Question

A set of exam scores is normally distributed and has a mean of 79.4 and a standard deviation of 11. What is the probability that a randomly selected score will be greater than 54.1? Answer = (round to four decimal places) Note: Be careful...only use the Z Table here and round z scores to two places since that’s what the table uses...do not use technology or the 68-95-99.7 Rule.

Answer 1

There is a 98.93% probability that a randomly selected score will be greater than 54.1.

Explanation:

Problems of normally distributed samples can be 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.

In this problem, we have that:

A set of exam scores is normally distributed and has a mean of 79.4 and a standard deviation of 11. This means that
`\mu = 79.4, \sigma = 11`.

What is the probability that a randomly selected score will be greater than 54.1?

This probability is 1 subtracted by the pvalue of Z when
`X = 54.1`.

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

`Z = (54.1 - 79.4)/(11)`

`Z = -2.3`

`Z = -2.3` has a pvalue of 0.0107.

This means that there is a 1-0.0107 = 0.9893 = 98.93% probability that a randomly selected score will be greater than 54.1.