Question

When Victoria goes bowling, her scores are normally distributed with a mean of 130 and a standard deviation of 11. What is the probability that the next game Victoria bowls, her score will be between 123 and 130, to the nearest thousandth?

Answer 1

Answer:

0.239 = 23.9% probability that the next game Victoria bowls, her score will be between 123 and 130

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.

Mean of 130 and a standard deviation of 11.

This means that $\mu = 130, \sigma = 11$

What is the probability that the next game Victoria bowls, her score will be between 123 and 130?

This is the p-value of Z when X = 130 subtracted by the p-value of Z when X = 123. So

X = 130

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

$Z = \frac{130 - 130}{11}$

$Z = 0$

$Z = 0$ has a p-value of 0.5

X = 123

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

$Z = \frac{123 - 130}{11}$

$Z = -0.64$

$Z = -0.64$ has a p-value of 0.2611

0.5 - 0.261 = 0.239

0.239 = 23.9% probability that the next game Victoria bowls, her score will be between 123 and 130