The number of automobiles entering a tunnel per 2-minute period follows a Poisson distribution. The mean number of automobiles entering a tunnel per 2-minute period is four. (A) Find the probability that the number of automobiles entering the tunnel during a 2- minute period exceeds one. (B) Assume that the tunnel is observed during four 2-minute intervals, thus giving 4 independent observations, X1, X2, X3, X4, on a Poisson random variable. Find the probability that the number of automobiles entering the tunnel during a 2-minute period exceeds one during at least one of the four 2-minute intervals.
Mean/Expected number of events of interest: λ = 4
poisson probability distribution |
P(X=x) = e-λλx/x! |
X | P(X) |
0 | 0.0183 |
1 | 0.0733 |
p(x >1) = 1- (p(0) +p(1))
= 1-( 0.0183 + 0.0733)
=0.9084
......
b)
p = 0.9084
n= 4
now it is binomial distribution :
P(X=x) = C(n,x)*px*(1-p)(n-x) |
P ( X = 0) = C (4,0) * 0.9084^0 * ( 1 - 0.9084)^4=
0.0001
p(x>= 1) = 1- p(0)
= 1- 0.0001
= 0.9999
................
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