Question

The average age of residents in a large residential retirement community is 69 years with standard deviation 5.8 years. A simple random sample of 100 residents is to be selected, and the sample mean age of these residents is to be computed. The probability that the average age of the 100 residents selected is less than 68.5 years is_______.

Answer 1

Answer: 0.1943

Step-by-step explanation:

Given : Mean : $\mu=69$

Standard deviation : $\sigma=5.8$

Sample size : $n=100$

The formula to calculate the z-score :-

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

For x = 68.5

$z=\dfrac{68.5-69}{\dfrac{5.8}{\sqrt{100}}}=-0.862$

The p-value = $P(z\leq-0.862)= 0.1943438\approx0.1943$

Hence,probability that the average age of the 100 residents selected is less than 68.5 years is 0.1943  .