According to a 2017 survey conducted by the technology market research firm The Radicati Group, U.S. office workers receive an average of 121 e-mails per day (Entrepreneur magazine website). Assume the number of e-mails received per hour follows a Poisson distribution and that the average number of e-mails received per hour is five.

a.  What is the probability of receiving no e-mails during an hour (to 4 decimals)?


b.  What is the probability of receiving at least three e-mails during an hour (to 4 decimals)? For this question, if calculating the probability manually make sure to carry at least 4 decimal digits in your calculations.


c.  What is the expected number of e-mails received during 15 minutes (to 2 decimals)?


d.  What is the probability that no e-mails are received during 15 minutes (to 4 decimals)?

### Given Information

- The average number of e-mails received per hour is 5.
- The number of e-mails received follows a Poisson distribution.

### Poisson Distribution Formula
The Poisson probability of observing \( k \) events in a given time period is:

\[
P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}
\]

Where:
- \( \lambda \) is the average number of events (e-mails in this case) in the given time period.
- \( k \) is the actual number of events observed.
- \( e \) is approximately 2.71828.

### a. Probability of Receiving No E-mails During an Hour

Here, \( \lambda = 5 \) e-mails per hour, and we need to find the probability of receiving no e-mails (\( k = 0 \)):

\[
P(X = 0) = \frac{e^{-5} \cdot 5^0}{0!} = e^{-5} \cdot 1 = e^{-5}
\]

Let's calculate this:

\[
P(X = 0) = e^{-5} \approx 0.0067
\]

So, the probability of receiving no e-mails during an hour is **0.0067**.

### b. Probability of Receiving At Least Three E-mails During an Hour

To find this, we first calculate the probabilities of receiving 0, 1, and 2 e-mails and subtract their sum from 1:

\[
P(X \geq 3) = 1 - P(X < 3) = 1 - (P(X = 0) + P(X = 1) + P(X = 2))
\]

We already calculated \( P(X = 0) = 0.0067 \).

Now, let's calculate \( P(X = 1) \) and \( P(X = 2) \):

\[
P(X = 1) = \frac{e^{-5} \cdot 5^1}{1!} = 5e^{-5} \approx 0.0337
\]

\[
P(X = 2) = \frac{e^{-5} \cdot 5^2}{2!} = \frac{25e^{-5}}{2} \approx 0.0842
\]

Now, sum these probabilities:

\[
P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) \approx 0.0067 + 0.0337 + 0.0842 = 0.1246
\]

So,

\[
P(X \geq 3) = 1 - 0.1246 \approx 0.8754
\]

Thus, the probability of receiving at least three e-mails during an hour is **0.8754**.

### c. Expected Number of E-mails Received During 15 Minutes

The rate of e-mails per hour is \( \lambda = 5 \).

To find the expected number of e-mails in 15 minutes, convert 15 minutes to hours:

\[
\lambda_{15 \text{ min}} = \lambda_{1 \text{ hour}} \times \frac{15}{60} = 5 \times \frac{1}{4} = 1.25
\]

Thus, the expected number of e-mails received during 15 minutes is **1.25**.

### d. Probability that No E-mails are Received During 15 Minutes

Here, \( \lambda = 1.25 \) e-mails in 15 minutes. The probability of receiving no e-mails (\( k = 0 \)):

\[
P(X = 0) = \frac{e^{-1.25} \cdot 1.25^0}{0!} = e^{-1.25}
\]

Let's calculate this:

\[
P(X = 0) = e^{-1.25} \approx 0.2865
\]

So, the probability that no e-mails are received during 15 minutes is **0.2865**.

Do you want more details, or do you have any questions?

Here are 5 related questions:
1. What is the probability of receiving exactly 5 e-mails in an hour?
2. How does the expected number of e-mails change if the average increases to 10 e-mails per hour?
3. What is the probability of receiving fewer than 2 e-mails in a 15-minute period?
4. What is the variance of the number of e-mails received during an hour?
5. How would the probabilities change if the time period was reduced to 10 minutes?

**Tip:** When dealing with Poisson distributions, remember that the time period and the average rate are crucial to determining probabilities.

Explore detailed probability calculations using Poisson distribution for analyzing e-mail reception rates. Learn the probabilities of receiving no e-mails, at least three e-mails, expected e-mails in 15 minutes, and more. Suitable for advanced high school students and beyond.

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