a basketball - average scoring score is 174, and it is normally distributed with a standard deviation of 13…

Explanation:

Step1: Calculate the z - score

The formula for the z - score is $z=\frac{x-\mu}{\sigma}$, where $x = 200$, $\mu=174$, and $\sigma = 13$. So $z=\frac{200 - 174}{13}=\frac{26}{13}=2$.

Step2: Find the proportion of values above the z - score

We know that the total area under the normal - distribution curve is 1. Looking up the z - score of 2 in the standard normal distribution table, the area to the left of $z = 2$ is approximately 0.9772. The area to the right of $z = 2$ (the proportion of values greater than 200) is $1-0.9772 = 0.0228\approx2.3%$.

Answer:

2.3% (closest to 2.28%, so if we assume the options are approximations, we need to know the options clearly, but based on the calculation the proportion of games with a score more than 200 is about 2.3%)