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Comparing with z-scores

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

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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.