Question

in a certain year, the batting averages of professional baseball players had a mean of 0.281 and standard deviation of 0.024. the distribution was approximately normal. what is the probability that a randomly chosen player has a batting average above 0.291?

Explanation:

Step1: Calculate the z - score

The formula for the z - score is \(z=\frac{x - \mu}{\sigma}\), where \(x = 0.291\), \(\mu=0.281\), and \(\sigma = 0.024\).

$$z=\frac{0.291 - 0.281}{0.024}=\frac{0.01}{0.024}\approx0.42$$

Step2: Find the probability using the standard normal table

Looking up \(z = 0.42\) in the standard normal table (the area to the left of \(z\) - value). The value from the standard normal table for \(z=0.42\) is \(0.6628\). But we want \(P(X>0.291)\), so \(P(X > 0.291)=1 - P(X\leq0.291)\)

$$P(X>0.291)=1 - 0.6628 = 0.3372$$

Answer:

The probability that a randomly chosen player has a batting average above \(0.291\) is approximately \(0.3372\)