The probability that, for any day, the number of special orders sent out will be exactly 3 is 0.1633.
The Poisson distribution is a discrete probability distribution that describes the likelihood of a specific number of events occurring in a specified span of time or space, if these events occur at a constant mean rate and regardless of the time since the last occurrence.
Let X ~ Pois(λ)
Then we have:
E(X) = λ = Var(X)
Since standard deviation is the square root (positive) of variance,
Thus,
The standard deviation of X = [tex]\sqrt{\lambda}[/tex]
Its probability function is given by
[tex]f(k; λ) = Pr(X = k) = \dfrac{\lambda^{k}e^{-\lambda}}{k!}[/tex]
Given the mean is 4.2.
We have to find the probability that on any day, the number of special orders sent out will be exactly 3. Therefore, the value of x will be 5.
Using the Poisson distribution,
[tex]P(x) = \dfrac{e^{-\lambda}\lambda ^x}{x!}\\\\P(5) =\dfrac{e^{-4.2}(4.2)^5}{5!}\\\\P(5) = 0.1633[/tex]
Hence, the probability that, for any day, the number of special orders sent out will be exactly 3 is 0.1633.
Learn more about Poisson distribution here:
https://brainly.com/question/7879375
#SPJ1
