Question

the overhead reach distances of adult females are normally distributed with a mean of 205 cm and a standard deviation of 8.9 cm.
a. find the probability that an individual distance is greater than 215.00 cm.
b. find the probability that the mean for 15 randomly selected distances is greater than 203.70 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 0.1314
(round to four decimal places as needed )
b. the probability is
(round to four decimal places as needed )

Explanation:

Step1: Calculate the z - score for part (b)

The formula for the z - score of a sample mean is \(z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\), where \(\mu = 205\) (population mean), \(\sigma = 8.9\) (population standard deviation), \(n = 15\) (sample size), and \(\bar{x}=203.70\) (sample mean).

$$z=\frac{203.70 - 205}{\frac{8.9}{\sqrt{15}}}$$

$$z=\frac{- 1.3}{\frac{8.9}{\sqrt{15}}}$$

$$z=\frac{-1.3}{2.299}\approx - 0.565$$

Step2: Find the probability

We want \(P(\bar{X}>203.70)\), which is equivalent to \(P(Z>- 0.565)\). Since \(P(Z > z)=1 - P(Z\leq z)\), and looking up \(P(Z\leq - 0.565)\) in the standard normal table (or using a calculator with a normal - distribution function). Using a calculator, \(P(Z\leq - 0.565)\approx0.2858\). Then \(P(Z>-0.565)=1 - 0.2858 = 0.7142\)

Answer:

\(0.7142\)