Researchers randomly select a married couple in which both spouses are employed. Let $X$be the income of the husband and $Y$be the income of the wife. Suppose that you know the means ${\mu }_X\mathrm{and}{\mu }_Y$and the variances ${{\mathrm{\sigma }}_{\mathrm{X}}}^2\mathrm{and}{{\mathrm{\sigma }}_{\mathrm{Y}}}^2$of both variables.

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

b. Can we calculate the variance of the total income to be${{\mathrm{\sigma }}_{\mathrm{X}}}^2+{{\mathrm{\sigma }}_{\mathrm{Y}}}^2$ ? Explain your answer.

Given :

$X$: The income of the husband

$Y$: The income of the wife.

The sum of the means of two random variables is the total of their means.

${\mu }_{X+Y}={\mu }_X+{\mu }_Y$

We may calculate the mean of total income $X+Y\mathrm{to}\mathrm{be}{\mu }_X+{\mu }_Y$since the property holds for any two random variables, and because the means for both $X\mathrm{and}Y$are known.

Given :

$X$: The income of the husband

$Y$: The income of the wife.

The variance of the total is equal to the sum of the variances of the two random variables.

However, this is only conceivable if two random variables are independent; otherwise, this is not true.

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

People who are intelligent tend to grow even more intelligent.

Furthermore, clever people are more likely to make more money.

As a result, it is reasonable to conclude that the husband's income $X$is unrelated to the wife's $Y$income.

As a result, the variance of total revenue cannot be calculated as${{\sigma }^2}_X+{{\sigma }^2}_Y$