Question

The number of accidents that occur at a busy intersection is Poisson distributed with a mean of 4.5 per week. Find the probability of the following events.
a. No accidents occur in one week.
b. 5 or more accidents occur in a week.
c. One accident occurs today.

Answer 1

Answer:

a) 0.01111

b) 0.4679

c) 0.33747

Step-by-step explanation:

We are given the following in the question:

The number of accidents per week can be treated as a Poisson distribution.

Mean number of accidents per week = 4.5

$\lambda= 4.5$

Formula:

$P(X =k) = \displaystyle\frac{\lambda^k e^{-\lambda}}{k!}\\\\ \lambda \text{ is the mean of the distribution}$

a) No accidents occur in one week.

$P(x =0)\\\\= \displaystyle\frac{4.5^0 e^{-4.5}}{0!}= 0.01111$

b) 5 or more accidents occur in a week.

$P( x \geq 5) = 1-\displaystyle \sum P(x$

c) One accident occurs today.

The mean number of accidents per day is given by

$\lambda = \dfrac{4.5}{7} = 0.64$

$P(x =1)\\\\= \displaystyle\frac{0.64^1 e^{-0.64}}{1!}= 0.33747$