91影视

Given :

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

For $F$:

Mean, ${渭}_F=120$

Standard deviation,${蟽}_F=28$

For$M$ :

Mean, ${渭}_M=105$

Standard deviation, ${蟽}_M=35$

$F$: 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 $F.$

Given :

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

For$F$ :

Mean, ${渭}_F=120$

Standard deviation,${蟽}_F=28$

For$M$ :

Mean, ${渭}_M=105$

Standard deviation, ${蟽}_M=35$

Mean difference is :

$$\begin{gathered} {渭}_{F-M}={渭}_F-{渭}_M \\ =120鈥�105 \\ =15 \end{gathered}$$

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

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

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

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

Given :

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

For$F:$

Mean, ${渭}_F=120$

Standard deviation,${蟽}_F=28$

For$M$ :

Mean,${渭}_M=105$

Standard deviation, ${蟽}_M=35$

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.