Question

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?

Short answer

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

The probability of no murder in a day is equal to 0.4025.

It can be said that many days have no murders as the probability is quite close to 0.5.

Step-by-step solution

Given information

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

Mean

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.

Probability

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.