Question

(Poisson/Exponential) The number of automobiles that arrive at a certain intersection per minute has a Poisson distribution with mean 3. Interest centers around the time that elapses before 9 automo- biles appear at the intersection.

What is the probability that more than 9 automobiles appear at the intersection during any given minute of time?

Answer 1

0.011

Step-by-step explanation:

Let X represents the number of automobiles that arrive at a certain intersection per minute

X follows a Poisson distribution with a mean of 3.

Calculating the probability that more than 9 automobiles appear at the intersection during any given minute of time;

P(X > 9) = 1 - P(X ≤ 9)

P(X>9) = 1 - ∑(e^-λ * λ^x)/x! For x = 0 to 9.

P(X>9) = 1 - (P(X=0) + P(X=1) +.....+P(X=9))

P(X>9) = 1 - ((e^-3 * 3^0)/0! + (e^-3 * 3^1)/1! + ...... + (e^-3 * 3^9)/9!))

P(X>9) = 1 - (0.049787068367863 + 0.149361205103591 + 0.224041807655387 + 0.224041807655387 + 0.168031355741540 + 0.100818813444924 + 0.050409406722462 + 0.021604031452483 + 0.008101511794681 + 0.002700503931560)

P(X>9) = 0.001102488130122

P(X>9) = 0.0011 --- Approximated

Answer 2

0.0110

Explanation:

This is a Poisson distribution problem

Poisson distribution formula is given as

P(X = x) = (e^-λ)(λˣ)/x!

where λ = mean = 3 automobiles per minute

x = variable whose probability is required = more than 9 automobiles in a minute

P(X>9) = 1 - P(X ≤ 9)

P(X ≤ x) = Σ (e^-λ)(λˣ)/x! (Summation From 0 to x)

P(X < 9) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5) + P(X=6) + P(X=7) + P(X=8) + P(X=9)

So, computing this one at a time, and summing it all up

P(X ≤ 9) = 0.99620

P(X>9) = 1 - P(X ≤ 9) = 1 - 0.99890 = 0.0110