Question

A ski resort gets an average of 2,000 customers per weekday with a standard deviation of 800 customers. Assume the underlying distribution is normal. Use an appropriate normal transformation to calculate the probability a ski resort averages more than 3,000 customers per weekday over the course of four weekdays.

Answer 1

0.62% probability that a ski resort averages more than 3,000 customers per weekday over the course of four weekdays

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 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.

In this problem, we have that:

`\mu = 2000, \sigma = 800, n = 4, s = (800)/(√(4)) = 400`

Trobability a ski resort averages more than 3,000 customers per weekday over the course of four weekdays.

This is 1 subtracted by the pvalue of Z when X = 3000. So

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

By the Central Limit Theorem

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

`Z = (3000 - 2000)/(400)`

`Z = 2.5`

`Z = 2.5` has a pvalue of 0.9938

1 - 0.9938 = 0.0062

0.62% probability that a ski resort averages more than 3,000 customers per weekday over the course of four weekdays