To find the probability that the author wrote at least one check on a randomly selected day using the Poisson distribution, follow these steps:
1. Calculate the Average Rate (λ):
First, we need to determine the average number of checks written per day. Given that the author wrote 189 checks in a year and there are 365 days in a year, the average rate (λ) is calculated as follows:
[tex]\[
\lambda = \frac{189 \text{ checks}}{365 \text{ days}} \approx 0.5178082191780822
\][/tex]
2. Poisson Distribution:
The Poisson distribution formula is:
[tex]\[
P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!}
\][/tex]
where [tex]\( P(X = k) \)[/tex] represents the probability of [tex]\( k \)[/tex] events occurring in a fixed interval.
3. Probability of No Checks Written (k = 0):
We need to find the probability of the author writing no checks on a given day (i.e., [tex]\( k = 0 \)[/tex]). Using the Poisson formula:
[tex]\[
P(X = 0) = \frac{0.5178082191780822^0 \cdot e^{-0.5178082191780822}}{0!}
\][/tex]
Simplifying further:
[tex]\[
P(X = 0) = e^{-0.5178082191780822} \approx 0.595825035758911
\][/tex]
4. Probability of At Least One Check:
To find the probability of writing at least one check, we use the complement rule. The complement of writing no checks ([tex]\( P(X = 0) \)[/tex]) gives us the probability of writing at least one check ([tex]\( P(X \geq 1) \)[/tex]):
[tex]\[
P(X \geq 1) = 1 - P(X = 0)
\][/tex]
Substituting the value of [tex]\( P(X = 0) \)[/tex]:
[tex]\[
P(X \geq 1) = 1 - 0.595825035758911 \approx 0.404
\][/tex]
5. Rounding to Three Decimal Places:
Finally, we round the result to three decimal places:
[tex]\[
P(X \geq 1) \approx 0.404
\][/tex]
Therefore, the probability that the author wrote at least one check on a randomly selected day is approximately [tex]\( \boxed{0.404} \)[/tex].
