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I've come across the following problem, and I am tempted to delve into order statistics to solve this. I would greatly appreciate any help!

Suppose you draw 6 independent samples from a continuous distribution. What is the approximate probability that the population median of the distribution lies between the smallest and largest observations?

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    $\begingroup$ What is the probability that a sample of 6 data points are all below the population median? Do you see the next steps? $\endgroup$ – soakley Jan 7 '19 at 22:53
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    $\begingroup$ RIght. What's the chance that 5 are below it and 1 is above? $\endgroup$ – Glen_b -Reinstate Monica Jan 8 '19 at 2:47
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    $\begingroup$ The analysis posted in reply to a similar question at stats.stackexchange.com/questions/122001 provides a full answer. Another analysis at stats.stackexchange.com/questions/45124 also answers this question. $\endgroup$ – whuber♦ Jan 8 '19 at 16:11
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    $\begingroup$ Foxah - You found the probability that all 6 data points will be below the population median. So you also know the probability that all 6 data points will be above the population median. There is only one other pertinent possibility - that the sample range contains the median. $\endgroup$ – soakley Jan 8 '19 at 18:52
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    $\begingroup$ @Foxah my hint was an attempt to get you to spot you were just doing binomial calculations (so giving a shortcut to computing all such quantile questions) $\endgroup$ – Glen_b -Reinstate Monica Jan 9 '19 at 0:37
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The probability that the population median does not lie between the largest and smallest observatrions is the probability that either all of them are larger than the median or all of them are smaller. This is $$ \left( \frac 1 2 \right)^6 + \left( \frac 1 2 \right)^6. $$

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