The average number of cars per minute is: `mu=300/60=5`

(a) `P(x_0)=(e^-5 5^0)/(0!)=6.7379xx10^-3`

(b) Expected number each 2 minutes = E(X) = 5 × 2 = 10

(c) Now, with μ = 10, we have: `P(x_10)=(e^-10 10^10)/(10!)=0.12511`

Histogram of Probabilities

Based on the function

`P(X)=(e^-10 10^x)/(x!)`

we can plot a histogram of the probabilities for the number of cars for each 2 minute period:

Poisson distribution histogram

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