The times that customers spend in a book store are normally distributed with a mean of 39.5 minutes and a standard deviation of 15.9 minutes. A random sample of 60 customers has a mean of 36.1 minutes or less. Would this outcome be considered unusual, so that the store should reconsider its displays?

Since |Z| = 1.66 < 2, this outcome should not be considered unusual.

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 and standard deviation , the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

If |Z| > 2, X is considered unusual.

Central Limit Theorem

The Central Limit Theorem estabilishes 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.

In this question, we have that:

A random sample of 60 customers has a mean of 36.1 minutes or less. Would this outcome be considered unusual, so that the store should reconsider its displays?

We have to find Z when X = 36.1.

By the Central Limit Theorem

So |Z| = 1.66

Since |Z| = 1.66 < 2, this outcome should not be considered unusual.

it 2

step-by-step explanation:

8^1/3=2