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

Answer:

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

Step-by-step 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:

$Z = \frac{X - \mu}{\sigma}$

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.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{190 - 197}{12}$

$Z = -0.58$

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

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