The number of views on a web site follows a Poisson distribution with an average of 1.5 views per minute. (a) What is the probability of no views in a minute? (b) What is the probability of two or fewer views in 10 minutes? (c) Determine the length of a time interval such that the probability of no views in an interval of this length is 0.001.

The Poisson distribution is typically used to model the number of events occurring within a fixed period of time. It is defined by the following probability mass function:

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

where:
- \( X \) is the random variable representing the number of events,
- \( k \) is the number of occurrences,
- \( \lambda \) is the average rate of occurrences (mean) over the specified interval,
- \( e \approx 2.71828 \) is Euler's number.

Given that the average number of views per minute is 1.5, we can proceed to solve each part:

### (a) Probability of no views in a minute
We are asked to find \( P(X = 0) \) when the average number of views per minute \( \lambda = 1.5 \).

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

### (b) Probability of two or fewer views in 10 minutes
First, determine the new rate \( \lambda \) over a 10-minute interval. Since \( \lambda = 1.5 \) views per minute, the total rate over 10 minutes is:

\[
\lambda = 1.5 \times 10 = 15
\]

Now, we want \( P(X \leq 2) \) for \( \lambda = 15 \).

\[
P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)
\]

Using the Poisson formula:

\[
P(X = k) = \frac{15^k e^{-15}}{k!}
\]

Calculating each term:

\[
P(X = 0) = \frac{15^0 e^{-15}}{0!} = e^{-15}
\]
\[
P(X = 1) = \frac{15^1 e^{-15}}{1!} = 15e^{-15}
\]
\[
P(X = 2) = \frac{15^2 e^{-15}}{2!} = \frac{225e^{-15}}{2}
\]

Now, summing them up:

\[
P(X \leq 2) = e^{-15} + 15e^{-15} + \frac{225e^{-15}}{2} = e^{-15}(1 + 15 + 112.5) = e^{-15}(128.5) \approx 3.61 \times 10^{-5}
\]

### (c) Time interval such that the probability of no views is 0.001
Let \( t \) be the length of the time interval in minutes. The probability of no views in this interval is given by:

\[
P(X = 0) = e^{-\lambda}
\]

where \( \lambda \) is the average number of views in \( t \) minutes. Since there are 1.5 views per minute, the rate over \( t \) minutes is:

\[
\lambda = 1.5t
\]

We want \( P(X = 0) = 0.001 \).

\[
e^{-1.5t} = 0.001
\]

Taking the natural logarithm on both sides:

\[
-1.5t = \ln(0.001)
\]

\[
t = -\frac{\ln(0.001)}{1.5} \approx \frac{6.907755}{1.5} \approx 4.6052 \text{ minutes}
\]

So, the length of time such that the probability of no views is 0.001 is approximately 4.6052 minutes.

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Would you like more details or have any questions? 

Here are 5 related questions:

1. How would the probabilities change if the average rate of views per minute increased?
2. What is the probability of getting exactly 5 views in a 10-minute interval?
3. How would you calculate the probability of getting more than 10 views in a 5-minute interval?
4. If the probability of at least one view in a minute is required to be 0.9, what should be the new average rate of views?
5. How would the results change if a different time interval, say 20 minutes, is considered?

**Tip:** When using the Poisson distribution, always ensure that the time interval and the average rate match. Adjust the rate based on the desired time period for accurate results.

Explore a detailed solution to a Poisson distribution problem involving website views. Learn how to calculate the probability of no views in a minute, two or fewer views in 10 minutes, and the time interval such that the probability of no views is 0.001. Suitable for students in Grades 11-12 or college-level courses.

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