4
$\begingroup$

If we select one random sample with 4 elements from a normal distribution, and we denote the minimum value among the sample with $a$, and denote the maximum value among the sample with $b$, what is the probability that distance between $a$ and $b$ (i.e. the range) includes the true population mean?

Could anyone help me how to solve this problem from the 2012 local-math-contest? The short solution of this question: $7/8$.

Improve this question
$\endgroup$

2 Answers 2

Reset to default
2
$\begingroup$

Here is an illustration of what @GregSnow suggested:

Tree

We start at the top and go down. At the top, there are two possibilities for the first value:

  1. The first value is below the median.
  2. The first value is above the median.

Because the median and mean are the same in a normal distribution, there is a 50% chance that the first value lies below and a 50% chance that it lies above the median.

For the second value, we have again two possibilities: either it lies above or below the median. Again, the chances are 50/50.

We are only interested in the paths where either all 4 values fall below or above the median. These paths are marked with green arrows in the graphic. The other paths are omitted for clarity.

As we go down, we multiply the probabilities. At the bottom, we add the probabilities.

Can you solve it from here?

Improve this answer
$\endgroup$
4
  • $\begingroup$ is it possible to complete it ? $\endgroup$
    – Johnatan Morian
    Commented Mar 17, 2015 at 10:25
  • $\begingroup$ @JohnatanMorian Were you able to finish the calculation yourself? If not, where are you stuck? $\endgroup$
    – COOLSerdash
    Commented Mar 17, 2015 at 13:13
  • $\begingroup$ i had no enough background.... if it's possible complete it. $\endgroup$
    – Johnatan Morian
    Commented Mar 17, 2015 at 14:46
  • $\begingroup$ @JohnatanMorian One more hint: The probability that all 4 values are below the median are: $(1/2)^4=1/16$. So what is the probability that all 4 values are above the median? Add the two probabilities and substract that from 1, as Greg Snow suggested. $\endgroup$
    – COOLSerdash
    Commented Mar 17, 2015 at 15:13
2
$\begingroup$

I would solve the inverse: what is the probability that the range does not include the mean, then subtract that from 1.

So that is the sum of the 2 cases: a is greater than the mean; b is less than the mean. Calculate the probability of these 2 cases (mutually exclusive) using the distribution of the min(max) from n=4, and subtract the 2 values (should be the same) from 1.

Instead of using the formula for order statistics you can also use the fact that the median and the mean are the same for the normal and therefore there is a 50% chance of each point being above(below) the mean, so what is the probability of all 4 values being greater(less) than the mean?

Improve this answer
$\endgroup$
7
  • $\begingroup$ 1/4 ? is correct? $\endgroup$
    – Johnatan Morian
    Commented Mar 16, 2015 at 20:23
  • $\begingroup$ infact I confused, is it possible to make it clearer? $\endgroup$
    – Johnatan Morian
    Commented Mar 16, 2015 at 20:26
  • $\begingroup$ @JohnatanMorian, What is the probability of 1 value being greater than the mean? what is the probability of 2 (independent) observations both being greater than the mean? now do 3, then 4. The same for all less than the mean. $\endgroup$
    – Greg Snow
    Commented Mar 16, 2015 at 21:23
  • $\begingroup$ you are expert, but my background is Geology. if I didnt waste your time, like a teacher, learn it me. I'm so glad. $\endgroup$
    – Johnatan Morian
    Commented Mar 16, 2015 at 21:28
  • $\begingroup$ @JohnatanMorian, click on or hover over the "self-study" tag that your question is tagged with. It states that the policy for this community is to "provide helpful hints", which I have done. If you really want to understand the answer (rather than having someone do your homework for you) then you should be able to work it out (google for basic rules of probability if you need help there) from my hints. $\endgroup$
    – Greg Snow
    Commented Mar 16, 2015 at 21:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.