Question

Note that $X\mathrm{and}Y$are independent random variables because the two students are randomly selected from each of the campuses. At the main campus, full-time students pay $\backslash (50$per unit. At the downtown campus, full-time students pay $\backslash )55$per unit. Suppose we randomly select one full-time student from each of the two campuses. Find the standard deviation of the difference $D$(Main − Downtown) in the number of units that the two randomly selected students take.

Short answer

The standard deviation and the units of data values are the same. The standard deviation is $\mathbf{3}\mathbf{.}\mathbf{0877}$.

Step-by-step solution

Given Information  

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

Such that

For $X$:

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

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

For $Y$:

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

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

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

Calculating standard deviation of the Main campus

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 :

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

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.

Finding the standard deviation of the difference D

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

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

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" $$\begin{gathered} {{\sigma }^2}_{(Main-Downtown)}={{\sigma }^2}_{Main}+{{\sigma }^2}_{Downtown} \\ ={(2.06)}^2+{(2.3)}^2 \\ =9.5336 \end{gathered}$$

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

As a result, :

${\sigma }_{(Main-Downtown)}=\sqrt{9.5336}\approx \mathbf{3}\mathbf{.}\mathbf{0877}$