Question
please help
Ship collisions in the Houston Ship Channel are rare but follow a Poisson distribution. Suppose the mean number of collisions is 1.2 for any two-month period of time. What is the probability of having exactly one collision in a two-month period? If you need to use a value of e^x, use one of the following: e^-9 = 0.0001 e^-8 = 0.0003 e^-6 = 0.0025 e^-5 = 0.0067 e^-4 = 0.0183 e^-3 = 0.0498 e^-2 = 0.1353 e^-1 = 0.3679 e^-0.1 = 0.9048 e^-0.2 = 0.8187 e^-0.3 = 0.7408 e^-0.4 = 0.6703 e^-0.5 = 0.6065 e^-0.6 = 0.5488 e^-0.8 = 0.4493 e^-0.9 = 0.4066 e^-1.2 = 0.3012 e^-1.5 = 0.2231 e^-1.6 = 0.2019 e^-1.8 = 0.1653 e^-2.1 = 0.1225 e^-2.4 = 0.907 e^-2.5 = 0.0821 e^-2.3 = 0.0672 e^-3.2 = 0.0407 e^-1.6 = 0.0273 e^-4.8 = 0.0082 e^-1/3 = 0.7165 e^-2/4 = 0.7788 e^-1/6 = 0.8465 e^-1/8 = 0.8825 e^-1/9 = 0.8948 e^-2/3 = 0.5134 e^-3/4 = 0.4724 e^-3/6 = 0.4346 e^-3/8 = 0.6873 e^-5/8 = 0.5353 e^-7/8 = 0.4169 e^-4/3 = 0.2636 e^-5/3 = 0.1889 a. 2177 b. 3148 c. 5488 d. 4800 e. 3614 f. 3293
Explanation / Answer
a) We have mean = 1.2 for any two month
P(X = 1)
Use Excel function
POISSON.DIST(1,1.2,FALSE ) = 0.3614
Correct option is E
