Question

The average time to run the 5K fun run is 20 minutes and the standard deviation is 2.4 minutes. 8 runners are randomly selected to run the 5K fun run. Round all answers to 4 decimal places where possible and assume a normal distribution.

What is the distribution of X X ? X X ~ N(,)

What is the distribution of ¯ x x¯ ? ¯ x x¯ ~ N(,)

What is the distribution of ∑ x ∑x ? ∑ x ∑x ~ N(,)

If one randomly selected runner is timed, find the probability that this runner's time will be between 20.3272 and 21.1272 minutes. For the 8 runners, find the probability that their average time is between 20.3272 and 21.1272 minutes.

Find the probability that the randomly selected 8 person team will have a total time more than 152.8.

For part e) and f), is the assumption of normal necessary? NoYes

The top 15% of all 8 person team relay races will compete in the championship round. These are the 15% lowest times. What is the longest total time that a relay team can have and still make it to the championship round? minutes

Answer 1

a) X~ N(20,2.4)

b) Xbar~ N(20,2.4/sqrt(8))

c) sum(X)~N(8*20,sqrt(8*2.4^2))

=N(160,6.79)

d) If one randomly selected runner is timed, find the probability that this runner's time will be between 20.3272 and 21.1272 minutes.

p(20.3272 < X < 21.1272)

=NORMDIST(21.1272,20,2.4,1)-NORMDIST(20.3272,20,2.4,1)

=0.1264

e) For the 8 runners, find the probability that their average time is between 20.3272 and 21.1272 minutes.

we have to find p(20.3272 < Xbar < 21.1272)

=NORMDIST(20.21.1272,20,(2.4/sqrt(8)),1)-NORMDIST(20.3272,20,(2.4/sqrt(8)),1)

=NORMDIST(21.1272,20,(2.4/SQRT(8)),1)-NORMDIST(20.3272,20,(2.4/SQRT(8)),1)

=0.2578

f)

Find the probability that the randomly selected 8 person team will have a total time more than 152.8.

p(sum(x)>152.8)

=1-NORMDIST(152.8,160,(2.4*SQRT(8)),1)

=0.8555

g) Yes,the assumption of normal is necessary.

h)Y=sum(x)

p(Y<=M)=0.15

p((Y-160)/sqrt(8)*2.4<(M-160)/(sqrt(8)*2.4))=0.15

in Normal table p(Z<-1.0364)=0.15

Hence M=160-sqrt(8)*2.4*1.0364

M=152.96