A population of values has a normal distribution with $\mu=6.8$ and $\sigma=54.8$. If a random sample of size $n=18$ is selected,
a. Find the probability that a single randomly selected value is less than 41.7 . Round your answer to four decimals.
\[
P(X< 41.7)=.9966
\]
b. Find the probability that a sample of size $n=18$ is randomly selected with a mean less than 41.7 . Round your answer to four decimals.
\[
P(M< 41.7)=
\]

Step 1 ：Given a population of values with a normal distribution where the mean \(\mu=6.8\) and the standard deviation \(\sigma=54.8\).

Step 2 ：For part a, we are asked to find the probability that a single randomly selected value is less than 41.7. To do this, we first calculate the z-score for the value 41.7 using the formula \(z = \frac{X - \mu}{\sigma}\), where X is the value we are interested in, \(\mu\) is the mean of the population, and \(\sigma\) is the standard deviation of the population. Substituting the given values, we get \(z = \frac{41.7 - 6.8}{54.8} = 2.701973722271224\).

Step 3 ：Next, we find the corresponding probability from the standard normal distribution table, which gives us \(P(X<41.7) = 0.7378924081414707\).

Step 4 ：For part b, we are asked to find the probability that a sample of size \(n=18\) is randomly selected with a mean less than 41.7. To do this, we first calculate the z-score for the sample mean using the formula \(z = \frac{M - \mu}{\sigma / \sqrt{n}}\), where M is the sample mean, \(\mu\) is the mean of the population, \(\sigma\) is the standard deviation of the population, and n is the sample size. Substituting the given values, we get \(z = \frac{41.7 - 6.8}{54.8 / \sqrt{18}} = 2.701973722271224\).

Step 5 ：Next, we find the corresponding probability from the standard normal distribution table, which gives us \(P(M<41.7) = 0.9965535395079758\).

Step 6 ：Rounding to four decimal places, we get the final answers: \(P(X<41.7) \approx \boxed{0.7379}\) and \(P(M<41.7) \approx \boxed{0.9966}\).