Computers and Technology, 07.05.2020 04:57 lanaasad7733

The number of hits on a certain website follows a Poisson distribution with a mean rate of 4 per minute. a. What is the probability that 5 messages are received in a given minute?b. What is the probability that 9 messages are received in 1.5 minutes?c. What is the probability that fewer than 3 messages are received in a period of 30 seconds?

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Answer from: andr8aa

A) The probability that 5 messages are received in a given minute is; 15.629%

B) The probability that 9 messages are received in 1.5 minutes is; 6.884%

C) The probability that fewer than 3 messages are received in a period of 30 seconds is; 67.668%

We are told that the number of hits follows a poisson distribution with mean of 4 per minutes.

Formula for poisson distribution is;

P(X = x) = $e^{-\lambda}(\frac{\lambda^{x}}{x!})$

where;

x is number of items in counting

λ is the mean rate

A) We want to find the probability that 5 messages are received in a given minute. Thus;

λ = 4 × 1 = 4

x = 5

Thus;

P(X = 5) = $e^{-4}(\frac{4^{5}}{5!})$

P(X = 5) = 0.15629

P(X = 5) = 15.629%

B) We want to find the probability that 9 messages are received in 1.5 minutes. Thus;

λ = 4 × 1.5 = 6

x = 9

Thus;

P(X = 9) = $e^{-6}(\frac{6^{9}}{9!})$

P(X = 9) = 0.06884

P(X = 9) = 6.884%

C) We want to find the probability that 3 messages are received in 30 seconds. 30 seconds is 0.5 minutes. Thus;

λ = 4 × 0.5 = 2

x = 3

Thus;

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)

From online Poisson distribution calculator, we have;

P(X < 3) = 0.13534 + 0.27067 + 0.27067

P(X < 3) = 0.67668

P(X < 3) = 67.668%

Read more about Poisson distribution at; link

Answer from: princessa15266

a) $P(X=5) = 15.63 \%$

b) $P(X=9) = 6.88 \%$

c) P(X

Explanation:

a) What is the probability that 5 messages are received in a given minute?

The Poisson distribution is given by

$P(X=x) = e^{-\lambda}\frac{\lambda^{x}}{x!}$

Where x is the number of messages received in some time interval t.

mean rate = λ = 4 per minute

number of messages = x = 5

$P(X=x) = e^{-\lambda}\frac{\lambda^{x}}{x!}\\\\P(X=5) = e^{-4}\frac{4^{5}}{5!}\\\\P(X=5) = 0.1563\\\\P(X=5) = 15.63 \%$

b) What is the probability that 9 messages are received in 1.5 minutes?

mean rate = λ = 4*1.5 = 6

number of messages = x = 9

$P(X=x) = e^{-\lambda}\frac{\lambda^{x}}{x!}\\\\P(X=9) = e^{-6}\frac{6^{9}}{9!}\\\\P(X=9) = 0.0688\\\\P(X=9) = 6.88 \%$

c) What is the probability that fewer than 3 messages are received in a period of 30 seconds?

mean rate = λ = 0.5*4 = 2

number of messages = x < 3

P(X

Answer from: Quest

the correct answer is: a.)it is never appropriate to use acronyms and emoticons in business communication.

Answer from: Quest

kuikmnbk; ujgkjb; u65416+98532

9+5

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