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.

1. 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.)   
2. 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)

Answer 1

1)here mean =5*60+17=317 seconds

and std deviation =12 seconds

for 5th percentile critical value of z=		-1.645
therefore corresponding value=mean+z*std deviation=			297.3 seconds

2)

for normal distribution z score =(X-μ)/σ
here mean=       μ=	322
std deviation   =σ=	12.0000

proportion of boys from this region who qualify to run in this event in the state meet. :

probability =

P(X<297.3)

=

P(Z<-2.06)=

0.0198