$\begingroup$

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?

$\endgroup$

2 Answers

$\begingroup$

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 $23$.

However, your answer is more accurate.

$\endgroup$ 2 $\begingroup$

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.

$\endgroup$

Your Answer

Sign up or log in

Sign up using Google Sign up using Facebook Sign up using Email and Password Submit

Post as a guest

Post Your Answer Discard

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy