Question

the overhead reach distances of adult females are normally distributed with a mean of 202.5 cm and a standard deviation of 8.9 cm.
a. find the probability that an individual distance is greater than 212.50 cm.
b. find the probability that the mean for 20 randomly - selected distances is greater than 200.70 cm.
c. why can the normal distribution be used in part (b), even though the sample size does not exceed 30?

a. the probability is
(round to four decimal places as needed.)

Explanation:

Step1: Calculate z - score for part (a)

The z - score formula for an individual value is $z=\frac{x-\mu}{\sigma}$, where $x = 212.50$, $\mu=202.5$, and $\sigma = 8.9$.
$z=\frac{212.50 - 202.5}{8.9}=\frac{10}{8.9}\approx1.1236$
We want $P(X>212.50)$, which is $1 - P(X\leq212.50)$. Looking up the z - score in the standard normal table, $P(Z\leq1.1236)\approx0.8697$. So $P(X > 212.50)=1 - 0.8697 = 0.1303$.

Step2: Calculate z - score for part (b)

The z - score formula for the sample mean $\bar{X}$ is $z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}$, where $\bar{x}=200.7$, $\mu = 202.5$, $\sigma=8.9$, and $n = 20$.
$z=\frac{200.7-202.5}{\frac{8.9}{\sqrt{20}}}=\frac{- 1.8}{\frac{8.9}{4.4721}}\approx\frac{-1.8}{1.99}= - 0.9045$
We want $P(\bar{X}>200.7)$, which is $1 - P(\bar{X}\leq200.7)$. Looking up the z - score in the standard normal table, $P(Z\leq - 0.9045)\approx0.1834$. So $P(\bar{X}>200.7)=1 - 0.1834=0.8166$.

Step3: Answer part (c)

The normal distribution can be used in part (b) because the population (the overhead - reach distances of adult females) is normally distributed. The sampling distribution of the sample mean $\bar{X}$ is normal for any sample size $n$ when the population is normally distributed.

Answer:

a. $0.1303$
b. $0.8166$
c. The population is normally distributed.