# Statistics question - comparing single samples from normal distribution

**URL:** <https://boards.straightdope.com/t/statistics-question-comparing-single-samples-from-normal-distribution/546146>\
**Category:** Factual Questions\
**Created:** [July 9, 2010, 8:42pm UTC](https://boards.straightdope.com/t/statistics-question-comparing-single-samples-from-normal-distribution/546146 "2010-07-09T20:42:18Z")\
**Posts on this page:** 5\
**Page:** 1

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**Author:** ![JR\_Brown](https://avatars.discourse-cdn.com/v4/letter/j/d6d6ee/32.png) [@JR\_Brown](https://boards.straightdope.com/u/JR_Brown)\
**Post date:** [July 9, 2010, 8:42pm UTC](https://boards.straightdope.com/t/statistics-question-comparing-single-samples-from-normal-distribution/546146/1 "2010-07-09T20:42:18Z")

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Hi guys;

My statistics education is old and faintly remembered; some kind soul please help me work this out.

Let us say we have population #1, which is normally distributed with mean X and standard deviation a. If we randomly select two individuals from this population, what is the probability that they will differ by at least 5%?

Let us also have population #2, which is normally distributed with mean Y and standard deviation b. If we randomly select one individual from population #1 and one from population #2, what is the probability that they will differ by at least 5%?

If the latter is difficult to calculate, how about for the case in which the means differ but the standard deviations are equal (i.e. a=b).

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**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\
**Post date:** [July 10, 2010, 2:50am UTC](https://boards.straightdope.com/t/statistics-question-comparing-single-samples-from-normal-distribution/546146/2 "2010-07-10T02:50:47Z")

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Saying that two individuals differ by 5% is slightly ambiguous because 5% of the smaller value is not the same as 5% of the larger value. What exactly do you mean by that?

The difference between two individuals from population #1 has mean 0 and standard deviation a \* sqrt(2), and the difference between an individual from population #1 and one from population #2 has mean X - Y and standard deviation sqrt(a^2 + b^2). For specific values of a and b, you can compute the probabilities that these differences will be larger than any given value, but there’s no closed form symbolic expression.

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**Author:** ![JR\_Brown](https://avatars.discourse-cdn.com/v4/letter/j/d6d6ee/32.png) [@JR\_Brown](https://boards.straightdope.com/u/JR_Brown)\
**Post date:** [July 10, 2010, 4:35am UTC](https://boards.straightdope.com/t/statistics-question-comparing-single-samples-from-normal-distribution/546146/3 "2010-07-10T04:35:11Z")

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> [@ultrafilter](#):
>
> Saying that two individuals differ by 5% is slightly ambiguous because 5% of the smaller value is not the same as 5% of the larger value. What exactly do you mean by that?

Eh, I just picked 5% out of a hat for representative purposes, so as not to throw in another variable.

> [@ultrafilter](#):
>
> The difference between two individuals from population #1 has mean 0 and standard deviation a \* sqrt(2)

I’m not looking for the average difference, I’m looking for the probability of randomly picking two individuals which are more than \<blah units\> apart. Is there any way to calculate this?

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**Author:** ![Oukile](https://avatars.discourse-cdn.com/v4/letter/o/13edae/32.png) [@Oukile](https://boards.straightdope.com/u/Oukile)\
**Post date:** [July 10, 2010, 2:51pm UTC](https://boards.straightdope.com/t/statistics-question-comparing-single-samples-from-normal-distribution/546146/4 "2010-07-10T14:51:31Z")

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> [@JR\_Brown](#):
>
> I’m not looking for the average difference, I’m looking for the probability of randomly picking two individuals which are more than \<blah units\> apart. Is there any way to calculate this?

Indeed there is, and in order to calculate it you need to compute the average difference 😃

If two individuals are more than \<blah\>, say z units for convenience, then the difference between them is more than z.

As **ultrafilter** said, the difference between the two individuals follows a Gaussian distribution (call it capital Z),

- whose average is EZ = X-Y
- whose standard distribution is sZ = sqrt(a^2 + b^2).

(I call them EZ and sZ for convenience)

Now the question is: what the probability than Z\>z (or Z\<-z) ?

At this point you can just use an online calculator such at [this one](http://davidmlane.com/hyperstat/z_table.html). Compute EZ, sZ and enter them in the boxes ‘Mean’ and ‘Sd’, then select ‘Ouside’ and enter z in the two boxes.  
Just to get a little deeper in it, the usual way is to ‘normalize Z’, i.e. to work on the distribution U = (Z-EZ)/sZ, which has a mean of 0 and a standard deviation of 1. (the Gaussian distribution with mean of 0 and std. dev. of 1 is generally written U)

The idea is that Z\>z (or Z\<-z) is equivalent to U\>(z-EZ)/sZ (or U\<-(z-EZ)/sZ ).

So, your procedure would be:

- You know X,Y,a,b as well as the minimal difference z
- You compute EZ, sZ, and then (z-EZ)/sZ
- Then you use Gaussian distribution table which has been computed for U and in which you can read these probabilities directly.

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**Author:** ![JR\_Brown](https://avatars.discourse-cdn.com/v4/letter/j/d6d6ee/32.png) [@JR\_Brown](https://boards.straightdope.com/u/JR_Brown)\
**Post date:** [July 10, 2010, 3:19pm UTC](https://boards.straightdope.com/t/statistics-question-comparing-single-samples-from-normal-distribution/546146/5 "2010-07-10T15:19:37Z")

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OK, cool! Thanks 🙂