Question

A restaurant's receipts show that the cost of customers' dinners has a skewed distribution with a mean of $54 and a standard deviation of $18.

Based on this information, what is the probability that a randomly selected 100 customers will spend an average of less than $50 on dinner?

Answer 1

1.32% probability that a randomly selected 100 customers will spend an average of less than $50 on dinner

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 = (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 random variable X, with mean
`\mu` and standard deviation
`\sigma`, a large sample size can be approximated to a normal distribution with mean
`\mu` and standard deviation
`s = (\sigma)/(√(n))`

In this problem, we have that:

`\mu = 54, \sigma = 18, n = 100, s = (18)/(√(100)) = 1.8`

Based on this information, what is the probability that a randomly selected 100 customers will spend an average of less than $50 on dinner?

This probability is the pvalue of Z when X = 50. So

`Z = (X - \mu)/(\sigma)`

By the Central Limit Theorem

`Z = (X - \mu)/(s)`

`Z = (50 - 54)/(1.8)`

`Z = -2.22`

`Z = -2.22` has a pvalue of 0.0132.

1.32% probability that a randomly selected 100 customers will spend an average of less than $50 on dinner