uppose that a category of world class runners are known to run a marathon (26 miles) in an average of 143 minutes with a standard deviation of 12 minutes. Consider 49 of the races.

Let
X
= the average times of the 49 races.

(a) Find the median of the average running times. ( That is, find the median of the sampling distribution of the average running times.)

Correct: Your answer is correct.
min

(b) Find the slowest 10 percent of the average running times. ( Round the answer to the second decimal place.)
min

(c) Find the fastest 10 percent of the average running times. ( Round the answer to the second decimal place.)
min

To find the answer to these questions, we need to use the concept of the sampling distribution of the mean. In this case, we are interested in the average running times of 49 races.

(a) The median of the sampling distribution of the average running times is equal to the mean of the population. Given that the average running time is 143 minutes, the median of the average running times is also 143 minutes.

(b) To find the slowest 10 percent of the average running times, we need to find the cutoff point below which 10 percent of the values lie.

First, we need to find the z-score corresponding to the 10th percentile. We can use the z-table or a calculator to find that the z-score for a cumulative probability of 0.10 is approximately -1.28.

Next, we can use the formula for z-score: z = (X - 𝜇) / (𝜎 / √n), where X is the value we want to find, 𝜇 is the population mean, 𝜎 is the population standard deviation, and n is the sample size (in this case, 49).

Rearranging the formula, we get X = 𝜇 + (z * 𝜎 / √n).

Plugging in the values, X = 143 + (-1.28 * 12 / √49) = 143 - 1.79 = 141.21.

So, the slowest 10 percent of the average running times is 141.21 minutes (rounded to the second decimal place).

(c) Similarly, to find the fastest 10 percent of the average running times, we need to find the cutoff point above which 90 percent of the values lie.

We can use the z-score for a cumulative probability of 0.90, which is approximately 1.28.

Using the same formula as above, X = 143 + (1.28 * 12 / √49) = 143 + 1.79 = 144.79.

So, the fastest 10 percent of the average running times is 144.79 minutes (rounded to the second decimal place).