Question

help me solve this please .! ..
The number of traffic accidents that occur on a particular stretch of road during a month follows a Poisson distribution with a mean of 8.9. Find the probability that less than three accidents will occur next month on this stretch of road.
Answer choices..>>>

0.022777

0.977223

0.006752

0.993248

Answer 1

You should calculate that. p(0) + p(1) + p(2)

Answer 2

?? im confused i have no idea what im doing

Answer 3

For the Poisson Distribution, \(p(n) = \dfrac{e^{-\lambda}\cdot \lambda^{n}}{n!}\).

Does that look familiar?

Answer 4

very familiar, but im not sure how to plug in

Answer 5

Simply evaluate p(0), p(1), and p(2). You'll be done after you add them up.

Answer 6

\(\lambda=8.9\) -- Just a property of this Poisson Distribution, since you are given the Mean.

Answer 7

hmm ok i got .90 something so im thinking its either B or D>?

Answer 8

e^{-8.9} = 0.000136388926482

\(\dfrac{\lambda^{0}}{0!} = 1\)

\(\dfrac{\lambda^{1}}{1!} = 8.9\)

\(\dfrac{\lambda^{2}}{2!} = 39.605\)

0.000136388926482*(1 + 8.9 + 39.605) = 0.00675

How did you manage ".90 something"? Frankly, the mean is 8.9. I sincerely hope we don't get 0.90 in the first three probabilities.