Question

The auto parts department of an automotive dealership sends out a mean of 3.3 special orders daily. What is the probability that, for any day, the number of special orders sent out will be no more than 5? Round your answer to four decimal places.

Answer 1

Answer: 0.5848

Step-by-step explanation:

The formula of probability for Poisson distribution for random variable x :-

$P(x)=\dfrac{e^{\lambda}\lambda^x}{x!}$, where $\lambda$ 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.

$\lambda=3.3$

$P(\leq5)=\dfrac{e^{-3.3}(3.3)^0}{0!}+\dfrac{e^{-3.3}(3.3)^1}{1!}\dfrac{e^{-3.3}(3.3)^2}{2!}+\dfrac{e^{-3.3}(3.3)^3}{3!}+\dfrac{e^{-3.3}(3.3)^4}{4!}+\dfrac{e^{-3.3}(3.3)^5}{5!}\\\\=0.5847772874\approx0.5848$

Hence, the probability that, for any day, the number of special orders sent out will be no more than 5 = 0.5848