Respuesta :
Answer:
a) P(0) = 0.6824
b) P(1) = 0.2608
c) P(2) = 0.0498
d) P(3) = 0.00635
Step-by-step explanation:
Number of units (n) in the data, n = 280
Total number of events, k = 107
The probability using Poisson probabilities is calculated using the formula;
P(x) = [tex]\frac{e^{-\lambda}\lambda^x}{x!}[/tex]
here,
Poisson parameter, λ = [tex]\frac{107}{280}[/tex]
Thus,
a) P(0) = [tex]\frac{e^{-\frac{107}{280}}(\frac{107}{208})^0}{0!}[/tex]
or
P(0) = 0.6824
b) P(1) = [tex]\frac{e^{-\frac{107}{280}}(\frac{107}{280})^1}{1!}[/tex]
or
P(1) = 0.2608
c) P(2) = [tex]\frac{e^{-\frac{107}{280}}(\frac{107}{280})^2}{2!}[/tex]
or
P(2) = 0.0498
d) P(3) = [tex]\frac{e^{-\frac{107}{280}}(\frac{107}{280})^3}{3!}[/tex]
or
P(3) = 0.00635
Using the Poisson distribution, we find that:
a) P(0) = 0.6824.
b) P(1) = 0.2608.
c) P(2) = 0.0498.
d) P(3) = 0.0064.
Poisson distribution:
[tex]P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}[/tex]
The parameters are:
- x is the number of successes.
- [tex]\mu[/tex] is the mean in the given interval.
In this problem, there are 107 homicides in 280 city-periods, thus:
[tex]\mu = \frac{107}{280} = 0.3821[/tex]
Item a:
[tex]P(0) = P(X = 0) = \frac{e^{-0.3821}(0.3821)^{0}}{(0)!} = 0.6824[/tex]
Item b:
[tex]P(1) = P(X = 1) = \frac{e^{-0.3821}(0.3821)^{1}}{(1)!} = 0.2608[/tex]
Item c:
[tex]P(2) = P(X = 2) = \frac{e^{-0.3821}(0.3821)^{2}}{(2)!} = 0.0498[/tex]
Item d:
[tex]P(3) = P(X = 3) = \frac{e^{-0.3821}(0.3821)^{3}}{(3)!} = 0.0064[/tex]
A similar problem is given at https://brainly.com/question/13971530
