Question

Thompson and Thompson is a steel bolts manufacturing company. Their current steel bolts have a mean diameter of 143 millimeters, and a variance of 64. If a random sample of 48 steel bolts is selected, what is the probability that the sample mean would differ from the population mean by greater than 0.2 millimeters? Round your answer to four decimal places.

Answer 1

Answer:

86.50% probability that the sample mean would differ from the population mean by greater than 0.2 millimeters

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

In a set with mean $\mu$ and standard deviation(which is the square root of the variance) $\sigma$, the zscore 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 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 random variable X, with mean $\mu$ and standard deviation $\sigma$, the sample means with size n of at least 30 can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 143, \sigma = sqrt{64} = 8, n = 48, s = \frac{8}{\sqrt{48}} = 1.15$

If a random sample of 48 steel bolts is selected, what is the probability that the sample mean would differ from the population mean by greater than 0.2 millimeters?

Either it differs by 0.2 millimeters or less, or it differs by more than 0.2 millimeters. The sum of the probabilities of these events is decimal 1.

Probability it differs by 0.2 millimeters or less:

pvalue of Z when X = 143+0.2 = 143.2 subtracted by the pvalue of Z when X = 143-0.2 = 142.8. So

X = 143.2

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

By the Central Limit Theorem

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

$Z = \frac{143.2 - 143}{1.15}$

$Z = 0.17$

$Z = 0.17$ has a pvalue of 0.5675

X = 142.8

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

$Z = \frac{142.8 - 143}{1.15}$

$Z = -0.17$

$Z = -0.17$ has a pvalue of 0.4325

0.5675 - 0.4325 = 0.1350

0.1350 probability it differs by 0.2 millimeters or less

Probability if differs by more than 0.2 millimeters:

p + 0.1350 = 1

p = 0.8650

86.50% probability that the sample mean would differ from the population mean by greater than 0.2 millimeters