Suppose \(p\) is the number of cars per minute passing through a certain road…
2013
Suppose \(p\) is the number of cars per minute passing through a certain road junction between 5 PM and 6 PM, and \(p\) has a Poisson distribution with mean 3. What is the probability of observing fewer than 3 cars during any given minute in this interval?
Answer: C. \(17/(2e^3)\) — We want P(X < 3) for X ~ Poisson(mean = 3), so compute P(X=0)+P(X=1)+P(X=2). P(X=0) = e^-3 * 3^0 / 0! = e^-3 ≈ 0.0497871 P(X=1) = e^-3 * 3^1 / 1! = 3 e^-3 ≈…
- A.
\(8/(2e^3)\) - B.
\(9/(2e^3)\) - C.
\(17/(2e^3)\) - D.
\(26/(2e^3)\)
Attempted by 57 students.
Show answer & explanation
Correct answer: C
We want P(X < 3) for X ~ Poisson(mean = 3), so compute P(X=0)+P(X=1)+P(X=2).
P(X=0) = e^-3 * 3^0 / 0! = e^-3 ≈ 0.0497871
P(X=1) = e^-3 * 3^1 / 1! = 3 e^-3 ≈ 0.1493613
P(X=2) = e^-3 * 3^2 / 2! = (9/2) e^-3 = 4.5 e^-3 ≈ 0.2240420
Sum: P(X<3) = e^-3(1 + 3 + 9/2) = (17/2) e^-3 = 17/(2 e^3) ≈ 0.42319