hroncits8y

2021-11-29

Suppose that the random variable X is defined as $X=0.6{X}_{1}+0.3{X}_{2}+0.1{X}_{3}$ , where the probability distributions of the independent random variables $X}_{1},{X}_{2},{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}{X}_{3$ are

$${f}_{X1}(x)=\left(\begin{array}{c}5\\ x\end{array}\right){0.1}^{x}\times {0.9}^{5-x}$$

$${f}_{X2}(x)=\left(\begin{array}{c}5\\ x\end{array}\right){0.4}^{x}\times {0.6}^{5-x}$$

$${f}_{X3}(3)=\left(\begin{array}{c}5\\ x\end{array}\right){0.6}^{x}\times {0.4}^{5-x}$$

$x=0,1,2,3,4,5$

a) Evaluate the mean of X.

b) Evaluate the variance of X.

a) Evaluate the mean of X.

b) Evaluate the variance of X.

Archie Griffin

Beginner2021-11-30Added 9 answers

Step 1

Given:

$X=0.6{X}_{1}+0.3{X}_{2}+0.1{X}_{3}$

Where all three are independent

$${f}_{X1}(x)=\left(\begin{array}{c}5\\ x\end{array}\right){0.1}^{x}\times {0.9}^{5-x}$$

$${f}_{X2}(x)=\left(\begin{array}{c}5\\ x\end{array}\right){0.4}^{x}\times {0.6}^{5-x}$$

$${f}_{X3}(3)=\left(\begin{array}{c}5\\ x\end{array}\right){0.6}^{x}\times {0.4}^{5-x}$$

$x=0,1,2,3,4,5$

Step 2

It can be seen that all the three distributions follows binomial distribution

So, Mean of binomial is

$n\times p$

Variance of binomial is

$n\times p\times q$

So,${X}_{1}\sim B\in (5,0.1)$

$E\left[{X}_{1}\right]=5\times 0.1=0.5$

$Var\left[{X}_{1}\right]=5\times 0.1\times 0.9=0.45$

${X}_{2}\sim B\in (5,0.4)$

$E\left[{X}_{2}\right]=5\times 0.4=2$

$Var\left[{X}_{2}\right]=5\times 0.4\times 0.6=1.2$

${X}_{3}\sim B\in (5,0.6)$

$E\left[{X}_{3}\right]=5\times 0.6=3$

$Var\left[{X}_{3}\right]=5\times 0.6\times 0.4=1.2$

Step 3

a) Mean of X

$E\left[X\right]=0.6\times E\left[{X}_{1}\right]+0.3\times E\left[{X}_{2}\right]+0.1\times E\left[{X}_{3}\right]$

$E\left[X\right]=0.6\times 0.5+0.3\times 2+0.1\times 3$

$E\left[X\right]=1.2$

Hence, the mean is 1.2

Step 4

b) If the events are independent, then the variance is

$Var[A+B]=Var\left[A\right]+Var\left[B\right]$

Here, all are independent, so the variance of X can be calculated as follows

$Var\left[X\right]={0.6}^{2}\times Var\left[{X}_{1}\right]+{0.3}^{2}\times Var\left[{X}_{2}\right]+{0.1}^{2}\times Var\left[{X}_{3}\right]$

$Var\left[X\right]={0.6}^{2}\times 0.45+{0.3}^{2}\times 1.2+{0.1}^{2}\times 1.2$

Given:

Where all three are independent

Step 2

It can be seen that all the three distributions follows binomial distribution

So, Mean of binomial is

Variance of binomial is

So,

Step 3

a) Mean of X

Hence, the mean is 1.2

Step 4

b) If the events are independent, then the variance is

Here, all are independent, so the variance of X can be calculated as follows

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