Current Attempt in Progress Let $x$ denote the time taken to run a road race. Suppose $x$ is approximately normally distributed with a mean of 190 minutes and a standard deviation of 21 minutes. If one runner is selected at random, what is the probability that this runner will complete this road race in 218 to 248 minutes? Round your answer to four decimal places.
Solution
Step 1 ：Let's denote the time taken to run a road race as \(x\). Suppose \(x\) is approximately normally distributed with a mean of 190 minutes and a standard deviation of 21 minutes. If one runner is selected at random, we want to find the probability that this runner will complete this road race in 218 to 248 minutes.
Step 2 ：We need to convert the time range (218 to 248 minutes) into z-scores. The z-score is a measure of how many standard deviations an element is from the mean. It is calculated as \((X - \mu) / \sigma\), where \(X\) is the value, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.
Step 3 ：Let's calculate the z-scores for 218 and 248 minutes. For 218 minutes, the z-score is \(z1 = (218 - 190) / 21 = 1.3333333333333333\). For 248 minutes, the z-score is \(z2 = (248 - 190) / 21 = 2.761904761904762\).
Step 4 ：We can use the standard normal distribution table (also known as the z-table) to find the probabilities corresponding to these z-scores. The probability that a randomly selected runner will complete the race in between 218 and 248 minutes is the difference between these two probabilities.
Step 5 ：From the z-table, the probability corresponding to \(z1 = 1.3333333333333333\) is \(prob1 = 0.9087887802741321\) and the probability corresponding to \(z2 = 2.761904761904762\) is \(prob2 = 0.9971267380880181\).
Step 6 ：The probability that a randomly selected runner will complete the race in between 218 and 248 minutes is the difference between these two probabilities, which is \(prob_{diff} = prob2 - prob1 = 0.08833795781388598\).
Step 7 ：Final Answer: The probability that a randomly selected runner will complete the race in between 218 and 248 minutes is approximately \(\boxed{0.0883}\).
