Question

The times of the finishers in the New York City 10km run are normally distributed with mean of 61 minutes and standard deviation ? minutes. It is known that 70% of finishers have a finish time greater than 60 minutes. Let ?denote the finishing time for finishers in this race.

Note: Show your R codes/output.

Note: You can use the functions pnorm() or qnorm() in R to help you in solving the following parts.

The function pnorm(), compute probabilities from known bounding values. The function qnorm() , aims to do the opposite: given an area, find the boundary value that determines this area.

a) (2 points) Find the standard deviation of the finishing time (?).
Note: Provide the R code and output for the z-value or finding area under the standard normal curve.

b) (2 points) In 2013 approximately 7748 individuals took part in the run. A random sample of 9 individuals is drawn and their finishing times are recorded. Assuming everyone finished the run. What is the probability that among the 9 finishers selected their average finishing time is greater than 59 minutes. Note: Mention the R code and output for the z-value or finding area under the standard normal curve. Do the calculations for 3 decimal points.

c) (2 points) A second, independent sample of individuals is drawn from this population. How large of a sample must be drawn if the probability that the average finishing time is less than 62 must be 80%? Note: Show the R code and output for the z-value or for finding area under the standard normal curve.

Answer 1

ao If t denote the time to complete the 10 km run then, we can approximate it's distribution by Normal distribution. here given mean off is 61 and standard deviation thas ~ Normal (61,6) a) given that PCX>60)=0-7 1-DX560)=0-> => PCX560103 • with using the command "anorm 10-3)" will find the 2-score corresponding to 0.3 which is -0.5244 and a-score corresponding to 60 is thate Ż = .60-6861 = -0.52 uu =) 6- ośmun = 1907 b) first will find the probability that X a) sa i. pcx35992 HP(xs 5a)= 0.853 than If y denotes averaverage time for the a selected runners, then y follows Normal (61, 1.907 ) thes, P(Y>5a)=t-Pryssa)= 0.999 upi-Poom (52,6), 1.907)" if Zdenotas c) het n be the sample size then ) the average finishing time, then Za Nonnal (6), We want Plz<62) = 0.8 =) Z-score corresponding this probability - "Enorm (0-8)" = 0.8416 and z-score corresponding to 62% is 62 -1 = 0.8416 * 0.8416*1.907 = n = 2.5723 2) Una 0.8416*1.907