52. Study habits The academic motivation and study habits of female students as a group are better than those of males. The Survey of Study Habits and Attitudes $\left(SSHA\right)$is a psychological test that measures these factors. The distribution of $SSHA$scores among the women at a college has mean$120$and standard deviation $28$, and the distribution of scores among male students has mean $105$and standard deviation $35.$You select a single male student and a single female student at random and give them the$SSHA$ test.
(a) Explain why it is reasonable to assume that the scores of the two students are independent.
(b) What are the expected value and standard deviation of the difference (female minus male) between their scores?
(c) From the information given, can you find the probability that the woman chosen scores higher than the man? If so, find this probability. If not,
explain why you cannot.

Given in the question that, the distribution of SSHA scores among the women at a college has mean $120$and standard deviation $28$, and the distribution of scores among male students has mean $105$and standard deviation $35$. We have to assume that the scores of the two students are independent.

According to the information, the Population mean for woman $\left(\mu \right)=120$
Population standard deviation $\left(\sigma \right)=28$

The Population mean for men $\left(\mu \right)=105$

Population standard deviation $\left(\sigma \right)=35$

The model was selected at random, according to the given statement. As a result, it is taught that a large number of men and women are self-sufficient.

The population mean for women is$120$and standard deviation is$28$

The population mean for men is $105$and standard deviation is $35$

We have to find the expected value and standard deviation of the difference (female minus male) between the scores.

The anticipated value and standard deviation of two students' scores are determined as follows:
$$\begin{gathered} \begin{array}{r}{\mu }_{F-M}={\mu }_F-{\mu }_M\end{array} \\ =120-105 \\ =15 \end{gathered}$$

$$\begin{gathered} {\sigma }_{F-M}=\sqrt{{\sigma }^{2}F-{\sigma }^{2}M} \\ =\sqrt{{28}^2+{35}^2} \end{gathered}$$

Then,

$$\begin{gathered} {\mu }_{F-M}={\mu }_F-{\mu }_M \\ =120-105 \\ =\sqrt{2009} \\ =44.8219 \end{gathered}$$

The probability of that the woman chosen scores higher than the man.

Because no data on the distribution has been provided, it is impossible to predict the likelihood that women have scored higher than males. As a result, it is impossible to determine the needed probability.