Question 8
A population of values has a normal distribution with $\mu=25.5$ and $\sigma=91.5$.
a. Find the probability that a single randomly selected value is between 33.1 and 34.7 . Round you answer to four decimal places.
\[
P(33.1< X< 34.7)=
\]
b. Find the probability that a randomly selected sample of size $n=36$ has a mean between 33.1 34.7. Round your answer to four decimal places.
\[
P(33.1< M< 34.7)=
\]
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Step 1 ：Given a population of values with a normal distribution where the mean \(\mu=25.5\) and the standard deviation \(\sigma=91.5\).

Step 2 ：We are asked to find the probability that a single randomly selected value is between 33.1 and 34.7.

Step 3 ：To solve this, we need to standardize the values 33.1 and 34.7 using the formula for z-score which is \((X - \mu) / \sigma\) where X is the value, \(\mu\) is the mean and \(\sigma\) is the standard deviation.

Step 4 ：Standardizing the values, we get \(z1 = 0.0830601092896175\) and \(z2 = 0.10054644808743173\).

Step 5 ：We can use the standard normal distribution (Z-distribution) to find the probabilities corresponding to the z-scores. The probability that a value is between 33.1 and 34.7 is the probability that the z-score is between the z-scores of 33.1 and 34.7.

Step 6 ：Using the cumulative distribution function (CDF), we find the probabilities corresponding to \(z1\) and \(z2\) to be \(p1 = 0.5330981278123774\) and \(p2 = 0.5400447453001012\) respectively.

Step 7 ：The probability that a single randomly selected value is between 33.1 and 34.7 is the difference between \(p2\) and \(p1\), which is \(p = 0.0069466174877237385\).

Step 8 ：Rounding to four decimal places, the final answer is \(\boxed{0.0069}\).