 Suppose that the number of drivers who travel between a particular origin and destination during a designated time period has a Poisson distribution with parameter 𝜇 = 20 (suggested in the article "Dynamic Ride Sharing: Theory and Practice"†). (Round your answer to three decimal places.)
(a)
What is the probability that the number of drivers will be at most 15?
(b)
What is the probability that the number of drivers will exceed 29?
(c)
What is the probability that the number of drivers will be between 15 and 29, inclusive?
What is the probability that the number of drivers will be strictly between 15 and 29?
(d)
What is the probability that the number of drivers will be within 2 standard deviations of the mean value? 

To solve these problems involving the Poisson distribution, let's first summarize the key information and recall some important formulas.

The Poisson distribution is defined by the probability mass function:

\[
P(X = k) = \frac{e^{-\mu} \mu^k}{k!}
\]
where:
- \( \mu \) is the average rate (mean) of occurrences (in this case, \( \mu = 20 \)),
- \( k \) is the number of occurrences (drivers in this case).

The standard deviation for a Poisson distribution is the square root of the mean, i.e., \( \sigma = \sqrt{\mu} \).

Here, \( \mu = 20 \), so \( \sigma = \sqrt{20} \approx 4.472 \).

Now we can address each part:

---

### (a) **What is the probability that the number of drivers will be at most 15?**

This is asking for the cumulative probability of the number of drivers being less than or equal to 15:

\[
P(X \leq 15) = \sum_{k=0}^{15} P(X = k)
\]

We'll use a Poisson cumulative distribution function (CDF) to compute this.

---

### (b) **What is the probability that the number of drivers will exceed 29?**

This is asking for the probability of more than 29 drivers:

\[
P(X > 29) = 1 - P(X \leq 29)
\]

We can compute \( P(X \leq 29) \) using the Poisson CDF and subtract that from 1.

---

### (c) **What is the probability that the number of drivers will be between 15 and 29, inclusive?**
This is the probability that \( X \) lies between 15 and 29:

\[
P(15 \leq X \leq 29) = P(X \leq 29) - P(X < 15)
\]

Similarly, we can use the CDF to find these values.

**What is the probability that the number of drivers will be strictly between 15 and 29?**

This excludes the boundaries, so:

\[
P(15 < X < 29) = P(X \leq 28) - P(X \leq 15)
\]

---

### (d) **What is the probability that the number of drivers will be within 2 standard deviations of the mean value?**

The mean is \( \mu = 20 \), and two standard deviations is \( 2\sigma = 2 \times 4.472 \approx 8.944 \).

Thus, we are interested in the range:

\[
20 - 8.944 \leq X \leq 20 + 8.944
\]
This gives us approximately \( 11 \leq X \leq 29 \).

So we want to calculate:

\[
P(11 \leq X \leq 29) = P(X \leq 29) - P(X < 11)
\]

Let me calculate these probabilities now.Here are the computed probabilities:

### (a) Probability that the number of drivers will be at most 15:
\[
P(X \leq 15) = 0.157 \text{ (rounded to three decimal places)}
\]

### (b) Probability that the number of drivers will exceed 29:
\[
P(X > 29) = 0.022
\]

### (c) Probability that the number of drivers will be between 15 and 29 (inclusive):
\[
P(15 \leq X \leq 29) = 0.873
\]

Probability that the number of drivers will be strictly between 15 and 29 (excluding the boundaries):
\[
P(15 < X < 29) = 0.809
\]

### (d) Probability that the number of drivers will be within 2 standard deviations of the mean value (between 11 and 29):
\[
P(11 \leq X \leq 29) = 0.955
\]

Would you like more details or have any questions on any of these?

Here are 5 related questions for further exploration:
1. What is the probability that exactly 20 drivers will travel between the origin and destination?
2. How does the Poisson distribution change if the mean number of drivers is doubled?
3. What is the variance of the number of drivers in this Poisson distribution?
4. How does the standard deviation of a Poisson distribution relate to the mean?
5. Can you find the median number of drivers given this Poisson distribution?

**Tip:** In a Poisson distribution, the variance is always equal to the mean.

Suppose that the number of drivers who travel between a particular origin and destination during a designated time period has a Poisson distribution with parameter 𝜇 = 20 (suggested in the article 'Dynamic Ride Sharing: Theory and Practice'). (Round your answer to three decimal places.)
(a) What is the probability that the number of drivers will be at most 15?
(b) What is the probability that the number of drivers will exceed 29?
(c) What is the probability that the number of drivers will be between 15 and 29, inclusive?
What is the probability that the number of drivers will be strictly between 15 and 29?
(d) What is the probability that the number of drivers will be within 2 standard deviations of the mean value?

Solve a Poisson distribution problem where the number of drivers between an origin and destination follows a Poisson distribution with mean μ = 20. Learn to calculate probabilities for various ranges of drivers, using key formulas and standard deviation concepts. Suitable for advanced high school or university students.

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