There is an average of four accidents per month at an intersection. Assume that

Question

There is an average of four accidents per month at an intersection. Assume that a month has 30 days. (a) What is the probability of three or more accidents next month? (b) What is the probability of two or more accidents in a 23-day period? (c) What is the probability of no accidents during the next seven days?.
There is an average of four accidents per month at an intersection. Assume that a month has 30 days. (a) What is the probability of three or more accidents next month? (b) What is the probability of two or more accidents in a 23-day period? (c) What is the probability of no accidents during the next seven days?.
There is an average of four accidents per month at an intersection. Assume that a month has 30 days. (a) What is the probability of three or more accidents next month? (b) What is the probability of two or more accidents in a 23-day period? (c) What is the probability of no accidents during the next seven days?.

Explanation / Answer

Ans:

a)

P(x>=3)=1-P(x<=2)=1-e-4*(40/0!+41/1!+42/2!)

=1-e-4*(1+4+8)

=1-0.2381=0.7619

b)on average there will be (4/30)*23=3.067 accidents in 23 days

P(x>=2)=1-e-3.067*(3.0670/0!+3.0671/1!)

=1-e-3.067*(1+3.067)

=1-0.1894=0.8106

c)there will be on average (4/30)*7=0.933 accidents in 7 days

P(x=0)=e-0.933*(0.9330/0!)=0.3934

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