I'm given the following problem:
Could anyone give some insight into how to solve this?
I understand that for the normal distribution, I will likely have to look the percentage up in a table. However, I'm not sure how to find the value that I need to be searching for.
I suspect that the uniform distribution will be 1, but I'm not sure how to work it out.
For the uniform distribution, the variance is $\frac{1}{12}(b-a)^2$. So $3$ standard deviation units up from the mean is $\frac{a+b}{2}+\frac{\sqrt{3}}{2}(b-a)$. We can get a similar expression for $3$ standard deviation units below the mean.
Since $\frac{\sqrt{3}}{2}(b-a)\gt \frac{b-a}{2}$, the interval from $3$ standard deviation units below the mean to $3$ standard deviation units above the mean covers the whole interval $[a,b]$, and more. So you are right, the probability is $1$.
The variance of the (continuous) uniform can be looked up. It is also not difficult to calculate. If $X$ is uniform on $[a,b]$, we need to calculate $E(X^2)-(E(X))^2$. Symmetry gives us $E(X)$, and $E(X^2)$ can be found by integrating $\frac{x^2}{b-a}$ over the interval $[a,b]$.
For the normal, use tables to find $\Pr(-3\le Z\le 3)$, where $Z$ is standard normal. How one does it depends on how the table is organized. The one I looked up on Wikipedia says that $\Pr(0\le Z\le 3)\approx 0.49865$. Double this number.