The average finishing time among all high school boys in a particular track even

Question

The average finishing time among all high school boys in a particular track event in a certain state is 5 minutes 17 seconds. Times are normally distributed with standard deviation 12 seconds.

a. The qualifying time in this event for participation in the state meet is to be set so that only the fastest 5% of all runners qualify. Find the qualifying time in seconds (round it to the closest second). (Hint: Convert minutes to seconds.)   

b. In the western region of the state the times of all boys running in this event are normally distributed with standard deviation 12 seconds, but with mean 5 minutes 22 seconds. Find the proportion of boys from this region who qualify to run in this event in the state meet. (Hint: normalcdf)

Explanation / Answer

Mean = 5min17sec = 317 s

Stdev = 12 s

a. Qualifying time is basically the top 5percetile cutoff

Z for .05 is -1.645 from the Z tables, So, Z = -1.645

Fastest 5% is given by Mean + Z*Stdev = 317-1.645*12 = 297.26 s or 297 seconds

Rounding off to the closest second = 297 seconds

b. So, 297.26 is the cutoff time. Now we have to see the area above the 336.74 seconds in the distribution with params of (5min22s, 12s)

So, Z = (297.26-322)/12 = -2.06

The area under curve for P(Z<=-2.06) = .0197

or 0.0197 of boys from west region will qualify to run in this event in the state meet

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