Answer: 0.5848
Step-by-step explanation:
The formula of probability for Poisson distribution for random variable x :-
[tex]P(x)=\dfrac{e^{\lambda}\lambda^x}{x!}[/tex], where [tex]\lambda[/tex] is the mean of the distribution .
Given : The auto parts department of an automotive dealership sends out a mean of 3.3 special orders daily.
[tex]\lambda=3.3[/tex]
[tex]P(\leq5)=\dfrac{e^{-3.3}(3.3)^0}{0!}+\dfrac{e^{-3.3}(3.3)^1}{1!}\dfrac{e^{-3.3}(3.3)^2}{2!}+\dfrac{e^{-3.3}(3.3)^3}{3!}+\dfrac{e^{-3.3}(3.3)^4}{4!}+\dfrac{e^{-3.3}(3.3)^5}{5!}\\\\=0.5847772874\approx0.5848[/tex]
Hence, the probability that, for any day, the number of special orders sent out will be no more than 5 = 0.5848
