# How To Solve for Percentage When The Only Given Values Are Mean and Standard Deviation

How To Solve for Percentage When The Only Given Values Are Mean and Standard Deviation

For Example: The scores of students in Mathematics examination is normally distributed with a mean of 60 and a standard deviation of 8.

Question: How many percent of the examinees got scores below 44?

Can you please explain how to solve this kind of problem and how to check if I got the right answer. You don't have to give the answer for the sample question if you don't want to. Just please enlighten us with this kind of problem. Thank you.

I will point out that contrary to the question title, not two, but three pieces of information have been communicated in the question: (1) the fact that the scores are normally distributed, (2) the mean of the scores, and (3) the standard deviation of the scores. By far the most important is that first piece of information, so don't overlook that. (Indeed, if only the mean and standard deviation were given, then the question could not be answered.)

For practical purposes, one turns to statistical tables or technology which give the area under selected parts of a normal curve, which is the same as the percent/probability of elements in that interval (i.e., the "integral" from calculus). It is convenient to first standardize the distribution in question, transforming it to the normal curve with mean 0 and standard deviation 1 (which has the same relative areas, and thus requires only a single table to look-up, or simplified programming within the computer application).

So for a percent-less-than question, one would standardize and then look up in a normal curve left-area table (or else use technology). For the question here, the standard score for the cut is $z = \frac{x - \mu}{\sigma} = \frac{44 - 60}{8} = -2$. Looking up this z-score in a normal table like here gives the area of 0.0228, that is 2.28%.

This is a very fundamental technique when dealing with common probability distributions (esp. the normal curve), and if you're in a class on the subject it should be well-laid out in a chapter on this subject. For this particular problem with $z = -2$, some will even be able to answer it mentally by remembering the Empirical Rule (or making a sketch); knowing that 95.44% of the area is within $z = \pm 2$, the remainder in the outside tails is 4.56%, with exactly half in the left-hand tail, that is, $4.56\% / 2 = 2.28\%$.

For a normal random variable X, with mean $\mu$ and standard deviation $\sigma$ the probabiliity that it smaller than some value $c$ is:

$P\left(X\leq c\right)=F(c;\mu,\sigma)$ where $F()$ is the cumulative distribution function of the random variable. Similarly, $P\left(X> c\right)=1-F(c;\mu,\sigma)$.

In order to solve that you need to evaluate $F(44;60,8)$. In e.g. excel, this can be done with normdist(44;60;8).