Respuesta :
Answer
a) Probability that there are 3 drug overdoses in a random night = 0.055
b) Probability that there are at most 3 drug overdoses in a random
night = 0.087
Explanation
The Poisson distribution will be used to solve this one.
The Poisson distribution is given as
[tex]P(X=x)=\frac{(e^{-\mu})(\mu^x)}{x!}[/tex]where
x = actual variable whose probability we want to find = 3 drug overdoses in one night
μ = mean of the distribution = 6.9 drug overdoses per night
[tex]\begin{gathered} P(X=x)=\frac{(e^{-\mu})(\mu^x)}{x!} \\ P(X=3)=\frac{(e^{-6.9})(6.9^3)}{3!}=0.055 \end{gathered}[/tex]b) For the probability of at most 3 drug overdoses in one night,
P (X ≤ 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
For each of them,
μ = 6.9
But for x, we will calculate for each of them using
x = 0, x = 1, x = 2, x = 3
[tex]\begin{gathered} P(X=x)=\frac{(e^{-\mu})(\mu^x)}{x!} \\ P(X=0)=\frac{(e^{-6.9})(6.9^0)}{0!}=0.00101 \\ P(X=1)=\frac{(e^{-6.9})(6.9^1)}{1!}=0.00695 \\ P(X=2)=\frac{(e^{-6.9})(6.9^2)}{2!}=0.02399 \\ P(X=3)=\frac{(e^{-6.9})(6.9^3)}{3!}=0.0552 \end{gathered}[/tex]P (X ≤ 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
P (X ≤ 3) = 0.00101 + 0.00695 + 0.02399 + 0.0552 = 0.08714 = 0.087
Hope this Helps!!!
