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I explain it like this: It's 3rd and 4. You have two running backs. Running back A averages 8 yards a carry with a standard deviation of 6. Running back B
by crikli 16y ago
I explain it like this:
It's 3rd and 4. You have two running backs.
Running back A averages 8 yards a carry with a standard deviation of 6.
Running back B averages 5 yards a carry with a standard deviation of half a yard.
Which guy do you choose? If you just look at averages, you probably would choose A. But if you think of the standard deviation as an indication of consistency you choose B, because his 5 yard gains are very consistent, whereas A might bust one for big yards, but he's also likely to get dropped for a loss.
- jasonlotito 16y agoAhh yes, but it's not that simple. You can also account for the plays these are on - some people crack under pressure. If you just add up the numbers without accounting for other conditions, results can change dramatically. This isn't to say your example is wrong, but simply to say that all data is not equal and to make sure you use the right data for whatever you're trying to accomplish.