Question

The number of messages that arrive at a Web site is a Poisson distributed random variable with a mean of 6 messages per hour. Round your answers to four decimal places (e.g. 98.7654). (a) What is the probability that 5 messages are received in 1 hour? (b) What is the probability that 10 messages are received in 1.5 hours? (c) What is the probability that less than 2 messages are received in 1/2 hour?

Answer 1

Answer:

a) There is a 16.0623% probability that 5 messages are received in 1 hour.

b) There is a 11.5880% probability that 10 messages are received in 1.5 hours.

c) There is a 22.4042% probability that 2 messages are received in 0.5 hours.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

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

In which

x is the number of sucesses

$e = 2.71828$ is the Euler number

$\mu$ is the mean in the given time interval.

In this problem, we have a mean of 6 messages per hour.

(a) What is the probability that 5 messages are received in 1 hour?

Find the value of P when $x = 5$ and $\mu = 6$

So

$P(X = 5) = \frac{e^{-6}*(6)^{5}}{(5)!} = 0.160623$

There is a 16.0623% probability that 5 messages are received in 1 hour.

(b) What is the probability that 10 messages are received in 1.5 hours?

The mean is 6 messages in one hour.

For 1.5 hours, the mean is 6*1.5 = 9 messages.

So

We have to find the value of P when $x = 10$ and $\mu = 9$.

$P(X = 10) = \frac{e^{-9}*(9)^{10}}{(10)!} = 0.115880$

There is a 11.5880% probability that 10 messages are received in 1.5 hours.

(c) What is the probability that less than 2 messages are received in 1/2 hour?

The mean is 6 messages in one hour.

For 0.5 hours, the mean is 6*0.5 = 3 messages.

So

We have to find the value of P when $x = 2$ and $\mu = 3$.

$P(X = 10) = \frac{e^{-3}*(3)^{2}}{(2)!} = 0.224042$

There is a 22.4042% probability that 2 messages are received in 0.5 hours.