Question

question 8 of 10 > in baseball, a player’s batting average is the proportion of times the player gets a hit out of the total number of times at bat. the distribution of batting averages in a recent season for major league baseball players with at least 100 plate appearances can be modeled by a normal distribution with mean μ = 0.261 and standard deviation σ = 0.034. what proportion of players have a batting average of 0.300 or higher? round your answer to 4 decimal places.

Explanation:

Step1: Calculate the z - score

The z - score formula is $z=\frac{x-\mu}{\sigma}$. Here, $x = 0.300$, $\mu=0.261$, and $\sigma = 0.034$. So, $z=\frac{0.300 - 0.261}{0.034}=\frac{0.039}{0.034}\approx1.15$.

Step2: Find the proportion

We want $P(X\geq0.300)$, which is equivalent to $P(Z\geq1.15)$ in the standard normal distribution. Since the total area under the standard - normal curve is 1, and $P(Z\geq z)=1 - P(Z < z)$. Looking up $P(Z < 1.15)$ in the standard normal table, we find $P(Z < 1.15)=0.8749$. Then $P(Z\geq1.15)=1 - 0.8749 = 0.1251$.

Answer:

$0.1251$