A population of values has a normal distribution with $\mu=75.9$ and $\sigma=9.4$.
a. Find the probability that a single randomly selected value is greater than 74.8. Round your answer to four decimal places. $P(X> 74,8)=.5466 \sqrt{0}$
b. Find the probability that a randomly selected sample of size $n=144$ has a mean greater than 74.8 . Round your answer to four decimal places. $P(M> 74.8)=.5466 \times$

Step 1 ：Given a population of values with a normal distribution where the mean (μ) is 75.9 and the standard deviation (σ) is 9.4.

Step 2 ：We are asked to find the probability that a single randomly selected value is greater than 74.8.

Step 3 ：To solve this, we need to standardize the value 74.8 using the formula for z-score which is \((X - μ) / σ\) where X is the value we are standardizing, μ is the mean and σ is the standard deviation.

Step 4 ：Substituting the given values into the formula, we get a z-score of approximately -0.117.

Step 5 ：We then use the z-table or a function that gives the cumulative distribution function of the standard normal distribution to find the probability that a value is less than the standardized value. This gives us a probability of approximately 0.4534.

Step 6 ：Since the table or function gives the probability that a value is less than the given value, we subtract the result from 1 to get the probability that a value is greater than the given value. This gives us a final probability of approximately 0.5466.

Step 7 ：Final Answer: The probability that a single randomly selected value is greater than 74.8 is \(\boxed{0.5466}\).