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

Answers

Answer 1

Using the normal distribution and the central limit theorem, it is found that there is a 0.0028 = 0.28% probability that the manufacturing line will be shut down unnecessarily.

In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

It 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, which is the percentile of X.By the Central Limit Theorem, the sampling distribution of sample means of size n have standard deviation given by [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

In this problem:

Mean of 0.75 inches, thus [tex]\mu = 0.75[/tex].Standard deviation of 0.04 inches, thus [tex]\sigma = 0.04[/tex]Sample of 36, thus [tex]n = 36, s = \frac{0.04}{\sqrt{36}} = 0.0067[/tex].

The probability of being shutdown is the probability of a sample mean less than 0.73 inches or more than 0.77 inches. They are the same distance to the mean(0.02), thus, considering the Central Limit Theorem:

[tex]z = \frac{0.02}{s}[/tex]

[tex]z = \frac{0.02}{0.0067}[/tex]

[tex]z = 2.99[/tex]

The probability is P(|z| > 2.99), which is 2 multiplied by the p-value of z = -2.99.

z = -2.99 has a p-value of 0.0014.

2(0.0014) = 0.0028

0.0028 = 0.28% probability that the manufacturing line will be shut down unnecessarily.

A similar problem is given at https://brainly.com/question/25212662


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