Question

The time to complete an exam in a statistics class is a normal random variable with a mean of 50 minutes and a standard deviation of 10 minutes. What is the probability, given a class size of 30 students, the average time to complete the test is less than 48.5 minutes

Answer 1

0.2061 = 20.61% probability, given a class size of 30 students, the average time to complete the test is less than 48.5 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 establishes 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.

Mean of 50 minutes and a standard deviation of 10 minutes.

This means that
`\mu = 50, \sigma = 10`

Class size of 30 students

This means that
`n = 30, s = (10)/(√(30))`

What is the probability, given a class size of 30 students, the average time to complete the test is less than 48.5 minutes.

This is the p-value of Z when X = 48.5. So

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

By the Central Limit Theorem

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

`Z = (48.5 - 50)/((10)/(√(30)))`

`Z = -0.82`

`Z = -0.82` has a p-value of 0.2061

0.2061 = 20.61% probability, given a class size of 30 students, the average time to complete the test is less than 48.5 minutes