Normal distribution probability calculation I am stuck with a stats question..... help please

A television manufacturer is studying television remote control usage. One of the criteria they are measuring is the distance at which people attempt to activate the television set with the remote control. They have discovered that the activation distances are normally distributed with a mean distance of 6 feet and a standard deviation of 2.75 feet. If a remote’s maximum range is 10 feet, what percentage of the time will users attempts to operate the remote outside of its operating limit?

 A: The trick to word problems like this in general (not just in stats) is extricating what's important from what's not important. Once you've done that, the trick to solving math and stats problems is to start with what you already know. This is true even in cutting-edge mathematical research. It sounds obvious but it's strangely an acquired skill.
So here's what we know:
"Attempted activation distance" is normally distributed random variable, with $\mu$=6 and $\sigma$=2.75. Let's call it $X$. You're looking for the probability that X is greater than 10. In math terms, this means you're looking for $\mathrm{Pr}(X > 10)$. This should be a sign that you're looking to plug $X$ into its cumulative distribution function (CDF).
However, the CDF is defined as $\mathrm{F}(X)=\mathrm{Pr}(X \leq 10)$. Fortunately, there are only two possible places where X can be: less than or equal to 10, or greater than 10. It has to be one or the other and it can't be both. Now think $X\leq 10$ as one "event" and $X>10$ as a second "event."  The probability that one event occurs or another even occurs is equal to the sum of their probabilities. $\mathrm{Pr}(X \leq 10 \mathrm{\ or\ } X > 10)=\mathrm{Pr}(X \leq 10)+\mathrm{Pr}(X > 10)$. And we know one of them has to occur, so $\mathrm{Pr}(X \leq 10 \mathrm{\ or\ } X > 10)=1$. Move around the equation and we get $\mathrm{Pr}(X > 10)=1-\mathrm{Pr}(X \leq 10)$.
We actually know enough about $X$ to tell a computer how to calculate $1-\mathrm{Pr}(X \leq 10)$. But you don't learn anything that way so I'll just put the code at the end. You can't calculate $\mathrm{Pr}(X \leq 10)$ by hand, but most textbooks have a table at the end that looks something like this. Find your $X$ using the edges (ones and 10ths on the left, 100th on the top), and that will point you to the probability $\mathrm{Pr}(X \leq 10)$. HOWEVER, those tables are constructed to assume $\mu =0$ and $\sigma =1$, which is not the case here.
A random variable with $\mu =0$ and $\sigma =1$ is said to be "standardized" and is usually denoted with $Z$. There is a well-known formula for "standardizing" an arbitrary random variable: $\frac{Y-\mu}{\sigma}=Z$, where $Z$ is what we'll call the standardized random variable and $Y$ is any random variable. In most cases, you can freely mess with an equation or inequality (such as $X\leq 10$) by applying the same formula to both sides. So let's plug in both sides of $X\leq 10$ to get $\frac{X-\mu}{\sigma}\leq \frac{10-\mu}{\sigma}$. Note that this is the same thing as saying $Z\leq \frac{10-\mu}{\sigma}$. In this case, we know $\mu$ and $\sigma$, so go ahead and plug them in to get $Z\leq \mathrm{number}$.
Now because $Z$ is standardized, we can just use the table to find $\mathrm{Pr}(Z\leq \mathrm{number})$. Finally, compute $1-\mathrm{Pr}(Z\leq \mathrm{number})$. That's your answer. Note that it's pretty close to zero. This is because $\mathrm{number}$ is quite far from the mean. If people on average usually stand 6 feet from the TV with the remote, and a standard deviation is 2.75, very few people will try to turn on the TV from more than 10 feet.
When you're done, go to WolframAlpha and type in 1-CDF[NormalDistribution[6,2.75],10] to see if you get a similar answer. Computers can compute any normal distribution -- that's where your Z table comes from in the first place. But you don't learn anything about statistics from typing stuff into a computer.
