Question

A company is interested in determining the average cost of lunch in its cafeteria, so it asks a sample of 121 randomly chosen employees what they spent. If the true average amount spent on lunch is $7.65 with a standard deviation of $2.15, what is the probability that the sample average will be less than $8?

Answer 1

96.33% probability that the sample average will be less than $8.

Explanation:

To solve this question, we have 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 = 7.65, \sigma = 2.15, n = 121, s = (2.15)/(√(121)) = 0.1955`

What is the probability that the sample average will be less than $8?

This is the pvalue of Z when X = 8. So

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

By the Central Limit Theorem

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

`Z = (8 - 7.65)/(0.1955)`

`Z = 1.79`

`Z = 1.79` has a pvalue of 0.9633.

96.33% probability that the sample average will be less than $8.