Mathematics, 25.06.2021 04:00 jbrooks4091

The fracture strength of tempered glass averages 14 (measured in thousands of pounds per square inch) and has standard deviation 2. (a) What is the probability that the average fracture strength of 100 randomly selected pieces of this glass exceeds 14.2

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0.1587 = 15.87% probability that the average fracture strength of 100 randomly selected pieces of this glass exceeds 14.2.

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.

The fracture strength of tempered glass averages 14 (measured in thousands of pounds per square inch) and has standard deviation 2.

This means that $\mu = 14, \sigma = 2$

Sample of 100:

This means that $n = 100, s = \frac{2}{\sqrt{100}} = 0.2$

What is the probability that the average fracture strength of 100 randomly selected pieces of this glass exceeds 14.2?

This is 1 subtracted by the p-value of Z when X = 14.2. So

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

By the Central Limit Theorem

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

$Z = \frac{14.2 - 14}{0.2}$

Z = 1

Z = 1 has a p-value of 0.8413.

1 - 0.8413 = 0.1587

0.1587 = 15.87% probability that the average fracture strength of 100 randomly selected pieces of this glass exceeds 14.2.

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