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.
