This problem can be solved using the Poisson distribution, which is commonly used to model the number of events happening within a fixed period of time when the events occur independently.

The formula for the Poisson distribution is:

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

where:

• $X$ is the random variable representing the number of typographical errors.
• $\lambda$ is the average number of events (in this case, the average number of errors per hour).
• $k$ is the actual number of events you want the probability for (in this case, 6 errors).
• $e$ is the mathematical constant approximately equal to 2.71828.

Given Data:

• The average number of errors per hour, $\lambda = 3$.
• We are looking for the probability of exactly 6 errors, so $k = 6$.

Now, we can plug these values into the Poisson formula:

$P(X = 6) = \frac{e^{-3} \cdot 3^6}{6!}$

Let's calculate this.

First, calculate $6!$ and $e^{-3}$.The probability that the court stenographer makes exactly 6 typographical errors during an hour-long court case is approximately 0.0504, or 5.04%.

Would you like to dive into more details about this calculation or ask further questions?

Here are 5 related questions to expand on this topic:

1. What is the probability that the stenographer makes fewer than 6 errors?
2. How would the probability change if the average number of errors per hour increased to 4?
3. What is the probability that the stenographer makes no errors during the hour?
4. How does the Poisson distribution differ from the binomial distribution?
5. What are the assumptions behind using the Poisson distribution?

Tip: The Poisson distribution is ideal for events happening independently and at a constant average rate over a fixed period.