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

Answer:

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

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 $\mu$ and standard deviation $\sigma$, the zscore 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 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 $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 54, \sigma = 18, n = 100, s = \frac{18}{\sqrt{100}} = 1.8$

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

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

By the Central Limit Theorem

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

$Z = \frac{50 - 54}{1.8}$

$Z = -2.22$

$Z = -2.22$ 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