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Let Y be a continuous random variable with cumulative distribution function. $$ F (x) = \begin{cases} 0 &\text{ if } x < 1 \\ 1 - \dfrac{1}{x^2} &\text{ if } x ≥ 1 \end{cases} $$

and probability density function $$ f(y) = \begin{cases} \dfrac{2}{y^3} &\text{ if } y ≥ 1 \\\\ 0 &\text{ otherwise} \end{cases} $$

(a) Find the median of Y.

(b) Find the interquartile range of Y.

(c) Why is the expected value of Y less than, equal to, or greater than its median?

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closed as off-topic by heropup, BruceET, iadvd, Leucippus, Namaste Feb 17 '17 at 0:53

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  • 1
    $\begingroup$ Just need to use the definition in each case, what have you tried? $\endgroup$ – Smurf Feb 16 '17 at 20:10
  • $\begingroup$ stuck in integeration! $\endgroup$ – tesla Feb 16 '17 at 20:19
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    $\begingroup$ All three of your posts so far have been asked without demonstrating any of your own effort, and more curiously, have specifically asked for help from a particular user on this site. Math.SE is not your personal question-answering service and users are not at your every beck and call. Your behavior suggests you created a second account after having your previous account locked for failing to adhere to community guidelines. $\endgroup$ – heropup Feb 16 '17 at 20:21
  • $\begingroup$ Pareto distribution.. Quartiles and median in Answer of @Measure_theory correct: In R: x = c(2, sqrt(2), sqrt(4/3)); 1 - 1 /x^2 returns 0.75, 0.50, 0.25. Try integration for mean again, note limits $\int_1^\infty.$ Check answ in Wilkipedia link. And, I agree with previous comment that you need to show more initiative before posting hwk problems. $\endgroup$ – BruceET Feb 16 '17 at 20:42
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For (a), note that the median is the point where $F(x) = 0.5$. So we need to solve for $x$ in $$0.50 = 1-\frac{1}{x^{2}}$$

multiplying through by $x^2$ and rearranging yields $$(1-0.50)x^2 = 1 = 0.5x^2$$

which gives us $ x = \sqrt2$

For (b): the interquartile range can be found in a similar manner by subtracting the third quantile (where $F(x) = 0.75$) from the second (where $F(x) = 0.25$)

Using the same approach as before, we have $$(1-0.75)x^2 = 1\Rightarrow x = 2$$ and $$(1-0.25)x^2 = 1\Rightarrow x = \sqrt{1.333}$$

which gives us $2 - \sqrt{1.333}$

For (c), I would first calculate the mean and variance and then use the skewness formula $\mathbb{E}\bigg[\bigg(\frac{X-\mu}{\sigma}\bigg)\bigg]^3$ where $\mu$ is the mean and $\sigma$ is the standard deviation.

The sign of the above will tell you whether the mean is below the medium or above. See here

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  • $\begingroup$ That's probably a good thing to check for in general but this definitely look likes a homework question and I presume so. This should be easy since the PDF is a polynomial. $\endgroup$ – measure_theory Feb 16 '17 at 20:27
  • $\begingroup$ Yes, need to check 3rd moment before recommending it: $\int_1^\infty x^3(2/x^3)\,dx = ??.$ Commonly, right skewed distributions have $\mu > \eta$ (pop mean exceeding pop median). Quartiles and median OK (+1). $\endgroup$ – BruceET Feb 16 '17 at 20:53
  • $\begingroup$ how to solve the integeration .. $\endgroup$ – tesla Feb 18 '17 at 22:36

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