Questions: The mean number of words per minute (WPM) read by sixth graders is 90 with a variance of 529. If 94 sixth graders are randomly selected, what is the probability that the sample mean would be greater than 87.78 WPM? Round your answer to four decimal places.

Solution Steps

Given the variance \( \sigma^2 = 529 \), the standard deviation \( \sigma \) is calculated as follows:

\[ \sigma = \sqrt{\sigma^2} = \sqrt{529} = 23.0 \]

To find the probability that the sample mean is greater than \( 87.78 \), we first calculate the Z-score for \( 87.78 \) using the formula:

\[ Z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}} \]

Where:

• \( \bar{x} = 87.78 \)
• \( \mu = 90 \)
• \( \sigma = 23.0 \)
• \( n = 94 \)

Calculating the standard error:

\[ \text{Standard Error} = \frac{\sigma}{\sqrt{n}} = \frac{23.0}{\sqrt{94}} \approx 2.37 \]

Now, substituting into the Z-score formula:

\[ Z = \frac{87.78 - 90}{2.37} \approx -0.9358 \]

Using the Z-score, we find the probability that the sample mean is less than \( 87.78 \):

\[ P(Z < -0.9358) \approx 0.1753 \]

Thus, the probability that the sample mean is greater than \( 87.78 \) is:

\[ P(Z > -0.9358) = 1 - P(Z < -0.9358) \approx 1 - 0.1753 = 0.8247 \]

Final Answer