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Volgvan
3 years ago
8

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?
Mathematics
1 answer:
Hitman42 [59]3 years ago
5 0

Answer:

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

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 \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 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}}.

In this problem, we have that:

\mu = 50, \sigma = 5, n = 25, s = \frac{5}{\sqrt{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 = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

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

Z = \frac{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

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