91Ó°ÊÓ

It is given that the author wrote 181 checks in a year.

The number of checks that the author wrote is equal to 181.

The number of days in a year is equal to 365.

The mean number of checks per year is computed below:

\(\begin{aligned}{c}\mu = \frac{{{\rm{Number}}\;{\rm{of}}\,{\rm{checks}}}}{{{\rm{Number}}\;{\rm{of}}\;{\rm{days}}}}\\ = \frac{{181}}{{365}}\\ = 0.49589\\ \approx 0.50\end{aligned}\)

Thus, the mean number of checks per day is equal to 0.50.

Let X be the number of checks written in a day.

Here, X follows a Poisson distribution with mean equal to\({\kern 1pt} \mu = 0.50\).

The formula of Poisson distribution is:

\[P\left( {X = x} \right) = \frac{{{\mu ^x}{e^{ - \mu }}}}{{x!}}\]

The probability that the author wrote at least 1 check per day is computed below:

\[\begin{aligned}{c}P\left( {X \ge 1} \right) = 1 - P\left( {X < 1} \right)\\ = 1 - P\left( {X = 0} \right)\\ = 1 - \frac{{{{\left( {0.50} \right)}^0}{{\left( {2.71828} \right)}^{ - 0.50}}}}{{0!}}\\ = 1 - 0.6065\end{aligned}\]

\[ = 0.3935\]

Therefore, the probability that the author wrote at least 1 check per day is equal to 0.3935.