Questions: A population of values has a normal distribution with μ=84.6 and σ=33.1. You intend to draw a random sample of size n=72. Find the probability that a sample of size n=72 is randomly selected with a mean between 78.7 and 85.8 . P(78.7<M<85.8)= Enter your answers as numbers accurate to 4 decimal places. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.

Solution Steps

To find the probability that the sample mean \( M \) falls between \( 78.7 \) and \( 85.8 \), we first calculate the Z-scores for the lower and upper bounds using the formula:

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

Where:

• \( \mu = 84.6 \)
• \( \sigma = 33.1 \)
• \( n = 72 \)

Calculating the Z-score for the lower bound \( 78.7 \):

\[ Z_{start} = \frac{78.7 - 84.6}{33.1 / \sqrt{72}} \approx -1.5125 \]

Calculating the Z-score for the upper bound \( 85.8 \):

\[ Z_{end} = \frac{85.8 - 84.6}{33.1 / \sqrt{72}} \approx 0.3076 \]

Next, we find the probability that the sample mean \( M \) is between these Z-scores using the cumulative distribution function \( \Phi \):

\[ P(78.7 < M < 85.8) = \Phi(Z_{end}) - \Phi(Z_{start}) \]

Substituting the Z-scores:

\[ P(78.7 < M < 85.8) = \Phi(0.3076) - \Phi(-1.5125) \]

Using the values from the standard normal distribution:

\[ P(78.7 < M < 85.8) \approx 0.5556 \]

Final Answer