Question

the overhead reach distances of adult females are normally distributed with a mean of 205.5 cm and a standard deviation of 7.8 cm. a. find the probability that an individual distance is greater than 218.90 cm. b. find the probability that the mean for 25 randomly selected distances is greater than 204.00 cm. c. why can the normal distribution be used in part (b), even though the sample size does not exceed 30?

Explanation:

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

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(x = 218.90\), \(\mu=205.5\), and \(\sigma = 7.8\).

$$z=\frac{218.90 - 205.5}{7.8}=\frac{13.4}{7.8}\approx1.72$$

We want to find \(P(X>218.90)\), which is \(1 - P(X\leq218.90)\). Looking up the z - score of \(1.72\) in the standard normal table, \(P(Z\leq1.72)=0.9573\). So \(P(X > 218.90)=1 - 0.9573=0.0427\)

Step2: Calculate the z - score for part (b)

For a sample of size \(n = 25\), the standard deviation of the sample mean is \(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}=\frac{7.8}{\sqrt{25}}=\frac{7.8}{5}=1.56\)
The formula for the z - score for the sample mean is \(z=\frac{\bar{x}-\mu}{\sigma_{\bar{x}}}\), where \(\bar{x}=204.00\), \(\mu = 205.5\), and \(\sigma_{\bar{x}}=1.56\)

$$z=\frac{204.00 - 205.5}{1.56}=\frac{- 1.5}{1.56}\approx - 0.96$$

We want to find \(P(\bar{X}>204.00)\), which is \(1 - P(\bar{X}\leq204.00)\). Looking up the z - score of \(-0.96\) in the standard normal table, \(P(Z\leq - 0.96)=0.1685\). So \(P(\bar{X}>204.00)=1 - 0.1685 = 0.8315\)

Step3: Explain part (c)

The central limit theorem states that if the population is normally distributed, then the sampling distribution of the sample mean \(\bar{X}\) is also normally distributed for any sample size \(n\). In part (b), since the population of adult female overhead reach distances is normally distributed, the sampling distribution of the sample mean (for \(n = 25\)) is also normally distributed.

Answer:

a. \(0.0427\)
b. \(0.8315\)
c. Because the population is normally distributed, the sampling distribution of the sample mean is also normally distributed for any sample size \(n\) (by the central limit theorem for normal populations).