A population of values has a normal distribution with $\mu=43.6$ and $\sigma=78$. If a random sample of size $n=20$ is selected,
a. Find the probability that a single randomly selected value is greater than 88.9 . Round your answer to four decimals.
\[
P(X> 88.9)=
\]

Step 1 ：We are given a population of values that follows a normal distribution with a mean (\(\mu\)) of 43.6 and a standard deviation (\(\sigma\)) of 78.

Step 2 ：We are asked to find the probability that a single randomly selected value is greater than 88.9.

Step 3 ：To solve this, we first calculate the Z score. The Z score is a measure of how many standard deviations an element is from the mean. It is calculated as \(Z = \frac{X - \mu}{\sigma}\), where X is the value from the dataset.

Step 4 ：Substituting the given values into the formula, we get \(Z = \frac{88.9 - 43.6}{78} \approx 0.5808\).

Step 5 ：The probability that a single randomly selected value is greater than 88.9 is given by \(P(X > 88.9) = 1 - P(X \leq 88.9)\).

Step 6 ：Using the Z score, we can find \(P(X \leq 88.9)\) from the standard normal distribution table. This gives us a probability of approximately 0.2807.

Step 7 ：Final Answer: \(P(X>88.9) \approx \boxed{0.2807}\)