A population of values has a normal distribution with $\mathrm{U}=25.5$ and $\mathrm{O}=91.5$ Find the probability that a randomly selected sample of size $n=36$ has a mean between 33.1 and 34.7. Round your answer to four decimal places.
Answer
Finally, we round this probability to four decimal places to get the final answer. The probability that a randomly selected sample of size 36 has a mean between 33.1 and 34.7 is approximately \(\boxed{0.0360}\).
Step 1 ：We are given a population of values with a normal distribution where the mean (\(\mu\)) is 25.5 and the standard deviation (\(\sigma\)) is 91.5. We are asked to find the probability that a randomly selected sample of size 36 (\(n\)) has a mean between 33.1 and 34.7.
Step 2 ：We can use the z-score formula to solve this problem. The z-score is a measure of how many standard deviations an element is from the mean. The formula for the z-score is \(z = \frac{x - \mu}{\sigma / \sqrt{n}}\), where \(x\) is the value from the sample, \(\mu\) is the mean, \(\sigma\) is the standard deviation, and \(n\) is the sample size.
Step 3 ：First, we calculate the z-score for 33.1: \(z_1 = \frac{33.1 - 25.5}{91.5 / \sqrt{36}} = 0.498360655737705\).
Step 4 ：Next, we calculate the z-score for 34.7: \(z_2 = \frac{34.7 - 25.5}{91.5 / \sqrt{36}} = 0.6032786885245903\).
Step 5 ：We then find the area under the normal distribution curve between these two z-scores. This area represents the probability we are looking for. The probability is approximately 0.035953277396927885.
Step 6 ：Finally, we round this probability to four decimal places to get the final answer. The probability that a randomly selected sample of size 36 has a mean between 33.1 and 34.7 is approximately \(\boxed{0.0360}\).
