Question

Question 6 of 21 5 Points Assume that students arrive at a professor's office following a Poisson distribution with a mean (u) = 5 per hour. What is the probability that during a 3 hour period we get more than 3 students in the first hour, less than 3 in the second hour and exactly 3 in the last hour? A. None of these multiple choice options are correct B. 1-poisson(3,15,1))*poisson(2,15,1) poisson(3,15,0) C. (1-poisson(3,5,1))*poisson(2,5,1)*poisson(3,5,0)*6 c D. (1-poisson(3,5,1))*poisson(2,5,1)*poisson(3,5,0) O E. (1-poisson(3,5,1))*poisson(2,5,1)*poisson(3,5,0)*3 OF. poisson(9,5,1) Reset Selection

Answer 1

In each hour the arrival of student is following Poisson distribution with paraeter (5).  

For Poisson distribution, the parameter = mean of variable.

P( more than 3 students arrival in first hour, less than 3 students arrival in second hour and exactly 3 students arrival in third hour) = P( more than 3 students arrival in first hour)P(less than 3 students arrival in second hour)P( exactly 3 students arrival in third hour) [assuming arrivals in each hour are independent]

In an hour if X denote number of students arrival,

$P(X=x)=\frac{e^{-5}5^{x}}{x!},x=0.1,2,...$

Hence the required probability is

P(X>3)P(X<3)P(X=3) =

=$P(X>3)P(X\leq 2)P(X=3)$

=

(1 - P(X <3)] P(X < 2)P(X = 3)

Let us denote,

$P(X\leq x)=poisson(x,5,1)$ , where 5 is the parameter of Poisson distribution and 1 for denoting the cumulative probability.

and

$P(X=x)=poisson(x,5,0)$, where 5 is the parameter of Poisson distribution and 0 for  denoting the probability mass at x.

Now the require probability,

P(X>3)P(X<3)P(X=3) which is

  

(1 - P(X <3)] P(X < 2)P(X = 3)

can be denoted as

$\left [1-poisson(3,5,1) \right ]poisson(2,5,1)poisson(3,5,0)$

Hence the answer is Option D.