To solve the problem, we need to use the Poisson distribution formula, which is given by:

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

Where:

• $P(X = k)$ is the probability of exactly $k$ events occurring.
• $\lambda$ is the average number of events in the given time period.
• $k$ is the number of events we're interested in (in this case, 7).
• $e$ is the base of the natural logarithm, approximately equal to 2.71828.

Step 1: Determine the effective rate ($\lambda$)

The problem states that the average number of persons arriving per hour is 6. We need to calculate the average number of arrivals during a 90-minute period. Since 90 minutes is 1.5 hours, we multiply the rate by 1.5:

$\lambda = 6 \times 1.5 = 9$

Step 2: Apply the Poisson distribution formula

Now, we calculate the probability of exactly 7 people arriving ($k = 7$):

$P(X = 7) = \frac{e^{-9} \times 9^7}{7!}$

Step 3: Compute the values

Let's calculate this step by step.

• $9^7 = 4782969$
• $7! = 5040$
• $e^{-9}$ (approximately 0.0001234098)

Now, substitute these into the formula:

$P(X = 7) = \frac{0.0001234098 \times 4782969}{5040} \approx 0.117$

Final Answer:

The probability that exactly 7 people will arrive during any 90-minute period is approximately 0.117.

Would you like more details or have any questions?

Here are some related questions:

1. What is the cumulative probability for at most 7 people arriving?
2. How does the Poisson distribution apply to other real-world scenarios?
3. What is the probability that fewer than 7 people arrive in 90 minutes?
4. How would the probability change if the period were only 30 minutes?
5. What assumptions underlie the use of the Poisson distribution in this context?

Tip: The Poisson distribution is particularly useful for modeling the number of events happening within a fixed interval of time or space when those events occur independently.