Questions: The mean number of words per minute (WPM) typed by a speed typist is 143 with a standard deviation of 12 WPM. What is the probability that the sample mean would be greater than 142.2 WPM if 76 speed typists are randomly selected? Round your answer to four decimal places.

Solution Steps

We are tasked with finding the probability that the sample mean of words typed per minute (WPM) by 76 speed typists is greater than \( 142.2 \) WPM. The population mean \( \mu \) is \( 143 \) WPM, and the population standard deviation \( \sigma \) is \( 12 \) WPM.

To find the probability, we first need to calculate the Z-scores for the sample mean. The Z-score is calculated using the formula:

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

Where:

• \( X \) is the sample mean (in this case, \( 142.2 \)),
• \( \mu = 143 \),
• \( \sigma = 12 \),
• \( n = 76 \).

Calculating the standard error:

\[ \sigma_{\text{sample}} = \frac{\sigma}{\sqrt{n}} = \frac{12}{\sqrt{76}} \approx 1.3764 \]

Now, we can calculate the Z-score for \( 142.2 \):

\[ Z_{start} = \frac{142.2 - 143}{1.3764} \approx -0.5812 \]

The Z-score for the upper bound (infinity) is:

\[ Z_{end} = \infty \]

Using the Z-scores, we can find the probability that the sample mean is greater than \( 142.2 \):

\[ P(X > 142.2) = P(Z > -0.5812) = 1 - P(Z < -0.5812) \]

Using the cumulative distribution function \( \Phi \):

\[ P(X > 142.2) = \Phi(\infty) - \Phi(-0.5812) \]

From the output, we have:

\[ P(X > 142.2) = 1 - 0.2806 = 0.7194 \]

Final Answer