2. Suppose p is number of cars per minute passing through a certain road junction between 5 PM and 6PM, 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?
A. 8/(2e^3)
B. 9/(2e^3)
C. 17/(2e^3)
D. 26/(2e^3)
Ans : (C)
Explanation :
Formula for Poisson distribution P(p = n) = (e^-u)(u^n)/n!
Here, Mean is 3. i.e., u = 3
Probability of observing fewer than 3 cars is = P(p < 3)
P(p < 3) = P(p = 0)+ P(p = 1) + P(p = 2)
= (e^-3) (3^0)/0! + (e^-3) (3^1)/1!+ (e^-3) (3^2)/2!
= (e^-3) + (e^-3)(3)/(1)+ (e^-3) (9)/(2)
= (1+3+(9/2)) (e^-3)
= 17/2e^3
Hence the answer is C.
A. 8/(2e^3)
B. 9/(2e^3)
C. 17/(2e^3)
D. 26/(2e^3)
Ans : (C)
Explanation :
Formula for Poisson distribution P(p = n) = (e^-u)(u^n)/n!
Here, Mean is 3. i.e., u = 3
Probability of observing fewer than 3 cars is = P(p < 3)
P(p < 3) = P(p = 0)
= (e^-3) (3^0)/0! + (e^-3) (3^1)/1!
= (e^-3) + (e^-3)(3)/(1)
= (1+3+(9/2)) (e^-3)
= 17/2e^3
Hence the answer is C.
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