Question

A couple of cyclists are racing down a path in adjacent and parallel lanes. At time t = 0, the first cyclist is ahead of the second one by 1 km. The mean speed of the first cyclist (km/hr) is 40 with standard deviation 1 and the mean speed of the second cyclist is 39 with standard deviation 1. Both cyclists’ speeds are normally distributed.

a) After 2 hours of cycling, what is the probability that the second cyclist has not caught up to the first?

b) What is the probability that the cyclists are separated by at most 1 km after 2 hours?

Come up with a linear combination for a difference between random variables using both speeds

Answer 1

a)

here let x1 and x2 is distance travelled by first and second cyclists,

as difference in distance after 2 hours =x1-x2+1

here expected mean difference =2*(40-39)+1=2+1 =3

and std deviation =sqrt((2*1)²+(2*1)²)=2.828

hence probability that the second cyclist has not caught up to the first =P(X>0)=P(Z>(0-3)/2.828)=P(Z>-1.06)=0.8554

b)

probability that the cyclists are separated by at most 1 km after 2 hours =P(-1<X<1)

=P((-1-3)/2.828<X<(1-3)/2.828)=P(-1.41 <Z<-0.71)=0.2389-0.0793 =0.1596