Question

The time it takes for a certain athlete to complete 500 jumps on the rope follows a normal distribution with an expected value of 2.5 minutes and a standard deviation of 0.16 minutes.

a) Calculate the probability that in the next attempt the athlete takes less than 2.5 minutes to complete the 500 jumps.

b) Assuming that the athlete can perform at least 10 repetitions with the same stamina, what is the probability that he can complete 2000 jumps in less than 9.5 minutes?

c) If 10 athletes with the same characteristics perform the exercise simultaneously, what is the probability that less than 4 complete the 500 jumps before 2.25 minutes?

d) If the athlete wants to have at least a 10% probability of completing the 500 jumps in less than 2.2 minutes, how much should he lower his average?

Answer 1

a)

for normal distribution z score =(X-μ)/σx
mean μ=	2.5
standard deviation σ=	0.16

probability that in the next attempt the athlete takes less than 2.5 minutes to complete the 500 jumps :

probability =P(X<2.5)=(Z<(2.5-2.5)/0.16)=P(Z<0)=0.5

b)

expected time for 2000 jumps =2000*2.5/500 =10 minutes

and standard deviation =0.16*√4 =0.32

probability that he can complete 2000 jumps in less than 9.5 minutes :

probability =P(X<9.5)=(Z<(9.5-10)/0.32)=P(Z<-1.56)=0.0591

c)

probability =P(X<2.25)=(Z<(2.25-2.5)/0.16)=P(Z<(-1.5625)=0.0591

here for 10 athletes , this is binomial with parameter n=10 and p=0.5

probability that less than 4 complete the 500 jumps before 2.25 minutes :

P(X<=3)=

∑x=0a     (nCx)px(1−p)(n-x)    =

0.9981

d)

for 10th percentile critical value of z=-1.28

therefore corresponding average =x-z*standard deviaiton =2.2-(-1.28)*0.16 =2.40 minutes