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# Comparing with z-scores

Use standardized scores—also called z-scores— to compare data points from different distributions.

## Want to join the conversation?

- Is it very scientific method? Because there is no way of knowing percentile.

Without percentile there is no way of knowing how he faired relatively in both exam.(6 votes)- If you know the z-score and it's a normal distribution, you can figure out the percentile using a z-table.(25 votes)

- Why did he do better on the LSAT if he's 2.1 SD above the mean? Doesn't this mean he's 2.1 SD far from the mean? Comparable with 1.9, shouldn't he do better on the MCAT since it's closer to the mean?(4 votes)
- The mean is the average result. Would you rather score average on a test, or above average?(27 votes)

- why can he estimate which test score is better based on only the z-score?

what exactly does the z-score show?(2 votes)- z-score is like percentile. It can be used to compare different data sets with different means and standard deviations. It is a universal comparer for normal distribution in statistics. Z score shows how far away a single data point is from the mean relatively. Lower z-score means closer to the meanwhile higher means more far away. Positive means to the right of the mean or greater while negative means lower or smaller than the mean.(9 votes)

- What if I am looking for the standard deviation and I have the Z Score and the Mean?(1 vote)
- You would just plug in the Z-score and mean into our formula, and then solve for the standard deviation using algebra. You also will need the number that the Z-score comes from (in our case, that is Juwan's test scores). Let's use the LSAT example from the video.

(172 - 151) / X = 2.1

21 / X = 2.1

21 = 2.1 * X

10 = X

We get that the standard deviation is 10. Hope this helps! :)(9 votes)

- Hi! I have a question. Why Khan writes 11.9 / 6.4 is less than 2?(1 vote)
- At that point he was just estimating and already had the answer to the question without calculating the exact answer(5 votes)

- Why do we assume it is a normal distribution/relative close to a normal distribution? Would comparing z-scores not apply if it was not a normal distribution?(2 votes)
- Comparing z-scores assumes a normal distribution or a distribution that is approximately normal because the concept of z-scores is based on the properties of the standard normal distribution. While z-scores can still provide some relative comparison in non-normal distributions, their interpretation might be less straightforward due to potential skewness or other deviations from normality.(1 vote)

- Why do we divide the difference between the Juwan's score on each test and the mean, by the standard deviation of that test?

To be more clear, how come we don't do something like this:

Let's say the LSAT's score range is 5x bigger than the MCAT's score range. This means that the LSAT mean should be 5x bigger than the MCAT mean. If Juwan's scores on the two tests are comparable, his score on the LSAT should be 5x his score on the MCAT.

Then, we can compare these two values:

1. ((Juwan's LSAT)-(LSAT mean)) ÷ 5

2. ((Juwan's MCAT)-(MCAT mean))

And whichever one is greater is the test he did better on.

I'm essentially wondering, what's the benefit of dividing the mean-to-data-point difference by the standard deviation? What's the benefit of comparing z-scores rather than just the difference (corrected for scale)? Also, if there's anything I might be misunderstanding, please let me know.

Thank you!(1 vote)- So the z-score has application involving the normal distribution and hypothesis testing.(2 votes)

- why dont the standard deviations be subtracted by the mean to be the final answer(1 vote)
- because the standard deviation is how far the number is from the group the average and if it subtracted by the mean the sd wont be able to be interpreted(2 votes)

- Does this also mean he did better by his own standards? Does percentile from the z-table show this?(1 vote)
- Doing better by his own standards would depend on the individual's expectations or goals. The z-score itself does not directly indicate whether an individual's performance meets their personal standards. Percentiles derived from the z-score, however, can provide information on how an individual's performance ranks relative to others in the distribution.(1 vote)

- What if we take (172-151)/151 for LSAT and (37-25.1)/25.1 for MCAT figures to compare ? Does it make these two results comparable?(1 vote)
- I think that in the (172-151)/151 the 151s would cancel out and leave you with 172, and for the (37-25.1)/25.1 the 25.1s would cancel out leaving you with 37. So I don’t think so.(1 vote)

## Video transcript

- [Instructor] Before applying
to law school in the US, students need to take
an exam called the LSAT. Before applying to medical school, students need to take
an exam called the MCAT. Here are some summary
statistics for each exam. So the LSAT, the mean score is 151 with a standard deviation of 10. And the MCAT, the mean score is 25.1 with a standard deviation of 6.4 Juwan took both exams. He scored 172 on the
LSAT and 37 on the MCAT. Which exam did he do relatively better on? So pause this video, and see
if you can figure it out. So the way I would think about it is you can't just look at the absolute score because they are on different scales and they have different distributions. But we can use this information. If we assume it's a normal distribution or relatively close to
a normal distribution with a meet, centered at this mean, we can think about, well,
how many standard deviations from the mean did he score
in each of these situations? In both cases, he scored above the mean. But how many standard
deviations above the mean? So let's see if we can figure that out. So on the LSAT, let's see,
let me write this down, on the LSAT, he scored 172. So how many standard
deviations is that going to be? Well, let's take 172, his
score, minus the mean, so this is the absolute number that he scored above the mean, and now let's divide that
by the standard deviation. So on the LSAT, this is what? This is going to be 21 divided by 10. So this is 2.1 standard deviations, deviations above the mean, above the mean. You could view this as a z-score. It's a z-score of 2.1. We are 2.1 above the
mean in this situation. Now, let's think about
how he did on the MCAT. On the MCAT, he scored a 37. The mean is a 25.1, and there is a standard deviation of 6.4. So let's see, 37.1 minus 25 would be 12, but now it's gonna be 11.9, 11.9 divided by 6.4. So without even looking at this, so this is going to be approximately, well, this is gonna be a
little bit less than two. This is going to be less than two. So based on this information, and we could figure out
the exact number here. In fact, let me get my calculator out. So you get the calculator. So if we do 11.9 divided by 6.4, that's gonna get us to one
point, I'll just say one point, I'll just say approximately 1.86, so approximately 1.86. So relatively speaking, he did
slightly better on the LSAT. He did more standard deviations,
although this is close. I would say they're comparable. He did roughly two standard deviations if we were to round to the
nearest standard deviation. But if you wanted to get precise, he did a little bit better, relatively speaking, on the LSAT. He did 2.1 standard deviations here while over here he did 1.86
or 1.9 standard deviations. But in everyday language, you would probably say,
well, this is comparable. If this was three standard deviations and this is one standard
deviation, then you'd be like, oh, he definitely did better on the LSAT.