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The intersection of Easy Street and 69th Avenue averages 3 accidents every two days. Assuming the number of accidents follows a Poisson distribution:

What is the probability that there are exactly 8 accidents at that intersection in a week (7 days)?

Using Poisson's formula:

e^-9 * 9^8 / 8!

I got 9 since 3 accidents happens in two days so 3 * 3 equals a lambda of 9 in 6 days.

Got it wrong, what is the correct answer and solution?

Thinking of 10.5 since you add 1.5 from so that's the lambda for 7 days

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    Given that a week has 7 days, why do you figure the lambda for 6 days? – 2017-02-28
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    So would the lambda be 10.5 since that's half of the lambda of 3 to get 1 day which adds on to a total of 7 days – 2017-02-28
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    Yeah, an average of 3 accidents in two days works out to an average of 10.5 in 7 days. – 2017-02-28
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    So everything I did for that question is correct except for the lambda? – 2017-02-28
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    Yes. That is so. – 2017-02-28
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    I wish I can give you both more than a upvote, appreciate you both! – 2017-02-28

1 Answers 1

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An crucial first step in using the Poisson distribution is to use a rate $\lambda$ that exactly matches the question.

You are given that the accident rate for 2 days is $\lambda_2 = 3.$ So the rate for a week (7 days) is $\lambda_7 = 7(\lambda_2/2) = 10.5.$

Then the random variable $X$ that counts weekly accidents has $X \sim \mathsf{Pois}(\lambda_7 = 10.5),$ and you seek $P(X = 8) = e^{-10.5}10.5^8/8!.$ I will leave it to you do finish.

Note: You are on the right track, but you used $\lambda_6 = 9$ instead of $\lambda_7 = 10.5.$