Question
Rosetta averages 148 points per bowling game with a standard deviation of 14 points. Suppose Rosetta's points per bowling game are normally distributed. Let $X=$ the number of points per bowling game. Then $X \sim N(148,14)$.
If necessary, round to three decimal places.
Provide your answer below:
Suppose Rosetta scores 110 points in the game on Thursday. The z-score when $x=110$ is $\square$. The mean is 148 .
This $z$-score tells you that $x=110$ is standard deviations to the left of the mean.

Step 1 ：Let's denote the number of points per bowling game as $X$. Given that Rosetta averages 148 points per game with a standard deviation of 14 points, we can say that $X \sim N(148,14)$.

Step 2 ：We are asked to find the z-score when Rosetta scores 110 points in a game. The z-score is a measure of how many standard deviations an element is from the mean.

Step 3 ：The formula for calculating the z-score is: \(z = \frac{X - \mu}{\sigma}\), where $X$ is the element (in this case, the score of 110 points), $\mu$ is the mean (in this case, 148 points), and $\sigma$ is the standard deviation (in this case, 14 points).

Step 4 ：Substituting these values into the formula, we get: \(z = \frac{110 - 148}{14} = -2.7142857142857144\)

Step 5 ：Rounding to three decimal places, we get a z-score of -2.714. This tells us that a score of 110 points is 2.714 standard deviations to the left of the mean.

Step 6 ：Final Answer: The z-score when $x=110$ is \(\boxed{-2.714}\). This $z$-score tells you that $x=110$ is 2.714 standard deviations to the left of the mean.