Question

A factory makes components used in jet engines, including reinforced steel washers. the washers are required to have a very precise thickness. the thickness of the washers follow a normal distribution with mean 1.59 millimeters and standard deviation 0.042 millimeters. a technician randomly samples n = 20 washers and calculates the mean of their thicknesses, which is x¯. what is the probability that x¯<1.57?

Answer 1

Answer:

1.66% probability that x¯<1.57.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$, a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$

Normal probability distribution

Problems of normally distributed samples can be 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:

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

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.

In this problem, we have that:

$\mu = 1.59, \sigma = 0.042, n = 20, s = \frac{0.042}{\sqrt{20}} = 0.0094$

What is the probability that x¯<1.57?

This probability is the pvalue of Z when $X = 1.57$. So:

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

$Z = \frac{1.57-1.59}{0.0094}$

$Z = -2.13$

$Z = -2.13$ has a pvalue of 0.0166.

So there is a 1.66% probability that x¯<1.57.

Answer 2

That probability is about 1.7% according to my calculator.