A survey of 30-year-old males provided data on the number of auto accidents in the previous 5 years. The sample mean is 1.3 accidents per male. Test the hypothesis that the number of accidents follows a Poisson distribution at the 5% level of significance. No. of accident No. of males
0 39
1 22
2 14
3 11
>=4 4

Required:
a. What's the Expected probability of finding males with 0 accidents?
b. What's the Expected probability of finding males with 4 or more accidents?

0.2725

0.0431

Step-by-step explanation:

The distribution here is a poisson distribution :

λ = 1.3

The poisson distribution :

p(x) = [(e^-λ * λ^x)] ÷ x!

Expected probability of finding male with 0 accident ; x = 0

p(0) = [(e^-1.3 * 1.3^0)] ÷ 0!

p(0) = [0.2725317 * 1] ÷ 1

p(0) = 0.2725317

= 0.2725

2.)

P(x ≥ 4) = 1 - P(x < 4)

P(x < 4) = p(x = 0) + p(x. = 1) + p(x = 2) + p(x = 3)

p(x = 0) =  p(0) = [(e^-1.3 * 1.3^0)] ÷ 0! = 0.2725

p(x = 1) = p(1) = [(e^-1.3 * 1.3^1)] ÷ 1! = 0.35429

p(x = 2) = p(2) = [(e^-1.3 * 1.3^2)] ÷ 2! = 0.23029 p(x = 3) = p(3) = [(e^-1.3 * 1.3^3)] ÷ 0! = 0.09979

P(x < 4) = 0.2725 + 0.35429 + 0.23029 + 0.09979 = 0.95687

P(x ≥ 4) = 1 - 0.95687 = 0.0431

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