# Quartiles in a legitimate pdf function

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

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

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

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

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

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