Question

Suppose a batch of metal shafts produced in a manufacturing company have a standard deviation of 1.9 and a mean diameter of 200 inches. If 78 shafts are sampled at random from the batch, what is the probability that the mean diameter of the sample shafts would differ from the population mean by less than .2 inches?

Answer 1

0.3524

Explanation:

I don't actually have one. I got the problem wrong and saw the correct answer. Hope it works for you.

Answer 2

64.76% probability that the mean diameter of the sample shafts would differ from the population mean by less than .2 inches

Explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

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 = 200, \sigma = 1.9, n = 78, s = (1.9)/(√(78)) = 0.2151`

What is the probability that the mean diameter of the sample shafts would differ from the population mean by less than .2 inches?

This is the pvalue of Z when X = 200 + 0.2 = 200.2 subtracted by the pvalue of Z when X = 200 - 0.2 = 199.8. So

X = 200.2

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

By the Central Limit Theorem

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

`Z = (200.2 - 200)/(0.2151)`

`Z = 0.93`

`Z = 0.93` has a pvalue of 0.8238

X = 199.8

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

`Z = (199.8 - 200)/(0.2151)`

`Z = -0.93`

`Z = -0.93` has a pvalue of 0.1762

0.8238 - 0.1762 = 0.6476

64.76% probability that the mean diameter of the sample shafts would differ from the population mean by less than .2 inches