Prof. Miller ran a marathon ten years ago. Given that the mean finishing time for all runners was 259 minutes with a standard deviation of 38 minutes, and that finishing times were approximately normally distributed, find the following:

(a) The percent of finishers who ran faster than Prof. Miller’s time of 3 hours 30 minutes.

(b) A time that was the 85th percentile (85% of finishers were under that time).

(c) The percent of runners who finished between 3 hours and 4 hours.

(d) Suppose there happened to be 10 other runners with the same last name as Prof. Miller (so 11 Millers total). What is the probability that the mean of the 11 Millers’ times is more than 240 minutes?

To find the percent of finishers who ran faster than Prof. Miller's time, calculate the z-score and find the area to the right. For the 85th percentile, find the z-score and use it to calculate the time. To find the percent of runners who finished between 3 hours and 4 hours, find the areas between the z-scores. Lastly, calculate the z-score for the mean of the 11 Millers' times and find the probability using the standard normal distribution table.

Explanation:

To find the answer to part (a), we need to calculate the z-score for Prof. Miller's time and find the area under the standard normal curve to the right of that z-score. The z-score formula is (x - μ) / σ, where x is the value, μ is the mean, and σ is the standard deviation. For part (b), we need to find the time corresponding to the 85th percentile. To do this, we need to find the z-score that corresponds to the 85th percentile and use the formula z = (x - μ) / σ to solve for x. Part (c) involves finding the area under the standard normal curve between two z-scores. Lastly, for part (d), we need to calculate the z-score for the given mean of 240 minutes and find the probability of getting a mean of the 11 Millers' times more than that using the z-score and standard normal distribution table.

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