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 0.1303 (round to four decimal places as needed.)

Explanation:

Step1: Calculate z - score for part (a)

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $\mu = 202.5$ cm, $\sigma=8.9$ cm and $x = 212.50$ cm. So $z=\frac{212.50 - 202.5}{8.9}=\frac{10}{8.9}\approx1.12$.

Step2: Find the probability

We want $P(X>212.50)$. Using the standard normal distribution table, $P(Z > 1.12)=1 - P(Z\leq1.12)$. From the table, $P(Z\leq1.12) = 0.8686$, so $P(Z>1.12)=1 - 0.8686=0.1314$.

Step3: For part (b) with sample mean

The mean of the sampling distribution of the sample mean $\mu_{\bar{x}}=\mu = 202.5$ cm and the standard deviation of the sampling distribution of the sample mean $\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}$, where $n = 20$, $\sigma = 8.9$ cm. So $\sigma_{\bar{x}}=\frac{8.9}{\sqrt{20}}\approx1.99$.
The z - score for $\bar{x}=200.70$ cm is $z=\frac{\bar{x}-\mu_{\bar{x}}}{\sigma_{\bar{x}}}=\frac{200.70 - 202.5}{1.99}=\frac{- 1.8}{1.99}\approx - 0.90$.
We want $P(\bar{X}>200.70)=1 - P(\bar{X}\leq200.70)$. From the standard - normal table, $P(Z\leq - 0.90)=0.1841$, so $P(Z>-0.90)=1 - 0.1841 = 0.8159$.

Step4: Answer for part (c)

The normal distribution can be used in part (b) even though $n<30$ because the population (overhead reach distances of adult females) is normally distributed.

Answer:

a. $0.1314$
b. $0.8159$
c. The population is normally distributed.