Suppose the number of emails in an hour from panicked students the day before ho

Question

Suppose the number of emails in an hour from panicked students the day before homework is due follows a Poisson process with an average of 2 per 10 minutes. How many emails should be expected in 10 minutes? Find the probability of no emails in a 10 minute period. Find the probability of 3 emails in a 10 minute period if there were no emails received in the prior 10 minutes. Find the probability of 3 emails in a 30 minute period. e. Suppose a random sample of 36, non-overlapping 10 minute periods is selected. What's the approximate distribution of the average number of emails received in a 10 minute period? What's the probability the average number of emails received in a 10 minute period is over 2.5 emails? What's the probability the average number of emails received in a 10 minute period is between 1 and 2 emails?

Explanation / Answer

Probability of Solution:-

a) The expected emails in 10 minutes = 2 emails

Average number of successes within a given region is

= 2 (For 10 minutes)

b) The probability of no emails in 10 minutes period = 0.1353

P(x; ) = (e-) (x) / x!

P(x = 0) = 0.1353

c) The probability of 3 emails in 10 minutes period if there were no emails recieved in the prior 10 minutes is 0.0244.

Probility of getting 0 mails in first 10 minutes = 0.1353

Probability of getting 3 mails in next 10 minutes
P(x; ) = (e-) (x) / x!

P(x = 3) = 0.1804
The probability of 3 emails in 10 minutes period if there were no emails recieved in the prior 10 minutes = 0.1804 × 0.1353 = 0.0244

d) The probability of 3 emails in 30 minutes period is 0.08923.
Average number of successes within a given region in 30 minutes = 3 × 2 = 6

P(x; ) = (e-) (x) / x!

P(x =3 ) = 0.08923

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