Question

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.

Answer #1

Mean/Expected number of events of interest: λ = 4

poisson probability distribution |

P(X=x) = e^{-λ}λ^{x}/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)*p^{x}*(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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