Question

# The number of automobiles entering a tunnel per 2-minute period follows a Poisson distribution. The mean...

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

................

thanks

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