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In the context of the problem, the mean number of events refers to the average number of births per day at NYU-Langone Medical Center. To find this mean, we divide the total number of births for the year (4221) by the number of days in a year (365).
This can be represented mathematically as: \[\text{Mean} = \frac{4221}{365} \lambda = 11.56\ \text{(rounded to 2 decimal places)}\ \] where \(\lambda\) is the mean number of births per day.

This value tells us, on average, how many births occur each day at the medical center. In real-world terms, it means that we expect around 11.56 births every day throughout the year.
Understanding the mean number of events is crucial as it forms the base of our Poisson distribution calculation, allowing us to predict specific probabilities for different numbers of events occurring in a given time period.

Once we have the mean number of events (\(\lambda\)), the next step is to calculate the probability of observing a specific number of events, such as 15 births in a single day.
In this case, our interest is to find the probability that there will be exactly 15 births on a particular day. This is where the Poisson distribution formula comes into play.

The formula is: \ \[P(X = k) = \frac{e^{-\text{\lambda}} \times \text{\lambda}^k}{k!}\ \] In this formula

• \(X\) is the random variable representing the number of events (births)
• \(k\) is the number of events we want to find the probability for (15 births)
• \(e\) is the base of the natural logarithm (approximately equal to 2.71828)
• \(k!\) is the factorial of \(k\)

With \(\lambda = 11.56\) and \(k = 15\), we can plug these values into the formula to find the desired probability.

Now that we have our formula and the necessary values, we apply the Poisson distribution to find the probability of 15 births in a day.
Substituting in our values: \[P(X = 15) = \frac{e^{-11.56} \times 11.56^{15}}{15!}\text{ = 0.0425 (approximately)}\ \] This can be broken down into manageable parts. First, calculate \(e^{-11.56}\), then \(11.56^{15}\), and finally divide by \(15!\) (15 factorial).

Using a calculator or software:

• \(e^{-11.56} \approx 9.69 \times 10^{-6}\)
• \(11.56^{15} \approx 67237492.21\)
• \(15! \approx 1307674368000\)

Combining these steps: \[P(X = 15) = \frac{9.69\times 10^{-6} \times 67237492.21}{1307674368000} \approx 0.0425\ \] Hence, the probability is approximately 0.0425 or 4.25%.

The final step is to interpret the calculated probability. Our result shows a probability of 0.0425 (or 4.25%) for exactly 15 births occurring on a given day at the medical center.
This small probability suggests that while it's not impossible, it's relatively unlikely that there will be precisely 15 births on any specific day.

It's essential to understand that a Poisson distribution gives us probabilities for counts of events over a fixed period. In this case, it helps medical professionals prepare for birth rates that vary from the average. They can expect most days to have around the mean of 11.56 births, with fewer days having much higher or lower birth counts.
This insight can influence staffing and resource planning at the medical center, ensuring they can manage both typical and atypical days more effectively.