Questions: The overhead reach distances of adult females are normally distributed with a mean of 195 cm and a standard deviation of 8 cm . a. Find the probability that an individual distance is greater than 207.50 cm . b. Find the probability that the mean for 25 randomly selected distances is greater than 192.80 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 0.0591 . (Round to four decimal places as needed.) b. The probability is 0.9154 . (Round to four decimal places as needed.) c. Choose the correct answer below. A. The normal distribution can be used because the original population has a normal distribution. B. The normal distribution can be used because the mean is large. C. The normal distribution can be used because the probability is less than 0.5 D. The normal distribution can be used because the finite population correction factor is small.

Solution Steps

To find the probability that an individual distance is greater than \( 207.50 \, \text{cm} \), we first calculate the Z-score using the formula:

\[ Z = \frac{X - \mu}{\sigma} \]

where:

• \( X = 207.50 \)
• \( \mu = 195 \)
• \( \sigma = 8 \)

Substituting the values:

\[ Z = \frac{207.50 - 195}{8} = 1.5625 \]

Next, we find the probability that an individual distance is greater than \( 207.50 \, \text{cm} \). This is given by:

\[ P(X > 207.50) = 1 - P(Z < 1.5625) \]

Using the Z-table or standard normal distribution, we find:

\[ P(Z < 1.5625) \approx 0.9409 \]

Thus,

\[ P(X > 207.50) = 1 - 0.9409 = 0.0591 \]

For the second part, we need to find the probability that the mean for \( n = 25 \) randomly selected distances is greater than \( 192.80 \, \text{cm} \). We use the standard error of the mean:

\[ \sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} = \frac{8}{\sqrt{25}} = \frac{8}{5} = 1.6 \]

Now, we calculate the Z-score for the sample mean:

\[ Z = \frac{\bar{X} - \mu}{\sigma_{\bar{X}}} = \frac{192.80 - 195}{1.6} = -1.375 \]

Next, we find the probability:

\[ P(\bar{X} > 192.80) = 1 - P(Z < -1.375) \]

Using the Z-table, we find:

\[ P(Z < -1.375) \approx 0.0846 \]

Thus,

\[ P(\bar{X} > 192.80) = 1 - 0.0846 = 0.9154 \]

The normal distribution can be used in part (b) because the original population has a normal distribution. This is valid even though the sample size does not exceed 30.

Final Answer