Question

Study habits : Refer to Exercise 51.

a. Assume that $F\mathrm{and}M$are independent random variables. Explain what this means in context.

b. Calculate and interpret the standard deviation of the difference $D=F-M$in their scores.

c. From the information given, can you find the probability that the randomly selected female student has a higher SSHA score than the randomly selected male student? Explain why or why not.

Short answer

a. The SSHA score of male student $M$has no bearing on the SSHA score of female student $F$.

b. The difference between female and male SSHA scores deviates from the mean difference of $15$points by approximately $\mathbf{44}\mathbf{.}\mathbf{8219}$points on average.

c. Determining the likelihood that a female student will receive a higher SSHA score than a male student is impossible.

Step-by-step solution

Part(a) Step 1 : Given Information 

Given :

$F\mathrm{and}M$are independent random variables.

For $F$:

Mean, ${\mu }_F=120$

Standard deviation,${\sigma }_F=28$

For$M$ :

Mean, ${\mu }_M=105$

Standard deviation, ${\sigma }_M=35$

Part(a) Step 2 : Simplification 

$F$: SSHA score of a female student at a large university who was chosen at random.

$M$: SSHA score of a female student at a large university who was chosen at random.

This means that the SSHA score of the female student $F$has no bearing on the SSHA score of the male student $M$in any way.

In the same way, the SSHA score of male student $M$has no bearing on the SSHA score of female student $F.$

Part(b) Step 1 : Given Information 

Given :

$F\mathrm{and}M$are independent random variables.

For$F$ :

Mean, ${\mu }_F=120$

Standard deviation,${\sigma }_F=28$

For$M$ :

Mean, ${\mu }_M=105$

Standard deviation, ${\sigma }_M=35$

Part(b) Step 2 : Simplification 

Mean difference is :

$$\begin{gathered} {\mu }_{F-M}={\mu }_F-{\mu }_M \\ =120–105 \\ =15 \end{gathered}$$

The variance of the difference is equal to the sum of the variances of the random variables when they are independent.

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

We also know that the standard deviation equals the variance squared:

$$\begin{gathered} {\sigma }_{F-M}=\sqrt{{{\sigma }^2}_{F-M}} \\ =\sqrt{2009} \\ \approx \mathbf{44}\mathbf{.}\mathbf{8219} \end{gathered}$$

Part(c) Step 1 : Given Information 

Given :

$F\mathrm{and}M$are independent random variables.

For$F:$

Mean, ${\mu }_F=120$

Standard deviation,${\sigma }_F=28$

For$M$ :

Mean,${\mu }_M=105$

Standard deviation, ${\sigma }_M=35$

Part(c) Step 2 : Simplification 

Part (b) revealed that the $F-M$difference had a mean of $15$points and a standard deviation of $44.8219$points.

Because the $M$distribution's distribution is unknown.

As a result, the difference $F-M$distribution cannot be calculated.

As a result, determining the likelihood that a female student will receive a higher SSHA score than a male student is impossible.