A superstar major league baseball player just signed a new deal that pays him a record amount of money. The star has driven in an average of 110 runs over the course of his career, with a standard deviation of 31 runs. An average player at his position drives in 80 runs. What is the probability the superstar bats in fewer runs than an average player next year? Assume the number of runs batted in is normally distributed.
A) 0.1666
B) 0.3340
C) 0. 8340
D) 0.9700

Question

Quizplus · Accepted Answer

To find the probability that the superstar bats in fewer runs than an average player, we use the Z-score formula: $Z = \frac{X - \mu}{\sigma}$, where $X$ is the value we're comparing to the mean ($\mu$) and $\sigma$ is the standard deviation. Here, $X = 80$ runs, $\mu = 110$ runs, and $\sigma = 31$ runs. Plugging these values into the formula gives us $Z = \frac{80 - 110}{31} = -0.967$. Looking up this Z-score in a standard normal distribution table or using a calculator gives us a probability of approximately 0.1666, indicating the likelihood that the superstar bats in fewer runs than an average player is about 16.66%.

A Superstar Major League Baseball Player Just Signed a New