Question

In a recent​ year, an author wrote 169 checks. Use the Poisson distribution to find the probability​ that, on a randomly selected​ day, he wrote at least one check.

Answer 1

We should first calculate the average number of checks he wrote per day.  To do that, divide 169 by 365 (the number of days in a year) and you get (rounded) 0.463.  This will be λ in our Poisson distribution.  Our formula is
$P(X=k)= \frac{ \lambda ^{k}-e^{-\lambda} }{k!}$.  We want to evaluate this formula for X≥1, so first we must evaluate our case at k=0.  
$P(X=0)= \frac{0.463 ^{0}-e ^{-0.463} }{0!} \\ = \frac{1-e ^{-0.463} }{1} =0.3706$
To find P(X≥1), we find 1-P(X<1).  Since the author cannot write a negative number of checks, this means we are finding 1-P(X=0).  Therefore we have 1-0.3706=0.6294.
There is a 63% chance that the author will write a check on any given day in the year.