Question 1196756: All runners in the 2005 Fox Cities Half Marathon had a mean finishing time of 137 minutes and a standard deviation of 30 minutes. All runners in the 2005 Chicago Marathon had a mean finishing time of 266 minutes and a standard deviation of 48 minutes.
1. Herschel ran in both races - completing the Fox Cities Half Marathon in 88 minutes and the Chicago Marathon in 214 minutes. In which race did he run better relative to the other finishers in each race?
2. The finishing times for the Chicago Marathon had a bell-shaped distribution. What percent of the finishers would be expected to have finishing times above 170 minutes?
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Problem 1
mu = mean
sigma = standard deviation
Fox Cities: mu = 137, sigma = 30
Chicago: mu = 266, sigma = 48
Let's find how well Herschel did relative to the other runners in the Fox Cities Marathon.
Compute the z score for the raw score x = 88
Use the mu and sigma values mentioned for this particular city's marathon.
z = (x-mu)/sigma
z = (88-137)/30
z = -1.633 approximately
The negative z score tells us that Herschel's finishing time is below average. How far below average? Roughly 1.633 standard deviations below. Each standard deviation representing 30 minutes. So if for instance he got z = -2, then he'd be two chunks of 30 minutes below average. Keep in mind that being below the average is actually a good thing in this context. The smaller the value, the faster the finish time, and hence the faster the runner. So we're aiming to get z as small as possible.
Now let's repeat the same idea for Chicago's marathon.
He finished with a time of x = 214 minutes. Use the mu and sigma values for this marathon.
z = (x - mu)/sigma
z = (214 - 266)/48
z = -1.083 approximately
He finished faster than the average runner in the Chicago marathon, but not as fast compared to the previous race. Once again, we're aiming to get z as small as possible.
Answer: Fox Cities Half Marathon
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Problem 2
Compute the z score for a raw score of x = 170
Use the mu and sigma values for Chicago mentioned in problem 1
z = (x - mu)/sigma
z = (170 - 266)/48
z = -2
We're exactly 2 standard deviations below the mean
So we're 2 chunks of 48, ie 2*48 = 96 minutes below the mean. Note how 266-96 = 170
The task of finding P(X > 170) is the same as P(Z > -2)
Use a Z table, such as this one,
https://www.ztable.net/
to find that P(Z < -2) = 0.02275 approximately
Then,
P(Z > -2) = 1 - P(Z < -2)
P(Z > -2) = 1 - 0.02275
P(Z > -2) = 0.97725
Roughly 97.725% of the runners finish above 170 minutes
If you were to use the Empirical Rule, then P(Z > -2) = 0.975 approximately which isn't too far from 0.97725
Answer: Either 97.725% or 97.5% (depending which method you use as mentioned earlier). These percentages are approximate.
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