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 \(\boxed{0.0069}\).