Questions: A population of values has a normal distribution with μ=77.1 and σ=19.9. a. Find the probability that a single randomly selected value is greater than 82. Round your answer to four decimal places. P(X>82)= b. Find the probability that a randomly selected sample of size n=161 has a mean greater than 82. Round your answer to four decimal places. P(M>82)=

Solution Steps

To find the probability that a single randomly selected value is greater than 82, we first calculate the Z-score using the formula:

\[ z = \frac{X - \mu}{\sigma} = \frac{82 - 77.1}{19.9} = 0.2462 \]

Next, we determine the probability \( P(X > 82) \) using the Z-score. This can be expressed as:

\[ P(X > 82) = \Phi(\infty) - \Phi(0.2462) = 0.4028 \]

where \( \Phi \) is the cumulative distribution function of the standard normal distribution.

For a sample of size \( n = 161 \), we need to calculate the Z-score for the sample mean. The standard error of the mean is given by \( \frac{\sigma}{\sqrt{n}} \):

\[ z = \frac{M - \mu}{\sigma/\sqrt{n}} = \frac{82 - 77.1}{\frac{19.9}{\sqrt{161}}} \approx 3.1243 \]

Now, we find the probability \( P(M > 82) \):

\[ P(M > 82) = \Phi(\infty) - \Phi(3.1243) = 0.0009 \]

Final Answer