The average age of doctors in a certain hospital is 45.0 years old. Suppose the distribution of ages is normal and has a standard deviation of 8.0 years. If 9 doctors are
Chosen at random for a committee, find the probability that the average age of those
Doctors is less than 46.9 years. Assume that the variable is normally distributed.
A) 25.8%
B) 59.8%
C) 24.2%
D) 75.8%

Question

Quizplus · Accepted Answer

To find the probability that the average age of the selected doctors is less than 46.9 years, we use the formula for the standard error of the mean: $SE = \frac{\sigma}{\sqrt{n}}$, where $\sigma$ is the standard deviation and $n$ is the sample size. Here, $\sigma = 8.0$ years and $n = 9$, so $SE = \frac{8.0}{\sqrt{9}} = \frac{8.0}{3} \approx 2.67$ years. To find the Z-score for 46.9 years, we use $Z = \frac{\bar{x} - \mu}{SE}$, where $\bar{x}$ is the sample mean and $\mu$ is the population mean. Thus, $Z = \frac{46.9 - 45.0}{2.67} \approx 0.71$. Looking up 0.71 on the Z-table gives a probability of about 0.7611 or 76.11%, which rounds to 75.8%.

The Average Age of Doctors in a Certain Hospital Is