Questions: The heights of ten-year-old males are normally distributed with a mean of 56.4 inches and a standard deviation of 6.5 inches. If a pediatrician selects a random sample of 46 ten-year-old males from his patient population, what is the probability that the mean height of this sample will be greater than 56 inches? Round your answer to at least three decimal places

Solution Steps

We are tasked with finding the probability that the mean height of a sample of ten-year-old males is greater than 56 inches, given that the heights are normally distributed with a mean (\( \mu \)) of 56.4 inches and a standard deviation (\( \sigma \)) of 6.5 inches. The sample size (\( n \)) is 46.

To find the probability, we first calculate the Z-score for the sample mean of 56 inches using the formula:

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

Substituting the values:

\[ Z = \frac{56 - 56.4}{6.5 / \sqrt{46}} \approx -0.417 \]

Next, we find the probability that the sample mean is greater than 56 inches. This is calculated using the cumulative distribution function (\( \Phi \)) of the standard normal distribution:

\[ P(X > 56) = 1 - P(X \leq 56) = 1 - \Phi(Z_{start}) \]

From the calculations, we have:

\[ P(X \leq 56) = \Phi(-0.417) \approx 0.662 \]

Thus, the probability that the sample mean is greater than 56 inches is:

\[ P(X > 56) = 1 - 0.662 \approx 0.338 \]

Final Answer