91Ó°ÊÓ

Both $X\mathrm{and}Y$are independent events.

Such that

For $X$:

The value of Mean, ${\mathrm{μ}}_X=732.50$

The value of Standard deviation, ${\mathrm{σ}}_X=103$

For $Y$:

The value of Mean, ${\mathrm{μ}}_Y=825$

The value of Standard deviation,${\mathrm{σ}}_Y=126.50$

The full-time student pays at the main campus. : $\$50$per unit.

When we divide the total amount spent by $50$, we get the number of units.

The standard deviation is a measure of spread, as we all know. When each data value is multiplied by $50$, the spread measure should be multiplied by $50$as well.

As a result, the standard deviation for the Main campus is :

${\mathrm{σ}}_{Main}=\frac{{\mathrm{σ}}_X}{50}=\frac{103}{50}=2.06.$

When the total amount spent is divided by $55$, we get the number of units.

The standard deviation is a measure of spread, as we all know.

The spread measure should be multiplied by $55$as well.

As a result, the standard deviation for the Downtown campus is :

${\mathrm{σ}}_{Downtown}=\frac{{\mathrm{σ}}_Y}{55}=\frac{126.50}{50}=2.3$

The variance of the difference of two random variables is the sum of their variances when the random variables are independent.

role="math" localid="1654184360394" $$\begin{gathered} {{\mathrm{σ}}^2}_{(Main-Downtown)}={{\mathrm{σ}}^2}_{Main}+{{\mathrm{σ}}^2}_{Downtown} \\ ={(2.06)}^2+{(2.3)}^2 \\ =9.5336 \end{gathered}$$

We know that the square root of the variance is the standard deviation.

As a result, :

${\mathrm{σ}}_{(Main-Downtown)}=\sqrt{9.5336}≈\mathbf{3}\mathbf{.}\mathbf{0877}$