# Recentering a Normal Curve with CAS

Sometimes, knowing how to ask a question in a different way using appropriate tools can dramatically simplify a solution.  For context, I’ll use an AP Statistics question from the last decade about a fictitious railway.

THE QUESTION:

After two set-up questions, students were asked to compute how long to delay one train’s departure to create a very small chance of delay while waiting for a second train to arrive.  I’ll share an abbreviated version of the suggested solution before giving what I think is a much more elegant approach using the full power of CAS technology.

BACKGROUND:

Initially, students were told that X was the normally distributed time Train B took to travel to city C, and Y was the normally distributed time Train D took to travel to C.  The first question asked for the distribution of Y-X if the mean and standard deviation of X are respectively 170 and 20, and the mean and standard deviation of Y are 200 and 10, respectively.  Knowing how to transform normally distributed variables quickly gives that Y-X is normally distributed with mean 30 and standard deviation $\sqrt{500}$.

Due to damage to a part of the railroad, if Train B arrived at C before Train D, B would have to wait for D to clear the tracks before proceeding.  In part 2, you had to find the probability that B would wait for D.  Translating from English to math, if B arrives before D, then $X \le Y$.  So the probability of Train B waiting on Train D is equivalent to $P(0 \le Y-X)$.  Using the distribution information in part 1 and a statistics command on my Nspire, this probability is

FIX THE DELAY:

Under the given conditions, there’s about a 91.0% chance that Train B will have to wait at C for Train D to clear the tracks.  Clearly, that’s not a good railway management situation, setting up the final question.  Paraphrasing,

How long should Train B be delayed so that its probability of being delayed is only 0.01?

FORMAL PROPOSED SOLUTION:

A delay in Train B says the mean arrival time of Train D, Y, will remain unchanged at 200, while the mean of arrival time of Train B, X, is increased by some unknown amount.  Call that new mean of X, $\hat{X}=170+delay$.  That makes the new mean of the difference in arrival times

$Y - \hat{X} = 200-(170+delay) = 30-delay$

As this is just a translation, the distribution of $Y - \hat{X}$ is congruent to the distribution of $Y-X$, but recentered.  The standard deviation of both curves is $\sqrt{500}$.  You want to find the value of delay so that $P \left(0 \le Y - \hat{X} \right) = 0.01$.  That’s equivalent to knowing the location on the standard normal distribution where the area to the right is 0.01, or equivalently, the area to the left is 0.99.  One way that can be determined is with an inverse normal command.

The proposed solution used z-scores to suggest finding the value of delay by solving

$\displaystyle \frac{0-(30-delay)}{\sqrt{500}} = 2.32635$

A little algebra gives $delay=82.0187$, so the railway should delay Train B just a hair over 82 minutes.

CAS-ENHANCED ALTERNATIVE SOLUTION:

From part 2, the initial conditions suggest Train B has a 91.0% chance of delay, and part 3 asks for the amount of recentering required to change that probability to 0.01.  Rephrasing this as a CAS command (using TI-Nspire syntax), that’s equivalent to solving

Notice that this is precisely the command used in part 2, re-expressed as an equation with a variable adjustment to the mean.  And since I’m using a CAS, I recognize the left side of this equation as a function of delay, making it something that can easily be “solved”.

Notice that I got exactly the same solution without the algebraic manipulation of a rational expression.

My big point here is not that use of a CAS simplifies the algebra (that wasn’t that hard in the first place), but rather that DEEP KNOWLEDGE of the mathematical situation allows one to rephrase a question in a way that enables incredibly efficient use of the technology.  CAS aren’t replacements for basic algebra skills, they are enhancements for mathematical thinking.

I DON”T HAVE CAS IN MY CLASSROOM.  NOW WHAT????

The CAS solve command is certainly nice, but many teachers and students don’t yet have CAS access, even though it is 100% legal for the PSAT, SAT, and all AP math exams.   But that’s OK.  If you recognize the normCdf command as a function, you can essentially use a graph menu to accomplish the same end.

Too often, I think teachers and students think of normCdf and invNorm commands as nothing more than glorified “lookup commands”–essentially nothing more than electronic versions of the probability tables they replaced.  But, when one of the parameters is missing, replacing it with X makes it graphable.  In fact, whenever you have an equation that is difficult (or impossible to solve), graph both sides and find the intersection, just like a solution to a system of equations.  Using this strategy, graphing $y=normCdf(0,\infty,30-X,\sqrt{500})$ and $y=0.01$ and finding the intersection gives the required solution.

CONCLUSION

Whether you can access a CAS or not, think more deeply about what questions ask and find creative alternatives to symbolic manipulations.