A company makes windows for use in homes and commercial buildings. The standards for glass thickness call for the glass to average 0.325 inch with a standard deviation equal to 0.065 inch. Suppose a random sample of n = 44 windows yields a sample mean of 0.337 inch. Required:
What is the probability of x >= 20.337 if the windows meet the standards?

0.1102 = 11.02% probability of x >= 0.337 if the windows meet the standards.

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 and standard deviation , the z-score of a measure X is given by:

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 and standard deviation , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean and standard deviation .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

The standards for glass thickness call for the glass to average 0.325 inch with a standard deviation equal to 0.065 inch.

This means

Sample of 44

This means that

What is the probability of x >= 0.337 if the windows meet the standards?

This is 1 subtracted by the p-value of Z when X = 0.337. So

By the Central Limit Theorem

has a p-value of 0.8898

1 - 0.8898 = 0.1102

0.1102 = 11.02% probability of x >= 0.337 if the windows meet the standards.

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