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I want to ckeck if I am correct in how to calculate quartiles (Q1, Q2, Q3) in a pdf graph function and ask one question. I am working on a function $(3/16)*(x+2)^2$ on $ [-4,0]$

I know that the second quartile is the Median and the correct integral has form $$\int_{-4}^{M} \frac3{16}(x+2)^2dx =\frac 12$$ And from that M is calculated. However, what would be the values of the integral be for Q1 and Q3 and why?

Thank you

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  • $\begingroup$ you are leaving out the normalization factor of 3/16 $\endgroup$
    – WW1
    Feb 5, 2019 at 22:55

2 Answers 2

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If $x_p$ is the $p^{th}$ percentile then

$$ \frac 3{16}\int_{-4}^{x_p}(x+2)^2dx = \frac p{100} $$

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Median (same as Q2) means that $P(X \leq x) = 1/2$.

Similarly, Q1 means that $P(X \leq x) = 1/4$ and Q3 means $P(X \leq x) = 3/4$

Therefore, Q1 = a where $$\int_{-4}^{a} \frac3{16}(x+2)^2dx =\frac 14$$

And, Q3 = b where $$\int_{-4}^{b} \frac3{16}(x+2)^2dx =\frac 34$$

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  • $\begingroup$ Thank you for the explanation, i thought so but did not know why are these values used. Now I can solve my exam correctly =) $\endgroup$
    – Mike Peretolchyn
    Feb 5, 2019 at 23:03

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