Chapter 6.2, Problem 59E

(a)

To determine

Meaning for F and M assumed as independent random variables in context.

Answer to Problem 59E

Randomly selected female student F and randomly selected male student M does not affect each other’s SSHA score.

Explanation of Solution

Given information:

F and M are independent random variables.

For F :

Mean,

  ${\mu }_F=120$

Standard deviation,

  ${\sigma }_F=28$

For M :

Mean,

  ${\mu }_M=105$

Standard deviation,

  ${\sigma }_M=35$

F : SSHA score of a randomly selected female student at a large university

M : SSHA score of a randomly selected female student at a large university

We know that

F and M are independent random variables.

This implies

In any way, the SSHA score of the female student F does not affect the SSHA score of the male student M .

Similarly,

In any way, the SSHA score of the male student M does not affect the SSHA score of the female student F .

(b)

To determine

Interpretation and calculation of standard deviation of the difference in scores.

Answer to Problem 59E

On an average, the difference between the female and male SSHA scores deviates by approx. 44.8219 points from the mean difference of 15 points.

Explanation of Solution

Given information:

F and M are independent random variables.

For F :

Mean,

  ${\mu }_F=120$

Standard deviation,

  ${\sigma }_F=28$

For M :

Mean,

  ${\mu }_M=105$

Standard deviation,

  ${\sigma }_M=35$

F : SSHA score of a randomly selected female student at a large university

M : SSHA score of a randomly selected female student at a large university

Now,

Meandifference of the mean of both male and female SSHA scores:

  ${\mu }_{F-M}={\mu }_F-{\mu }_M=120-105=15$

When the random variables are independent, the variance of the difference is equal to the sum of their variances.

  ${\sigma }_{F-M}^2={\sigma }_F^2+{\sigma }_M^2={\left(28\right)}^2+{\left(35\right)}^2=2009$

We also know that

The standard deviation is the square root of the variance:

  ${\sigma }_{F-M}=\sqrt{{\sigma }_{F-M}^2}=\sqrt{2009}=7\sqrt{41}\approx 44.8219$

On an average, the difference between the female and male SSHA scores deviates by approx. 44.8219 points from the mean difference of 15 points.

(c)

To determine

Whether it is possible to find the probability for the randomly selected female student has a higher SSHA score than the randomly selected male student.

Answer to Problem 59E

No, because the distributions of F and M are not known.

Explanation of Solution

Given information:

F and M are independent random variables.

For F :

Mean,

  ${\mu }_F=120$

Standard deviation,

  ${\sigma }_F=28$

For M :

Mean,

  ${\mu }_M=105$

Standard deviation,

  ${\sigma }_M=35$

From Part (b),

We came to know that

The difference F - M had mean of 15 points and standard deviation of approx. 44.8219 points.

Since the distribution of F or the distribution M is not known.

Thus,

The distribution of the difference F - M cannot be determined.

Hence,

It is impossible to determine the probability that the female student scores a higher SSHA score than the male student.