Question

Data from 14 cities were combined for a​ 20-year period, and the total 280 ​city-years included a total of 107 homicides. After finding the mean number of homicides per​ city-year, find the probability that a randomly selected​ city-year has the following numbers of​ homicides, then compare the actual results to those expected by using the Poisson​ probabilities:

P(0)=
P(1)=

Answer 1

P(0) = 0.6825

P(1) = 0.2607

Explanation:

From the given information, the number of homicide is 107 and the total number of homicides per city –year is 280.

Let us denote the number of homicides per city-year as X.

The mean value, X is calculated as:

\begin{array}{c}\\{\rm{Mean}} = \frac{{107}}{{280}}\\\\ = 0.382\\\end{array}

Mean= 107/280 = 0.382

The mean number of homicides per city- year \left( {\lambda = \mu } \right)(λ=μ) is 0.382.

a. The probability that zero homicides is obtained is as below:

\begin{array}{c}\\P\left( {X = 0} \right) = \frac{{{e^{ -0.382}}{{\left( {0.382} \right)}^0}}}{{0!}}\\\\ = \frac{{\left( {0.6825 \right)\left( {\rm{1}} \right)}}{1}\\\\ = 0.6825\\\end{array}

P(X=0) = e −0.382 (0.382)⁰​/1

= (0.6825)(1) /1

P(X=0) = 0.6825

Thus, the probability that zero homicides P(0) is 0.6825.

b. The probability that one homicides is obtained is as below:

\begin{array}{c}\\P\left( {X = 0} \right) = \frac{{{e^{ -0.382}}{{\left( {0.382} \right)}^0}}}{{0!}}\\\\ = \frac{{\left( {0.6825 \right)\left( {\rm{1}} \right)}}{1}\\\\ = 0.6825\\\end{array}

P(X=1) = e −0.382 (0.382)¹​/1

= (0.6825)(0.382)/1

P(X=1) = 0.2607

Thus, the probability that zero homicides P(0) is 0.2607.

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