Question

Imagine that we randomly select a day from the past $10$years. Let $X$be the recorded rainfall on this date at the airport in Orlando, Florida, and $Y$be the recorded rainfall on this date at Disney World just outside Orlando. Suppose that you know the means data-custom-editor="chemistry" ${\mathrm{\mu }}_{\mathrm{X}}\mathrm{and}{\mathrm{\mu }}_{\mathrm{Y}}$and the variances data-custom-editor="chemistry" ${{\mathrm{\sigma }}_{\mathrm{X}}}^2\mathrm{and}{{\mathrm{\sigma }}_{\mathrm{Y}}}^2$of both variables.

a. Can we calculate the mean of the total rainfall $X+Y\mathrm{to}\mathrm{be}{\mu }_X+{\mu }_Y$? Explain your answer.

b. Can we calculate the variance of the total rainfall to be ${{\sigma }_X}^2+{{\sigma }_Y}^2$? Explain your answer

Short answer

a. Yes, it is reasonable.

b.No, it's not logical.

Step-by-step solution

Part(a) Step 1 : Given Information 

Given :

$X$: The recorded rainfall at the airport in Orlando

$Y$: The recorded rainfall at Disney World just outside Orlando

Part(a) Step 2 : Simplification  

Yes, it is reasonable.

The reason is because the mean has the condition ${\mu }_{X+Y}={\mu }_X+{\mu }_Y$for all conceivable pairs of variables.

As a result, the mean of $X+Y$ is always the same as ${\mu }_X+{\mu }_Y$.

Part(b) Step 1 : Given Information 

Given :

$X$: The recorded rainfall at the airport in Orlando

$Y$: The recorded rainfall at Disney World just outside Orlando

Part(b) Step 2 : Simplification  

No, it's not logical.

The reason for this is that the variance property applies only when the variables $X\mathrm{and}Y$are independent.

${{\sigma }^2}_{X+Y}={{\sigma }^2}_X+{{\sigma }^2}_Y$