Question

Suppose that X is the time it takes a randomly chosen professor in a university department to type and send a standard letter of recommendation for a past student who is applying for a graduate degree program. Suppose X has a normal distribution, and assume the mean is 10.5 minutes and the standard deviation 3 minutes. If you were to take a random sample of 50 professors and measure the time it takes each one of them to type and send such a standard letter of recommendation, what is the probability that their mean time is less than 9.5 minutes?

Answer 1

0.0091 = 0.91% probability that their mean time is less than 9.5 minutes.

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.

The mean is 10.5 minutes and the standard deviation 3 minutes.

This means that
`\mu = 10.5, \sigma = 3`

Sample of 50:

This means that
`n = 50, s = (3)/(√(50))`

What is the probability that their mean time is less than 9.5 minutes?

This is the pvalue of Z when X = 9.5. So

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

By the Central Limit Theorem

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

`Z = (9.5 - 10.5)/((3)/(√(50)))`

`Z = -2.36`

`Z = -2.36` has a pvalue of 0.0091

0.0091 = 0.91% probability that their mean time is less than 9.5 minutes.