91Ó°ÊÓ

X and Y are two random variables that are independent of each other.

Further X;

The mean is ${\mathrm{μ}}_{\mathrm{X}}=732.50$and standard deviation is ${\mathrm{σ}}_{\mathrm{X}}=103$

Finding Y:

The mean is ${\mathrm{μ}}_{\mathrm{Y}}=825$ and standard deviation is${\mathrm{σ}}_{\mathrm{Y}}=126.50$

We know that X is the amount spent on tuition at Main campus by a randomly picked student.

Y: the amount spent on tuition by a student at the Downtown campus who was chosen at random.

The full-time student on the main campus pays $ 50 per unit.

When the entire amount spent is divided by 50, we get the number of units.

Mean is a central measure, as we all know.

When each data value is multiplied by 50, the measure of centre should be multiplied by 50 as well.

Thus,

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

Mean is a central measure, as we all know.

When each data value is multiplied by 50, the measure of centre should be multiplied by 50 as well.

As a result, for the main campus,

${\mathrm{μ}}_{\text{Main}}=\frac{{\mathrm{μ}}_{\mathrm{X}}}{50}=\frac{732.50}{50}=14.65\text{units}$

At the moment, a full-time student at the downtown campus pays $ 55 per unit.

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

Mean is a central measure, as we all know.

When every data value is multiplied by 55, the centre measure should also be multiplied by $55.

As a result, when it comes to the Downtown campus,

${\mathrm{μ}}_{\text{Downtown}}=\frac{{\mathrm{μ}}_{\mathrm{Y}}}{55}=\frac{825}{55}=15\text{units}$

The mean of the difference between two random variables is equal to the difference between their means.

width="311">μD=μMaintown-Downtown=μMain−μDowntown=14.65−15=−0.35units

On average, students at the Main campus take fewer $0.35$units than students at the Downtown campus.