Data from 14 cities were combined for a​ 20-year period, and the total 280 ​city-years included a total of 107 homicides. After

Answer:

P(0) = 0.6825

P(1) = 0.2607

Step-by-step explanation:

Based on the provided data, there were 107 homicide incidents and a total of 280 city-years recorded.

Let us denote the variable representing homicides per city-year as X.

The mean value of X can be calculated with the following formula:

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

Mean=   107/280 = 0.382

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

a. To find the probability of no homicides occurring:

\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 chance of experiencing zero homicides P(0) is 0.6825.

b. To determine the probability of one homicide occurring:

\begin{array}{c}\\P\left( {X = 1} \right) = \frac{{{e^{ -0.382}}{{\left( {0.382} \right)}^1}}}{{1!}}\\\\ = \frac{{\left( {0.6825 \right)\left( {0.382} \right)}}{1}\\\\ = 0.2607\\\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 of having one homicide P(1) is 0.2607.

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