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Given independent random variables with means and standard deviations as shown, find the mean and standard deviation of X+Y.

 

Mean

Standard Deviation

$X$

$80$

$12$

$Y$$12$

$3$

The purpose of this question is to find the mean and standard deviation of the given expression using the expected values and standard deviations of the random variables given in the table.

A random variable numerically represents the result of a trial. Two types of random variables include a discrete random variable, which takes a finite number or an unbounded pattern of values. The second kind is a continuous random variable that takes the values in an interval.

Let $X$ be a discrete random variable. Its mean can be regarded as the weighted sum of its potential values. The central tendency or the position of a random variable is indicated by its mean. A measure of dispersion for a random variable distribution that specifies how far the values deviate from the mean is said to be the standard deviation.

Consider a discrete random variable: its standard deviation can be obtained by squaring the difference between the random variable’s value and the mean and adding them up along with the corresponding probability of all the random variable’s values, and in the end obtaining its square root.

Expert Answer

From the table:

$E(X)=80$  and $E(Y)=12$

Now since $E(X+Y)=E(X)+E(Y)$

Substitute the given values:

$E(X+Y)=80+12$

$E(X+Y)=92$

Now as $Var(X+Y)=Var(X)+Var(Y)$, also:

$Var(X)=[SD(X)]^2$  and  $Var(Y)=[SD(Y)]^2$

therefore, $Var(X)=[12]^2$  and  $Var(Y)=[3]^2$

$Var(X)=144$  and  $Var(Y)=9$

So that:

$Var(X+Y)=144+9$

$Var(X+Y)=153$

Finally, $SD(X+Y)=\sqrt{Var(X+Y)}$

$SD(X+Y)=\sqrt{153}$

$SD(X+Y)=12.37$

Example 1

Assume the same data as in the given question, and find the expected value and the variance of $3Y+10$.

Solution

Using the property of expected value:

$E(aY+b)=aE(Y)+b$

Here, $a=3$ and $b=10$, so that:

$E(3Y+10)=3E(Y)+10$

From the table, $E(Y)=12$ therefore:

$E(3Y+10)=3(12)+10$

$E(3Y+10)=36+10$

$E(3Y+10)=46$

Using the property of variance:

$Var(aY+b)=a^2Var(Y)$

Here $a=3$ and $b=10$, so that:

$Var(3Y+10)=(3)^2Var(Y)$

Now  $Var(Y)=[SD(Y)]^2$

$Var(Y)=(3)^2$

$Var(Y)=9$

Therefore,  $Var(3Y+10)=(3)^2(9)$

$Var(3Y+10)=(9)(9)$

$Var(3Y+10)=81$

Example 2

Find the expected value, variance, and standard deviation of $2X-Y$ assuming the data given in the table.

Solution

Using the property of expected value:

$E(aX-Y)=aE(X)-E(Y)$

Here $a=2$,  so that:

$E(2X-Y)=2E(X)-E(Y)$

From the table, $E(X)=80$ and $E(Y)=12$, therefore:

$E(2X-Y)=2(80)-12$

$E(2X-Y)=160-12$

$E(2X-Y)=148$

Using the property of variance:

$Var(aX)=a^2Var(X)$  and  $Var(X-Y)=Var(X)-Var(Y)$, we have:

$Var(aX-Y)=a^2Var(X)-Var(Y)$

Since  $Var(X)=144$  and $Var(Y)=9$ so that:

$Var(2X-Y)=(2)^2(144)-9$

$Var(2X-Y)=(4)(144)-9$

$Var(2X-Y)=576-9$

$Var(2X-Y)=567$

Also, $SD(2X-Y)=\sqrt{Var(2X-Y)}$, therefore:

$SD(2X-Y)=\sqrt{567}$

$SD(2X-Y)=23.81$

Example 3

Find $E(2.5X)$ and $E(XY)$ if $E(X)=0.2$ and $E(Y)=1.3$.

Solution

Since $E(aX)=aE(X)$, therefore:

$E(2.5X)=2.5E(X)$

$E(2.5X)=2.5(0.2)$

$E(2.5X)=0.5$

And $E(XY)=E(X)E(Y)$, therefore:

$E(XY)=(0.2)(1.3)$

$E(XY)=0.26$

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