# Intuition and the math behind normalization

What exactly is the purpose of normalization. From what I read, it is to adjust two different sets of values so you can compare them, but I don't understand why, nor the math behind it. Could anyone answer this by also giving a concrete example? Thanks!

• Normalization in what context? Commented Apr 29, 2014 at 20:31
• taking values and normalizing them. like what is the purpose of it and why use it? Commented Apr 30, 2014 at 14:42

Note: I'm assuming you are talking about statistical normalization.

Let's use an example. So colleges accept the SAT scores and ACT scores for their admissions processes. How do you compare the SAT scores to the ACT scores, when the two tests use two different scales? The answer is to normalize the data. That is, we're interested in averages and distances from the averages.

So the SAT has a mean score of say, 1700 (out of 2400) with a standard deviation of 150 points. The ACT has a mean score of 25 with a standard deviation of 1.3. Student A scores a 2000 on the SAT while student B scores a 27 on the ACT. Who is the better candidate for admissions?

Hence, we normalize. We take a z-score: $z_{SAT} = \dfrac{2000 - 1700}{150}$, and $z_{ACT} = \dfrac{27 - 25}{1.3}$. This allows us to see where the two points fall on a standard normal distribution $N(0, 1)$ (a mean of $0$ and standard deviation of $1$). We see that the student taking the SAT is two standard deviations above the mean, while the person who took the ACT is $1.54$ standard deviations above the mean. And so, assuming the two tests are equally valid measures of qualifications, then Student A would be the better candidate for admissions.

Does that make sense?

• Good explanation, though with test scores they often use percentiles rather than normalized measures. Commented Apr 29, 2014 at 21:11
• Of course, the percentiles come from said normalized measures. :-) Commented Apr 29, 2014 at 21:17
• I am pretty sure that's not true. Normalizing a draw from a standard normal distribution is the z-score, which is the $\Phi(z)$ percentile. Normalizing a draw from some asymmetric distribution must necessarily for some "z" score yield another percentile, because normalizing is symmetric about the mean while the cdf isn't. Commented Apr 29, 2014 at 21:23
• That's a good point about normalizing from the given distribution. Commented Apr 29, 2014 at 21:26
• This makes sense, but it to me it seems redundant, because isn't taking the average of each score giving you a percentage you can compare? What makes normalization a better method of comparing two separate pieces of data, than taking the average of said data? Commented Apr 30, 2014 at 13:14