Question

A small hair salon in Denver, Colorado, averages about 30 customers on weekdays with a standard deviation of 6. It is safe to assume that the underlying distribution is normal. In an attempt to increase the number of weekday customers, the manager offers a $2 discount on 5 consecutive weekdays. She reports that her strategy has worked since the sample mean of customers during this 5 weekday period jumps to 35.

Required:
What is the probability to get a sample average of 35 or more customers if the manager had not offered the discount?

Answer 1

0.0314 = 3.14% probability to get a sample average of 35 or more customers if the manager had not offered the discount

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
`\mu` and standard deviation
`\sigma`, the z-score 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 p-value, 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 normally distributed random variable X, with mean
`\mu` and standard deviation
`\sigma`, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
`\mu` and standard deviation
`s = (\sigma)/(√(n))`.

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

A small hair salon in Denver, Colorado, averages about 30 customers on weekdays with a standard deviation of 6.

This means that
`\mu = 30, \sigma = 6`

5 consecutive weekdays.

This means that
`n = 5, s = (6)/(√(5)) = 2.6832`

What is the probability to get a sample average of 35 or more customers if the manager had not offered the discount?

This is 1 subtracted by the pvalue of Z when X = 35.

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

By the Central Limit Theorem

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

`Z = (35 - 30)/(2.6832)`

`Z = 1.86`

`Z = 1.86` has a pvalue of 0.9686

1 - 0.9686 = 0.0314

0.0314 = 3.14% probability to get a sample average of 35 or more customers if the manager had not offered the discount