Question

A bowling leagues mean score is 197 with a standard deviation of 12. The scores are normally distributed. What is the probability a given player averaged less than 190?m

Answer 1

0.281 = 28.1% probability a given player averaged less than 190.

Explanation:

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.

A bowling leagues mean score is 197 with a standard deviation of 12.

This means that
`\mu = 197, \sigma = 12`

What is the probability a given player averaged less than 190?

This is the p-value of Z when X = 190.

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

`Z = (190 - 197)/(12)`

`Z = -0.58`

`Z = -0.58` has a p-value of 0.281.

0.281 = 28.1% probability a given player averaged less than 190.