Can this statistics question be answered using a ti-84 calculator? I know how to solve this question going the long route, using the formula y=a+bx.  But finding a and b is time consuming.  If we have some inputs given, can I solve this problem using a ti-84 calculator without having to do everything by hand.
The question comes from a stats book assigned to me.  I do not own the material and make no claims.

I cut and pasted an image of the question.  It looks like we have enough info to solve this.  I just don't know how to use my calculator to do it.
 A: All you need to know is that the regression line has a particularly simple formula when the variables are expressed in terms of their natural units of measure.
What "natural" means here is that (a) their mean should be recorded as zero and (b) their standard deviation should be one unit.  When this is done to produce numbers $x$ for the independent variable and $y$ for the response, the regression equation is
$$y = r x$$
where $r$ is the sample correlation coefficient.
Setting the mean to zero is sometimes called "recentering."  It is done by subtracting the mean from all values, thereby producing residuals.
Making one standard deviation the unit of measure is sometimes called "rescaling".  It is done by dividing the residuals by the standard deviation.
The combination of these two operations is usually termed "standardization."

Example
In this question, the data tell us:

*

*The mean of $\text{SAT}$ is $649.5$ and its standard deviation (SD) is $66.3$.  Therefore, $$x = \frac{\text{SAT} - 649.5}{66.3}.$$


*The mean of $\text{GPA}$ is $2.63$ and its SD is $0.58$.  Therefore, $$y = \frac{\text{GPA} - 2.63}{0.58}.$$


*The correlation is $$r = 0.194.$$
Consequently, the regression equation is
$$\frac{\text{GPA} - 2.63}{0.58} = y = rx = 0.194 \left(\frac{\text{SAT} - 649.5}{66.3}\right).$$
If you really want to you could algebraically "simplify" this to express $\text{GPA}$ more directly in terms of $\text{SAT}$.  (I'm pretty sure the TI-84 is up to the task. :-)  It is a pretty and useful equation in its present form, though, because it has a wonderfully simple interpretation:

For each standard deviation by which the SAT differs from the mean, the predicted GPA will be $0.194$ times as many standard deviations from its mean.

For instance, in the second half of the question an SAT score of $650$ is essentially right at the mean of $649.5$, so the predicted GPA is essentially its mean of $2.63$.
