Question

Suppose the Bolt9000 robot was built to run the 100m dash. Its finishing times of the race are normally distributed, with a mean of 10 seconds, and a standard deviation of 0.1 seconds.
(a) What is the range in which the middle 80% of finishing times of the robot would fall?

(b) What is the probability that the robot’s finishing times of its next two races will be within 0.2 seconds of each other?

(c) If you enter a 100 m race with seven Bolt9000 robots and you finish with a time of 10.20 seconds,what is the probability that you will finish last (i.e. in 8th place)?

Answer 1

a)

for middle 80% values critical z =-/+1.28

therefore corresponding values =mean -/+ z*std deviation =10-/+ 1.28*0.1 =9.872 ; 10.128

b)

let Y is time difference ; therefore mean of Y =10-10 =0

and std deviation =sqrt(0.1²+0.1²) =0.1414

hence P(Y is within 0.2 seconds)=P(-0.2 <Y<0.2)=P((-0.2-0)/0.1414<Z,<(0.2-0)/0.1414)

=P(-1.41<Z<1.41)=0.9207-0.0793 =0.8414

c)

probability that one finsh before 10.20 seconds =P(X<10.2)=P(Z<(10.2-10)/0.1)=P(Z<2)=0.9772

hence  probability that you will finish last =P(all other 7 finish before 10.2 seconds)=(0.9772)⁷

=0.8509