To solve this, we can model the scenario using a Poisson distribution because it describes the probability of a given number of events happening in a fixed interval of time when these events occur with a known constant mean rate.

For a Poisson distribution, the probability of observing exactly $k$ events (in this case, 4 special orders) when the average rate is $\lambda$ (mean = 6.7 orders) is given by:

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

where:

• $\lambda = 6.7$
• $k = 4$
• $e$ is the base of the natural logarithm (approximately 2.71828).

Step-by-Step Solution

1. Substitute the values into the Poisson formula: $P(X = 4) = \frac{{e^{-6.7} \cdot 6.7^4}}{{4!}}$

2. Calculate $e^{-6.7}$: $e^{-6.7} \approx 0.0012337$

3. Calculate $6.7^4$: $6.7^4 = 2015.1361$

4. Calculate $4!$: $4! = 24$

5. Combine all parts to get the probability: $P(X = 4) = \frac{{0.0012337 \times 2015.1361}}{{24}} \approx 0.1036$

Answer:

The probability that exactly 4 special orders are sent out on a given day is approximately 0.1036.

Would you like further details on the Poisson distribution or additional examples?

Here are 5 related questions to deepen your understanding:

1. What is the probability of receiving exactly 5 special orders on a day?
2. How would the probability change if the mean increased to 7 orders per day?
3. What is the probability of receiving more than 4 special orders on a given day?
4. If the mean rate of special orders changes daily, would the Poisson distribution still apply?
5. How can the Poisson distribution be used to model events other than special orders?

Tip: The Poisson distribution is particularly useful for events that occur independently and at a constant average rate over time or space.