Question

the overhead reach distances of adult females are normally distributed with a mean of 195 cm and a standard deviation of 7.8 cm.
a. find the probability that an individual distance is greater than 204.30 cm.
b. find the probability that the mean for 25 randomly selected distances is greater than 193.50 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 \(x = 204.30\), \(\mu=195\), and \(\sigma = 7.8\).

$$ z=\frac{204.30 - 195}{7.8}=\frac{9.3}{7.8}=1.19 $$

We want \(P(X>204.30)\), which is equivalent to \(P(Z > 1.19)\). Using the property \(P(Z>z)=1 - P(Z\leq z)\), and from the standard normal table \(P(Z\leq1.19) = 0.8830\).

$$ P(Z>1.19)=1 - 0.8830=0.1170 $$

Step2: 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 \(\bar{x}=193.50\), \(\mu = 195\), \(\sigma=7.8\), and \(n = 25\).

$$ z=\frac{193.50-195}{\frac{7.8}{\sqrt{25}}}=\frac{- 1.5}{\frac{7.8}{5}}=\frac{-1.5}{1.56}\approx - 0.96 $$

We want \(P(\bar{X}>193.50)\), which is equivalent to \(P(Z>-0.96)\). Using the property \(P(Z > z)=1 - P(Z\leq z)\), and from the standard normal table \(P(Z\leq - 0.96)=0.1685\).

$$ P(Z>-0.96)=1 - 0.1685 = 0.8315 $$

Step3: Answer for part (c)

The original population (overhead - reach distances of adult females) is normally distributed. According to the Central Limit Theorem, if the population is normally distributed, then the sampling distribution of the sample mean \(\bar{X}\) is also normally distributed for any sample size \(n\).

Answer:

a. \(0.1170\)
b. \(0.8315\)
c. The original population is normally distributed.