Leah's answering machine receives about twelve telephone calls between 8 a.m. and 10 a.m. What is the probability that Leah receives more than one call in the next 15 minutes?
Question 5 options:

.4422

.2034

.1834

.1734

To solve this problem, we can model the number of telephone calls as a **Poisson process** since the calls are likely occurring independently and randomly over time. The Poisson distribution is typically used for such problems where we are counting events that happen at a constant average rate in a fixed period of time.

### Step 1: Define the problem
- Leah receives 12 calls in 2 hours (120 minutes), so the rate \(\lambda\) of calls per minute is:
  \[
  \lambda = \frac{12 \text{ calls}}{120 \text{ minutes}} = 0.1 \text{ calls per minute}.
  \]
- For a 15-minute interval, the expected number of calls (mean) is:
  \[
  \lambda_{\text{15 minutes}} = 0.1 \times 15 = 1.5 \text{ calls}.
  \]
  
Thus, the number of calls in 15 minutes follows a Poisson distribution with a mean of 1.5.

### Step 2: Calculate the probability
We are asked to find the probability that Leah receives **more than one call** in the next 15 minutes. This is equivalent to calculating \( P(X > 1) \), where \( X \) is the number of calls, which follows a Poisson distribution with mean 1.5.

The probability that \( X \) takes a certain value in a Poisson distribution is given by:
\[
P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}
\]
where \( \lambda = 1.5 \) and \( k \) is the number of calls.

We need to find \( P(X > 1) \), which is:
\[
P(X > 1) = 1 - P(X = 0) - P(X = 1).
\]

First, calculate \( P(X = 0) \) and \( P(X = 1) \):

- \( P(X = 0) \):
  \[
  P(X = 0) = \frac{e^{-1.5} (1.5)^0}{0!} = e^{-1.5} \approx 0.2231.
  \]

- \( P(X = 1) \):
  \[
  P(X = 1) = \frac{e^{-1.5} (1.5)^1}{1!} = e^{-1.5} \times 1.5 \approx 0.3347.
  \]

### Step 3: Compute \( P(X > 1) \)
\[
P(X > 1) = 1 - P(X = 0) - P(X = 1)
= 1 - 0.2231 - 0.3347
= 0.4422.
\]

Thus, the probability that Leah receives more than one call in the next 15 minutes is approximately **0.4422**.

The correct answer is **0.4422**.

---

Do you want further details on this or have any other questions?

Here are 5 related questions to consider:
1. How would the probability change if Leah received more calls on average per hour?
2. What if we extended the time period to 30 minutes instead of 15 minutes?
3. How is the Poisson distribution different from a normal distribution?
4. Can we calculate the probability of receiving exactly 2 calls in the next 15 minutes?
5. What assumptions make this a valid use of the Poisson distribution?

**Tip**: When using a Poisson distribution, always check if the event rate is constant and events are independent over time.

Leah's answering machine receives about twelve telephone calls between 8 a.m. and 10 a.m. What is the probability that Leah receives more than one call in the next 15 minutes?

Explore the solution to finding the probability that Leah receives more than one call in the next 15 minutes using the Poisson distribution. This problem highlights key probability concepts, including event occurrence rates and the Poisson process. Suitable for students in Grades 10-12.

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