Answer to Question #141025 in Statistics and Probability for Margaret

Question

For a discrete random variable that is poission distributed with a mean i=20 which one of the following statement is incorrect
1. P (X=0)
2. P (X-0 )
3. P (X-1)
4. P (X-7)
5. The variance a is 4

Answer 1

$X\sim Po(\lambda)$

$$P(X=x)=\dfrac{e^{-\lambda}\cdot\lambda^x}{x!}$$

$mean=\lambda=Var(x)$

Given $mean=20$

Then $\lambda=20$

1.

$$P(X=0)=\dfrac{e^{-20}\cdot20^0}{0!}=$$

$$=e^{-20}\approx0.0000000021\approx0$$

2.

$$P(X>0)=1-P(X=0)=1-\dfrac{e^{-20}\cdot20^0}{0!}=$$

$$=1-e^{-20}\approx0.9999999979\approx1$$

3.

$$P(X>1)=1-P(X=0)-P(X=1)=$$

$$=1-\dfrac{e^{-20}\cdot20^0}{0!}-\dfrac{e^{-20}\cdot20^1}{1!}=$$

$$=1-21\cdot e^{-20}\approx0.9999999567\approx1$$

4.

$$P(X>7)=1-P(X=0)-P(X=1)-$$

$$-P(X=2)-P(X=3)-P(X=4)-$$

$$-P(X=5)-P(X=6)-P(X=7)=$$

$$=1-\dfrac{e^{-20}\cdot20^0}{0!}-\dfrac{e^{-20}\cdot20^1}{1!}-\dfrac{e^{-20}\cdot20^2}{2!}-$$

$$-\dfrac{e^{-20}\cdot20^3}{3!}-\dfrac{e^{-20}\cdot20^4}{4!}-\dfrac{e^{-20}\cdot20^5}{5!}-$$

$$-\dfrac{e^{-20}\cdot20^6}{6!}-\dfrac{e^{-20}\cdot20^7}{7!}\approx$$

$$\approx0.9992214099$$

5.

$Var(X)=\lambda=20\not=4$