A company is preparing to launch a new web site to sell its products. From past

Question

A company is preparing to launch a new web site to sell its products. From past experience, they know that purchases at the site will occur as a Poisson process with on average 20 purchases every 10 minutes.

a) What is the expected time in minutes before the first purchase after the site opens? Give your answer to four decimal places

b) What is the probability that more than 0.27 minutes pass before the first purchase? Give your answer to four decimal places.

c) What is the expected time in minutes until the eighth purchase? Give your answer to four decimal places. Correct: Your answer is correct.

d) What is the probability that the eighth purchase will occur between 3.5 minutes and 4.5 minutes after the site opens? Give your answer to four decimal places.

Explanation / Answer

a) the expected time in minutes before the first purchase after the site opens

=1/(20/10) = 10/20= 1/2 = 0.5000

b)the probability that more than 0.27 minutes pass before the first purchase

= e^-rt = e^-20/10(.27) = e^-.55 = 0.5827

C)the expected time in minutes until the eighth purchase  

1/(20/10) * 8= 80/20 = 4.0000 minutes

d) the probability that the eighth purchase will occur between 3.5 minutes and 4.5 minutes after the site opens

Note that the chi-squared distribution with 2 degrees of freedom has the exponential distribution with rate 1/2
Thus, the fifth purchase has the shape of the chi-squared distribution with 10 degrees of freedom.
Then, as the rate is 20/10, and we must transform this to a rate of 1/2, we multiply our times by 20/10/(1/2) = 4.0
Then, Chisq.dist(3.5*4.0,10,1) - chisq.dist(4.5*4.0,10,1) = 0.1729 -0.0549

= 0.1180

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