Question

A process is producing a particular part where the thickness of the part is following a normal distribution with a µ = 50 mm and σ = 5 mm. If a random sample of 25 parts were taken, what is the probability that this selected sample has an average thickness greater than 53?

Answer 1

0.13% probability that this selected sample has an average thickness greater than 53

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))`.

In this problem, we have that:

`\mu = 50, \sigma = 5, n = 25, s = (5)/(√(25)) = 1`

What is the probability that this selected sample has an average thickness greater than 53?

This is 1 subtracted by the pvalue of Z when X = 53. So

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

By the Central Limit Theorem

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

`Z = (53 - 50)/(1)`

`Z = 3`

`Z = 3` has a pvalue of 0.9987

1 - 0.9987 = 0.0013

0.13% probability that this selected sample has an average thickness greater than 53