# Probability that median falls between largest and smallest sample drawn

by ee8291   Last Updated July 09, 2019 02:19 AM

I have no starting point on which formula to use:

One draws 5 samples from a cont. distribution. Now, what is the approx. probability p that the median of the Population falls within the smallest and largest samples drawn?**

Can anybody guide me on which formula to use or if any other approach is better suited?

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#### Answers 2

I'm not sure about the general answer but if the continuous distribution is the standard normal then the probability that the median lies between the min and the max of a sample of size $$n$$ equals

$$\frac{ 2^{n-1} - 1 }{2^{n-1}}$$

You can work this out based on the distribution of normal order statistics, which I will leave to you as an exercise since this appears to be a self study question.

logistic
July 09, 2019 02:43 AM

Let $$m$$ denote the true median of the distribution. Since the distribution is continuous, by definition there is a probability of one-half that a given value will fall above/below the median. Hence, you have:

$$\mathbb{P}(X_{(1)} > m) = \prod_{i=1}^n \mathbb{P}(X_i > m) = \prod_{i=1}^n \frac{1}{2} = \frac{1}{2^n},$$

$$\mathbb{P}(X_{(n)} < m) = \prod_{i=1}^n \mathbb{P}(X_i < m) = \prod_{i=1}^n \frac{1}{2} = \frac{1}{2^n}.$$

Putting this together and using the inclusion-exclusion principle you get:

\begin{aligned} \mathbb{P}(X_{(1)} \leqslant m \leqslant X_{(n)}) &= 1 - \mathbb{P}(X_{(1)} > m \ \text{ or } \ X_{(n)} < m) \\[6pt] &= 1 - \mathbb{P}(X_{(1)} > m) - \mathbb{P}(X_{(n)} < m) + \mathbb{P}(X_{(n)} < m < X_{(1)}) \\[6pt] &= 1 - \mathbb{P}(X_{(1)} > m) - \mathbb{P}(X_{(n)} < m) + 0 \\[6pt] &= 1 - \frac{1}{2^n} - \frac{1}{2^n} \\[6pt] &= 1 - \frac{1}{2^{n-1}} \\[6pt] &= \frac{2^{n-1} - 1}{2^{n-1}}. \\[6pt] \end{aligned}

Note that this result holds for any continuous distribution. Unsurprisingly, as $$n \rightarrow \infty$$ we have $$\mathbb{P}(X_{(1)} \leqslant m \leqslant X_{(n)}) \rightarrow 1$$, so the range of the sample will eventually come to encompass the true median, with probability approaching one.

Ben
July 09, 2019 02:58 AM