Question 7
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.
\[
P(X> 74.8)=
\]
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.
\[
P(M> 74.8)=
\]
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Subrnit Question

Step 1 ：Given a population of values with a normal distribution, where the mean \(\mu\) is 75.9 and the standard deviation \(\sigma\) 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 Z-score formula, which is \(Z = \frac{X - \mu}{\sigma}\), where \(X\) is the value we are interested in, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.

Step 4 ：Substituting the given values into the Z-score formula, we get \(Z = \frac{74.8 - 75.9}{9.4} = -0.117\).

Step 5 ：We can find the probability that a value is greater than 74.8 by subtracting the cumulative probability of the Z-score from 1, because the total probability under a normal distribution curve is 1.

Step 6 ：Using the Z-table or a calculator, we find that the cumulative probability of -0.117 is approximately 0.4534.

Step 7 ：Subtracting this from 1 gives us the probability that a single randomly selected value is greater than 74.8, which is \(1 - 0.4534 = 0.5466\).

Step 8 ：Final Answer: The probability that a single randomly selected value is greater than 74.8 is approximately \(\boxed{0.5466}\).