Answer in Statistics and Probability for Swati malik #87028

The probability distribution of the Poisson random variable X, representing the number of outcomes occurring

in a given time interval or specified region denoted by t, is

“p(x; lambda t)={e^{-lambda t}(lambda t)^x over {x!}}”

Use the Poisson distribution with 

“lambda t=2”

i. No car will arrive 

“P(X=0)={e^{-2}(2)^0 over {0!}}=e^{-2}approx0.135335”

ii. At least two cars will arrive 

“P(Xge2)=1-(P(X=0)+P(X=1))=”

“=1-({e^{-2}(2)^0 over {0!}}+{e^{-2}(2)^1 over {1!}})=1-e^{-2}-2e^{-2}=1-3e^{-2}approx0.593994”

iii. At the most 3 cars will arrive 

“P(Xle3)=P(X=0)+P(X=1)+P(X=2)+P(X=3)=”

“={e^{-2}(2)^0 over {0!}}+{e^{-2}(2)^1 over {1!}}+{e^{-2}(2)^2 over {2!}}+{e^{-2}(2)^3 over {3!}}=”

“={19 over 3}e^{-2}approx0.857123”

iv. Between 1 and 3 cars will arrive

“P(1le Xle3)=P(X=1)+P(X=2)+P(X=3)=”

“={e^{-2}(2)^1 over {1!}}+{e^{-2}(2)^2 over {2!}}+{e^{-2}(2)^3 over {3!}}={16 over 3}e^{-2}approx0.721788”