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Chapter 6: Problem 25

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 a. in less than 160 minutes? b. in 215 to 245 minutes?

Short Answer

Expert verified
a. The probability is around 0.036.\nb. The probability is around 0.135. These probabilities may vary slightly depending on the specific standard normal distribution table used.

Step by step solution

01

Calculate Z-score

First, calculate the Z-score for the given times. The Z-score is a measure of how many standard deviations an element is from the mean. The formula to calculate the Z-score is \(Z = \frac{x - \mu}{\sigma}\) where \(x\) is the time, \(\mu\) is the mean time and \(\sigma\) is the standard deviation. For 160 minutes, \(Z = \frac{160 - 190}{21}\). For 215 and 245 minutes, calculate their Z-scores similarly.
02

Find probabilities using Z-table

Once the Z-scores have been calculated, use the standard normal distribution table (Z-table) to find the probabilities. The value in the table corresponding to a particular Z-score gives the probability that a randomly selected runner will take that amount of time or less to complete the race.
03

Calculate probability for each part

a. For less than 160 minutes, find the probability using the Z-score calculated before. The standard normal distribution table will show the relevance field.\nb. For between 215 to 245 minutes, find probabilities for each Z-score and take the difference. Remember, the table gives the probability of being less than a given time. To find probability between two times, one needs to subtract the probability of being less than 215 minutes from the probability of being less than 245 minutes.

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