Question

Data collected by emergency responders in a large city suggest that on averagethere are 6.9 drug overdoses per night.(a) What is the probability that there are 3 drug overdoses in a random night(please round to 3 decimal places)?(b) What is the probability that there are at most 3 drug overdoses in a randomnight (please round to 3 decimal places)?Quartion HelpMessage instructor

Answer 1

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

Step-by-step explanation

The Poisson distribution will be used to solve this one.

The Poisson distribution is given as

`P(X=x)=((e^(-\mu))(\mu^x))/(x!)`

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

`\begin{gathered} P(X=x)=((e^(-\mu))(\mu^x))/(x!) \\ P(X=3)=((e^(-6.9))(6.9^3))/(3!)=0.055 \end{gathered}`

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

`\begin{gathered} P(X=x)=((e^(-\mu))(\mu^x))/(x!) \\ P(X=0)=((e^(-6.9))(6.9^0))/(0!)=0.00101 \\ P(X=1)=((e^(-6.9))(6.9^1))/(1!)=0.00695 \\ P(X=2)=((e^(-6.9))(6.9^2))/(2!)=0.02399 \\ P(X=3)=((e^(-6.9))(6.9^3))/(3!)=0.0552 \end{gathered}`

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!!!