Chapter 6.2, Problem 61E

To determine

Standard deviation of the difference D (Main − Downtown) in the number of units taken by two randomly selected students.

Answer to Problem 61E

Standard deviation of the difference is approx. 3.0877 units.

Explanation of Solution

Given information:

X and Y are independent events.

Such that

For X :

Mean,

  ${\mu }_X=732.50$

Standard deviation,

  ${\sigma }_X=103$

For Y :

Mean,

  ${\mu }_Y=825$

Standard deviation,

  ${\sigma }_Y=126.50$

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

The standard deviation is a measure of spread.

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

Thus,

Standard deviation for Main campus,

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

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

The standard deviation is a measure of spread.

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

Thus,

Standard deviation for Downtown campus,

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

When random variables are independent, the variance of the difference of the two random variables equals the sum of their variances.

  ${\sigma }_{\text{Main}-\text{Downtown}}^2={\sigma }_{\text{Main}}^2+{\sigma }_{\text{Downtown}}^2={\left(2.06\right)}^2+{\left(2.3\right)}^2=9.5336$

We know that

Standard deviation is the square root of the variance.

Thus,

  ${\sigma }_{\text{Main}-\text{Downtown}}=\sqrt{9.5336}\approx 3.0877$

The units of data values and standard deviation are the same.

Thus,

The standard deviation is 3.0877 units.