Question

in baseball, a players 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 $mu = 0.261$ and standard deviation $sigma = 0.034$. 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 formula for the z - score is $z=\frac{x-\mu}{\sigma}$, where $x = 0.300$, $\mu=0.261$, and $\sigma = 0.034$.
$z=\frac{0.300 - 0.261}{0.034}=\frac{0.039}{0.034}\approx1.15$

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 $P(Z\geq z)=1 - P(Z\lt z)$. Looking up $P(Z\lt1.15)$ in the standard - normal table, we find $P(Z\lt1.15)=0.8749$. So $P(Z\geq1.15)=1 - 0.8749 = 0.1251$.

Answer:

0.1251