Bob's golf score at his local course follows the normal distribution with a mean of 92.1 and a standard deviation of 3.8. What is the probability that the score on his next round of golf will be more than 90?

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Quizplus · Accepted Answer

To find the probability that Bob's score will be more than 90, we first calculate the z-score for a score of 90 using the formula $z = \frac{x - \mu}{\sigma}$, where $x$ is the score, $\mu$ is the mean, and $\sigma$ is the standard deviation. Plugging in the values, we get $z = \frac{90 - 92.1}{3.8} = -0.5526$. Looking up this z-score in a standard normal distribution table or using a calculator, we find the area to the left of this z-score is approximately 0.2912. Since we're interested in the probability of scoring more than 90, we subtract this value from 1, resulting in approximately 0.7088.

Bob's Golf Score at His Local Course Follows the Normal