Z = (0.77 - 0.75)/(0.0067)

Z = 3

Z = 3 has a pvalue of 0.9987

1 - 0.9987 = 0.0013

2*0.0013 = 0.0026

0.0026 = 0.26% probability that the manufacturing line will be shut down unnecessarily

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