Question

According to a 2017 survey conducted by the technology market research firm The Radicati Group, U.S. office workers receive an average of 121 emails per day. Suppose for a particular office the number of emails received per hour follows a Poisson distribution and that the average number of emails received per hour is nine. (Round your answers to four decimal places.)
A. What is the probability of receiving no emails during an hour?B. What is the probability of receiving at least three emails during an hour?C. What is the expected number of emails received during 15 minutes?D. What is the probability that no emails are received during 15 minutes?

Answer 1

Answer:

a) 0.0065 = 0.65% probability of receiving no emails during an hour.

b) 0.1212 = 12.12% probability of receiving at least three emails during an hour

c) The expected number of emails received during 15 minutes is 1.2604.

d) 0.2835 = 28.35% probability that no emails are received during 15 minutes

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 question.

121 emails per day.

A day has 24 hours, so per hour, 121/24 = 5.0417, which means that $\mu = 5.0417n$, in which n is the number of hours.

A. What is the probability of receiving no emails during an hour?

n = 1, so $\mu = 5.0417$

This is P(X = 0).

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

$P(X = 0) = \frac{e^{-5.0417}*(5.0417)^{0}}{(0)!} = 0.0065$

0.0065 = 0.65% probability of receiving no emails during an hour

B. What is the probability of receiving at least three emails during an hour?

Either you receive less than three emails during an hour, or you receive at least 3. The sum of the probabilities of these events is decimal 1. So

$P(X < 3) + P(X \geq 3) = 1$

We want $P(X \geq 3)$

Then

$P(X \geq 3) = 1 - P(X < 3)$

In which

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

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

$P(X = 0) = \frac{e^{-5.0417}*(5.0417)^{0}}{(0)!} = 0.0065$

$P(X = 1) = \frac{e^{-5.0417}*(5.0417)^{1}}{(1)!} = 0.0326$

$P(X = 2) = \frac{e^{-5.0417}*(5.0417)^{2}}{(2)!} = 0.0821$

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0065 + 0.0326 + 0.0821 = 0.1212$

0.1212 = 12.12% probability of receiving at least three emails during an hour

C. What is the expected number of emails received during 15 minutes?

15 minutes is a fourth of an hour, so n = 1/4 and $\mu = \frac{5.0417}{4} = 1.2604$

The expected number of emails received during 15 minutes is 1.2604.

D. What is the probability that no emails are received during 15 minutes?

$P(X = 0) = \frac{e^{-1.2604}*(1.2604)^{0}}{(0)!} = 0.2835$

0.2835 = 28.35% probability that no emails are received during 15 minutes