A manufacturer knows that their items have a normally distributed length, with a mean of 18.2 inches, and standard deviation of 3.9 inches. If 2 items are chosen at random, what is the probability that their mean length is less than 21.9 inches

0.9099 = 90.99% probability that their mean length is less than 21.9 inches.

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.

Mean of 18.2 inches, and standard deviation of 3.9 inches.

This means that

2 itens:

This means that

What is the probability that their mean length is less than 21.9 inches?

This is the p-value of Z when X = 21.9. So

has a p-value of 0.9099.

0.9099 = 90.99% probability that their mean length is less than 21.9 inches.

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step-by-step explanation:

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