Question

1. Assume the number of births in a local hospital follows a Poisson distribution and averages 2.6 per day. (a) What is the probability that no births will occur today? (b) What is the probability that less than four births will occur today? (c) What is the probability that no more than one birth will occur in two days? 2. A particular intersection in Delaware is equipped with surveillance camera. The number of traffic tickets issued to drivers passing through the intersection follows the Poisson distribution and averages 4.5 per month (a) What is the probability that seven traffic tickets will be issued at the intersection next month? (b) What is the probability that less than three traffic tickets will be issued at the intern section next month? (c) What is the probability that five traffic tickets will be issued in two months?

Answer 1

Solution :

Given That :

The number of births in a local hospital follows a position distribution and averages 2.6 per day.

a) :

The probability that no births :

$The\ provided\ means\ is\ \lambda =2.6$

$We\ need\ to\ compute\ Pr(X\leq 0).Therefore,\ the\ following\ is\ obtained$

    $Pr(X\leq 0)=\sum_{i=0}^{0}Pr(X=i)$

  $=Pr(X=0)$

  $=0.0743$

b) :

The probability that less than the four births :

$The\ provided\ mean\ is\lambda =2.6$

$We\ need\ to\ compute\ Pr(X\leq 3).Therefore,\ the\ following\ is\ obtained$

  $Pr(X\leq 3)=\sum_{i=0}^{3}Pr(X=i)$

  $=Pr(X=0)+Pr(X=1)+Pr(X=2)+Pr(X=3)$

   $=0.0743+0.1931+0.251+0.2176$

  $=0.0736$

c) :

The probability that no more that one birth in two days :

$The\ provided\ mean\ is\lambda =2.6$

$We\ need\ to\ compute\ Pr(X\geq 2).Therefore,\ the\ following\ is\ obtained$

  $Pr(X\geq 2)=1-Pr(X<2)$

  $=1-Pr(X<1)$

  $=0.7326$