Question

Let X denote the number of tornadoes occurring in a specific region in 2016. Assume X has a Poisson distribution with variance 7 (a) Calculate P(X26) (b) Caleulate the probability that there will be 8 or more tornadoes given that there are at least 3 tornadoes, e. P(X8X23)

Answer 1

We are given that $X \sim Poisson (\lambda)$ .

Now we are given that :

$Var(X)=7 \; \; \Rightarrow \; \lambda =7$

Thus, X ~ Poisson(7)

(a)

$P(X\geq 6) =\sum_{k=6}^\infty P(X=k) =\sum_{k=6}^\infty \frac{e^{-\lambda}\lambda^k}{k !} =e^{-\lambda}\sum_{k=6}^\infty \frac{\lambda^k}{k !}$ (Using the formula for the pmf of a Poisson distibution)

$\Rightarrow P(X\geq 6) =e^{-\lambda}(\sum_{k=0}^\infty \frac{\lambda^k}{k !} - \sum_{k=0}^5 \frac{\lambda^k}{k !} )$

Now, we observe that : $\sum_{k=0}^\infty \frac{\lambda^k}{k !} =1 + \lambda + \frac{\lambda^2}{2!} + \frac{\lambda^3}{3!} + ... = e^\lambda$ . Thus :

$\Rightarrow P(X\geq 6) =e^{-\lambda}(e^{\lambda} - \sum_{k=0}^5 \frac{\lambda^k}{k !} )=1-e^{-\lambda}*\sum_{k=0}^5 \frac{\lambda^k}{k !}$

Now, $\sum_{k=0}^5 \frac{\lambda^k}{k !} =1 + \lambda + \frac{\lambda^2}{2!} + \frac{\lambda^3}{3!} + \frac{\lambda^4}{4!}+\frac{\lambda^5}{5!}$

$\Rightarrow \sum_{k=0}^5 \frac{\lambda^k}{k !} =1 + 7 + \frac{7^2}{2!} + \frac{7^3}{3!} + \frac{7^4}{4!}+\frac{7^5}{5!} = 329.76667$

Thus, $P(X \geq 6 )=1- e^{-7}*329.76667=0.69929$

(b)

$P(X \geq 8 | X \geq 3)=\frac{P(X \geq 8 \cap X\geq 3)}{ P(X \geq 3)} = \frac{P(X \geq 8 )}{ P(X \geq 3)} = \frac{1-P(X \leq 7)}{1-P(X \leq 2)}$

Now, $P(X \leq 7) = \sum_{k=0}^7\frac{e^{-\lambda}\lambda^k}{k!} = e^{-\lambda}\sum_{k=0}^7\frac{\lambda^k}{k!} = e^{-\lambda} *(1+\lambda+\frac{\lambda^2}{2!}+...+\frac{\lambda^7}{7!})$

$\Rightarrow P(X \leq 7) = e^{-7} *(1+7+\frac{7^2}{2!}+...+\frac{7^7}{7!}) = 0.00091188*656.569444$ $\Rightarrow P(X \leq 7) =0.598714$

Also, $P(X \leq 2) = \sum_{k=0}^2\frac{e^{-\lambda}\lambda^k}{k!} = e^{-\lambda}\sum_{k=0}^2\frac{\lambda^k}{k!} = e^{-\lambda} *(1+\lambda+\frac{\lambda^2}{2!})$

$\Rightarrow P(X \leq 2) = e^{-7} *(1+7+\frac{7^2}{2!}) = 0.00091188*32.5=0.029636$

Thus, $P(X \geq 8 | X \geq 3) = \frac{1-P(X \leq 7)}{1-P(X \leq 2)} =\frac{1-0.598714}{1-0.029636} =0.413542$