# Percentage greater than 2 standard deviations from the mean

A question reads: "The weights of $910$ young deer tagged and weighed in a research study are normally distributed with a mean of $86$ pounds and a standard deviation of $2.5$ pounds."

Approximately how many deer weigh more than $91$ pounds?

Since $34.1$% fall within the first standard deviation and $13.6$% fall within the second standard deviation, then $2.3$% will be greater than two standard deviations. $2.3$% of $910$ equals $20.93$.

Possible multiple choice answers are $21$ or $23$. I picked $21$. The test guide says $23$. I assume this is a typo. Anyone think I went awry?

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The answer key may be using the rougher guide ('empirical rule') that about $95\%$ of the area under a normal curve is within $2$ standard deviations of the mean. So about $2.5\%$ of the data is more than $2$ standard deviations above the mean. And $2.5\%$ of $910$ is $22.75$, close to their answer of $23$.

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+1: Nice. ${}{}{}$ – copper.hat Feb 9 at 16:25
@copper.hat: Thanks! – paw88789 Feb 9 at 16:29
As someone who reads and writes these kind of answer guides on the regular, I can confirm that this is the case. Whoever wrote the answer choice of 21 did not think about the exact value and just relied on the empirical rule. – Jared Goguen Feb 9 at 23:16

I think paw88789's answer is closer to the spirit of the question.

Let $W$ be the weight of the deer, then ${W - 86 \over 2.5}$ has distribution ${\cal N}(0,1)$.

You want to compute $p [W > 91]$, since ${91 - 86 \over 2.5} = 2$, we see that this is $p [W > 91] = 1 - p[W \le 91] = 1 - \Phi(2) \approx 0.02775$.

Hence you would expect to see $910 \,p [W > 91] \approx 21$ deer over 91 pounds.

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