Question

Suppose a batch of metal shafts produced in a manufacturing company have a standard deviation of 2 and a mean diameter of 200 inches.

If 83 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 0.2 inches? Round your answer to four decimal places.

Answer 1

Answer:

0.6372 = 63.72% probability that the mean diameter of the sample shafts would differ from the population mean by less than 0.2 inches.

Step-by-step explanation:

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

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

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

$Z = \frac{X - \mu}{\sigma}$

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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes 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 $s = \frac{\sigma}{\sqrt{n}}$.

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

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

Standard deviation of 2 and a mean diameter of 200 inches.

This means that $\sigma = 2, \mu = 200$

83 shafts

This means that $n = 83, s = \frac{2}{\sqrt{83}}$

What is the probability that the mean diameter of the sample shafts would differ from the population mean by less than 0.2 inches?

Mean between 200 - 0.2 = 199.8 inches and 200 + 0.2 = 200.2 inches, which is the p-value of Z when X = 200.2 subtracted by the p-value of Z when X = 199.8.

X = 200.2

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{200.2 - 200}{\frac{2}{\sqrt{83}}}$

$Z = 0.91$

$Z = 0.91$ has a p-value of 0.8186

X = 199.8

$Z = \frac{X - \mu}{s}$

$Z = \frac{199.8 - 200}{\frac{2}{\sqrt{83}}}$

$Z = -0.91$

$Z = -0.91$ has a p-value of 0.1814

0.8186 - 0.1814 = 0.6372

0.6372 = 63.72% probability that the mean diameter of the sample shafts would differ from the population mean by less than 0.2 inches.