Question

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.

Answer 1

Answer: 0.7619

Step-by-step explanation:

Given : Mean : $\mu=45.0$

Standard deviation : $\sigma =8.0$

Sample size : $n=9$

We assume that the variable is normally distributed.

The value of z-score is given by :-

$z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}$

a) For x= 46.9 years

$z=\dfrac{46.9-45.0}{\dfrac{8}{\sqrt{9}}}=0.7125$

The p-value : $P(z$

Hence, the  probability that the average age of those doctors is less than 46.9 years =0.7619