In Exercises 9–16, use the Poisson distribution to find the indicated probabilities.

Murders In a recent year, there were 333 murders in New York City. Find the mean number of murders per day, then use that result to find the probability that in a day, there are no murders. Does it appear that there are expected to be many days with no murders?

The total number of murders in a year in New York Cityis equal to 333.

The total number of murders in a year is given to be equal to 333.

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

The mean number of murders per day is equal to:

\(\begin{aligned}{c}\mu = \frac{{{\rm{Number}}\;{\rm{of}}\;{\rm{murders}}\;{\rm{in}}\;{\rm{the}}\;{\rm{year}}}}{{{\rm{Number}}\;{\rm{of}}\;{\rm{days}}\;{\rm{in}}\;{\rm{a}}\;{\rm{year}}}}\\ = \frac{{333}}{{365}}\\ = 0.91\end{aligned}\)

The mean number of murders per day is equal to 0.91.

Let X be the number of murders per day.

Here, X follows a Poisson distribution with mean equal to\({\kern 1pt} \mu = 0.91\).

The probability of no murderin day is computed below:

\[\begin{aligned}{c}P\left( x \right) = \frac{{{\mu ^x}{e^{ - \mu }}}}{{x!}}\\P\left( 0 \right) = \frac{{{{\left( {0.91} \right)}^0}{{\left( {2.71828} \right)}^{ - 0.91}}}}{{0!}}\\ = 0.4025\end{aligned}\]

Therefore, the probability ofnomurder in a day is equal to 0.4025.

Since the probability value is quite close to 0.5 (50% of the days in a year), it can be said that are many days that have no murders.