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 215.90 cm. b. find the probability that the mean for 15 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.) b. the probability is . (round to four decimal places as needed.) c. choose the correct answer below. a. the normal distribution can be used because the mean is large. b. the normal distribution can be used because the finite population correction factor is small. c. the normal distribution can be used because the probability is less than 0.5. d. the normal distribution can be used because the original population has a normal distribution.

Explanation:

Step1: Calculate z - score for part a

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $\mu = 202.5$, $\sigma=8.9$, and $x = 215.90$.
$z=\frac{215.90 - 202.5}{8.9}=\frac{13.4}{8.9}\approx1.5056$
We want $P(X>215.90)=1 - P(X\leq215.90)$. Looking up $P(Z\leq1.5056)$ in the standard - normal table, $P(Z\leq1.5056)\approx0.9345$. So $P(X>215.90)=1 - 0.9345 = 0.0655$.

Step2: Calculate z - score for part b

For the sample mean of $n = 15$ samples, the mean of the sampling distribution of the sample mean is $\mu_{\bar{X}}=\mu = 202.5$ and the standard deviation of the sampling distribution of the sample mean is $\sigma_{\bar{X}}=\frac{\sigma}{\sqrt{n}}=\frac{8.9}{\sqrt{15}}\approx2.30$.
The z - score is $z=\frac{\bar{x}-\mu_{\bar{X}}}{\sigma_{\bar{X}}}$, where $\bar{x}=200.70$.
$z=\frac{200.70 - 202.5}{2.30}=\frac{-1.8}{2.30}\approx - 0.7826$.
$P(\bar{X}>200.70)=1 - P(\bar{X}\leq200.70)$. Looking up $P(Z\leq - 0.7826)$ in the standard - normal table, $P(Z\leq - 0.7826)\approx0.2177$. So $P(\bar{X}>200.70)=1 - 0.2177 = 0.7823$.

Step3: Answer part c

The normal distribution can be used in part (b) because the original population has a normal distribution.

Answer:

a. $0.0655$
b. $0.7823$
c. D. The normal distribution can be used because the original population has a normal distribution.