Question

the overhead reach distances of adult females are normally distributed with a mean of 205 cm and a standard deviation of 8.6 cm. a. find the probability that an individual distance is greater than 214.30 cm. b. find the probability that the mean for 15 randomly selected distances is greater than 202.80 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 the z - score for part (a)

The formula for the z - score is $z=\frac{x-\mu}{\sigma}$, where $x = 214.30$, $\mu=205$, and $\sigma = 8.6$.
$z=\frac{214.30 - 205}{8.6}=\frac{9.3}{8.6}\approx1.08$

Step2: Find the probability for part (a)

We want $P(X>214.30)$, which is equivalent to $P(Z > 1.08)$. Since $P(Z>z)=1 - P(Z\leq z)$, and from the standard - normal table $P(Z\leq1.08) = 0.8599$, then $P(Z>1.08)=1 - 0.8599 = 0.1401$.

Step3: Calculate the z - score for part (b)

The formula for the z - score of the sample mean is $z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}$, where $\bar{x}=202.80$, $\mu = 205$, $\sigma = 8.6$, and $n = 15$.
$z=\frac{202.80 - 205}{\frac{8.6}{\sqrt{15}}}=\frac{- 2.2}{\frac{8.6}{\sqrt{15}}}\approx\frac{-2.2}{2.22}\approx - 0.99$

Step4: Find the probability for part (b)

We want $P(\bar{X}>202.80)$, which is equivalent to $P(Z>-0.99)$. Since $P(Z > - 0.99)=1 - P(Z\leq - 0.99)$, and from the standard - normal table $P(Z\leq - 0.99)=0.1611$, then $P(Z>-0.99)=1 - 0.1611 = 0.8389$.

Step5: Answer part (c)

The normal distribution can be used in part (b) because the population (overhead reach distances of adult females) is normally distributed. According to the Central Limit Theorem, if the population is normally distributed, the sampling distribution of the sample mean $\bar{X}$ is also normally distributed for any sample size $n$.

Answer:

a. $0.1401$
b. $0.8389$
c. The population is normally distributed.