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2. The number of hits on a popular Web page follows a Poisson process with an arrival rate of 5 per minute. One begins observ
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ANSWER:

Question 2 (a) Here

X(t) be the number of hits on the webpage by time t

so here the distribution of the number of hits in the first minute is poisson distribution with parameter \lambda = 5 * 1 = 5

so here p(x) = e-5 5x/x!  ; x >0

Pr(x < 4) = POISSON (x =< 4 ; 5)

by using poissoncdf function we can get this probability.

Pr(x < 4) = POISSONCDF (x =< 4 ; 5) = 0.4405

(b) Here the distribution of the time untile the first hit is an exponential distribution with the parameter \lambda = 1/5 = 0.2 min-1

so here the pdf is

f(x) = 0.2 e-0.2x ; x > 0

cdf of the distribution is

F(x) = 1 - e-0.2x ; x > 0

so here we have to find the probability that the time till the first hit exceeds 10 seconds

Pr(x > 10 second) = 1 - P(x < 10 seconds) = 1- (1 - e-0.2* 1/6) = 0.9672

(c) Here the appropriate distribution for the time untile he 4th hit has gamma distribution with parameter is \alpha = 4 and \beta = 1/ 5 = 0.2

so here as x ~ Gamma( 4, 0.2)

Now we have to find the probability with the help of GAMMADIST function

P(x > 24 seconds) = 1 - GAMMADIST(x < 0.4 mins; 4 ; 0.2) = 1- 0.1429 = 0.8571

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