Chapter 6.2, Problem 53E

To determine

Mean of the difference D in the number of units that the two randomly selected students take.

Answer to Problem 53E

Mean of the difference D ,

  ${\mu }_D=-0.35\text{units}$

Explanation of Solution

Given information:

X and Y are independent random variables.

For X :

Mean,

  ${\mu }_X=732.50$

Standard deviation,

  ${\sigma }_X=103$

For Y :

Mean,

  ${\mu }_Y=825$

Standard deviation,

  ${\sigma }_Y=126.50$

Calculations:

We know that

X: the amount spent on tuition by a randomly selected student at Main campus

Y: the amount spent on tuition by a randomly selected student at Downtown campus

At the main campus, the full − time student pays $50 per unit.

Thus,

We get the number of units when the total amount spent is divided by 50.

We know that

Meanis a measure of center.

When every data value is divided by 50, the measure of center should also be divided by 50.

Thus,

Mean for the Main campus,

  ${\mu }_{\text{Main}}=\frac{{\mu }_X}{50}=\frac{732.50}{50}=14.65\text{units}$

Now,

At downtown campus, the full − time student pays $55 per unit.

Thus,

We get the number of units when the total amount spent is divided by 55.

We know that

Mean is a measure of center.

When every data value is divided by 55, the measure of center should also be divided by 55.

Thus,

Mean for the Downtown campus,

  ${\mu }_{\text{Downtown}}=\frac{{\mu }_Y}{55}=\frac{825}{55}=15\text{units}$

For two random variables, the mean of the difference is equal to the difference of their means.

  ${\mu }_D={\mu }_{\text{Main}-\text{Downtown}}={\mu }_{\text{Main}}-{\mu }_{\text{Downtown}}=14.65-15=-0.35\text{units}$

On an average, the student at Main campus takes 0.35 units less than the students at Downtown campus.