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Let a distribution function be defined as the following:

$$F(x) = \begin{cases} 1 - \dfrac{16}{x^2} & x\geq 4 \\ 0 & x<4 \end{cases}$$

I was hoping for some confirmation on the median I calculated. I am very new to this material and do not have a lot of examples to work from. What I did was the following:

$$\left(1-\dfrac{16}{m^2}\right) - \left(1-\dfrac{16}{4^2}\right) = 0.5$$

Where $m$ is the median. I solved to get $m = \sqrt{32}$

Chinnapparaj R
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user3753
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1 Answers1

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Median of distribution is a point $x$ such that

$$P(X \le x) = P(X \ge x)$$ $$\implies P( X \le x) = 0.5$$ $$\implies F_X(x) = 0.5$$

Edit: Since the given distribution given is continuous, we don't have to worry about $P(X=x)$. If it were a discrete distribution, we would have to adjust accordingly for the value of $P(X=x)$. In particular, if $X$ is a discrete RV,

$$P(X \le x) = P(X < x) + P(X=x)$$

Vizag
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