Q.7.64 - A First Course in Probability

Question

Type$i$ light bulbs function for a random amount of time having mean$\mu i$ and standard deviation$\sigma i, i = 1, 2$. A light bulb randomly chosen from a bin of bulbs is a type$1$bulb with probability$p$ and a type$2$bulb with probability$1 - p$. Let $X$ denote the lifetime of this bulb. Find 

(a) $E\left[X\right]$; 

(b) $Var\left(X\right)$. 

Step-by-Step Solution

Answer

The values of $E\left[X\right]$ and $Var\left(X\right)$ is

a)$E\left[X\right]=p{\mu }_1+(1-p){\mu }_2$

b)$\mathrm{Var}\left(X\right)=\left[p{\sigma }_1^2+(1-p){\sigma }_2^2\right]+p(1-p){\left({\mu }_1-{\mu }_2\right)}^2$

Step 1:Given Information(part a)

Given that $X$denotes the lifetime of the bulb.

$Y=\left\{\begin{array}{l}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{ type }1\text{ bulb with probability }\mathrm{p}\\ 2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{ type }2\text{ bulb with probability }(1-\mathrm{p})\end{array}\right.$

$P(Y=1)=p$,$P(Y=2)=(1-p)$

Step 2:Explanation(part a)

$E\left[X\right]=E(X\mid Y=1)\cdot P(Y=1)+E(X\mid Y=2)\cdot P(Y=2)$

$=p{\mu }_1+(1-p){\mu }_2$

Step 3:Final Answer(part a)

The $E\left[X\right]$ is

$\begin{array}{r}E\left[X\right]=p{\mu }_1+(1-p){\mu }_2\end{array}$

Step 4:Given Information(part b)

Given that $X$ as the lifetime of this bulb.

$\mathrm{Var}\left(X\right)=E\left[X^2\right]-\left(E\right[X]{)}^2$

Step 5:Explanation(Step 5)

$E\left[X^2\right]=E\left(X^2\mid Y=1\right)\cdot P(Y=1)+E\left(X^2\mid Y=2\right)\cdot P(Y=2)$

$=p\left({\mu }_1^2+{\sigma }_1^2\right)+(1-p)\left({\mu }_2^2+{\sigma }_2^2\right)$

$=\left[p{\sigma }_1^2+(1-p){\sigma }_2^2\right]+\left[p{\mu }_1^2+(1-p){\mu }_2^2\right]$

$\mathrm{Var}\left(X\right)=\left[p{\sigma }_1^2+(1-p){\sigma }_2^2\right]+\left[p{\mu }_1^2+(1-p){\mu }_2^2\right]-p{\mu }_1+{\left[(1-p){\mu }_2\right]}^2$

$=\left[p{\sigma }_1^2+(1-p){\sigma }_2^2\right]+p(1-p)\left[{\mu }_1^2+{\mu }_2^2+2{\mu }_1{\mu }_2\right]$

$=\left[p{\sigma }_1^2+(1-p){\sigma }_2^2\right]+p(1-p){\left({\mu }_1-{\mu }_2\right)}^2$

Step 6:Final Answer(part b)

The $Var\left(X\right)$ is

$\mathrm{Var}\left(X\right)=\left[p{\sigma }_1^2+(1-p){\sigma }_2^2\right]+p(1-p){\left({\mu }_1-{\mu }_2\right)}^2$