Question

Let X₁, ... , X₂₀ be independent Poisson random variables with mean 1.

(a) Use the Markov inequality to obtain a bound on

$P\left\{\underset{1}{\overset{20}{\sum X_i>15}}\right\}$

(b) Use the central limit theorem to approximate

$P\left\{\underset{1}{\overset{20}{\sum X_i>15}}\right\}$

Short answer

a) $P\{\underset{i=1}{\overset{20}{\sum X_i}}\ge 15\}\le \frac{4}{3}$

b)$P\{\underset{i=1}{\overset{20}{\sum X_i}}\ge 15\}=0.86$

Step-by-step solution

Step 1. Given information

X₁, ... , X₂₀ be independent Poisson random variables.

Mean =$\lambda =1$

Step 2. a) By Markov's inequality

$P\{\underset{i=1}{\overset{20}{\sum X_i}}\ge 15\}\le \frac{E\left({\displaystyle \underset{i}{\sum X_i}}\right)}{15}=\frac{20}{15}=\frac{4}{3}$

Step 3. b) By central limit theorem 

$$\begin{gathered} P\{\underset{i=1}{\overset{20}{\sum X_i}}\ge 15\}=P\{\underset{i}{\sum X_i}-20\ge 15-20\} \\ P\{\underset{i=1}{\overset{20}{\sum X_i}}\ge 15\}=P\{\underset{i}{\sum X_i}-20\ge -5\} \end{gathered}$$

Step 4. Simplification

$$\begin{gathered} P\{\underset{i=1}{\overset{20}{\sum X_i}}\ge 15\}=P\{\frac{{\displaystyle \stackrel{-}{X}}-1}{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\sqrt{20}$}\right.}}\ge -.25\sqrt{20}\} \\ P\{\underset{i=1}{\overset{20}{\sum X_i}}\ge 15\}=P\{Z\ge -1.18\}=0.86 \end{gathered}$$

Step 5. Final answer 

a)$P\{\underset{i=1}{\overset{20}{\sum X_i}}\ge 15\}\le \frac{4}{3}$

b)$P\{\underset{i=1}{\overset{20}{\sum X_i}}\ge 15\}=0.86$