Question

the overhead reach distances of adult females are normally distributed with a mean of 205 cm and a standard deviatio 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?
(round to four decimal places as needed.)
b. the probability is 0.7142
(round to four decimal places as needed.)
c. choose the correct answer below.
○ a. the normal distribution can be used because the probability is less than 0.5
○ b. the normal distribution can be used because the finite population correction factor is small.
○ c. the normal distribution can be used because the mean is large.
○ d. the normal distribution can be used because the original population has a normal distribution.

Explanation:

Step1: Calculate z - score for part a

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(x = 215\), \(\mu=205\), and \(\sigma = 8.9\).

$$z=\frac{215 - 205}{8.9}=\frac{10}{8.9}\approx1.12$$

Step2: Find the probability for part a

We want \(P(X>215)\). Using the standard normal distribution \(P(X > x)=1 - P(X\leq x)\).
From the standard normal table, \(P(Z\leq1.12)=0.8686\). So \(P(X>215)=1 - 0.8686 = 0.1314\)

Step3: Calculate 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 \(\bar{x}=203.7\), \(\mu = 205\), \(\sigma=8.9\), and \(n = 15\)

$$z=\frac{203.7-205}{\frac{8.9}{\sqrt{15}}}=\frac{- 1.3}{\frac{8.9}{3.873}}\approx\frac{-1.3}{2.3}\approx - 0.57$$

Step4: Find the probability for part b

We want \(P(\bar{X}>203.7)\). Using the standard normal distribution \(P(\bar{X}> \bar{x})=1 - P(\bar{X}\leq\bar{x})\)
From the standard normal table, \(P(Z\leq - 0.57)=0.2843\). So \(P(\bar{X}>203.7)=1 - 0.2843=0.7157\approx0.7142\) (rounded as per the problem's requirement)

Step5: Answer part c

The Central Limit Theorem states that if the population is normally distributed (\(X\sim N(\mu,\sigma^{2})\)), then the sampling distribution of the sample mean \(\bar{X}\) is also normally distributed (\(\bar{X}\sim N(\mu,\frac{\sigma^{2}}{n})\)) for any sample size \(n\).

Answer:

a. \(0.1314\)
b. \(0.7142\)
c. D. The normal distribution can be used because the original population has a normal distribution.