Question

2.[34] During an ultra-marathon, 20 runners cross the finish line between 11 am and 5 pm on race day, The number of competitors that finish at any given time during that period can be approximated using the Poisson distribution a) Suppose a photographer who is taking photos as runners finish the race wants to take a 20-minute break. What is the probability that she will miss at least one athlete finishing? [8] What is the expected number of runners who will cross the finish line during her break? [4] c) The photographer has a What is the probability that assuming she photogra Pher has a 15% chance of photographing someone smiling as they cross the finish line. ability that she will get 3 or more photos of smiling competitors during her shift ng she photographs all 20 runners)? [10]

Answer 1

There are 6 hours between 11am-5pm

This means that there are 6*3 = 18 slots of 20 mins

20 runners cross the finish line during this time

So in 20 mins, (20/18) runners cross the finish line.

PDF Poisson distribution is: P(X=x) = e^-(20/18)*(20/18)^x/x!

a) Now using Poisson distribution,

We want, P(X>=1) = 1- P(X=0) = 1- e^-(20/18) = 0.6708

b) Expected value of a Poisson distribution is the parameter lamda here it is 20/18 = 1.111

So approximately 1 person will cross the finish line during the break.

c) Let X=no of smiling photographs that the photographer gets.

We have n= 20

p= 0.15

So, X~Binomial(20,0.15)

PDF of this is, P(X=x) = 20Cx(0.15)^x*(0.85)^20-x

Where x= 0,1,2,....20

We want, P(X>=3) = 1- P(X=0) - P(X=1) - P(X=2)

Putting the respective values in the PDF above and calculating we get, P(X>=3) = 0.5951

d) Probability that a person is smiling in their photograph = 0.15

Probability of a smiling person buying their photograph = 0.9

Probability of a person not smiling in their photograph = 0.85

Probability of a person who's not smiling buying = 0.75

So the probability of a person buying their photograph = 0.15*0.9 + 0.85*0.75 = 0.7725

Probability that the p