(a) X has a probabiliity density of
[tex]f_X(x) = \begin{cases}\dfrac1{b-a}&\text{if }a<x<b\\0&\text{otherwise}\end{cases}[/tex]
If the lower quartile is 5 and the upper quartile is 9, then
[tex]\displaystyle \int_a^5 f_X(x)\,\mathrm dx = 0.25 \text{ and } \int_9^b f_X(x)\,\mathrm dx = 0.25[/tex]
Computing the integrals gives the following system of equations:
[tex]\displaystyle \int_a^5\frac{\mathrm dx}{b-a} = \frac{5-a}{b-a} = 0.25 \\\\ \int_9^b \frac{\mathrm dx}{b-a} = \frac{b-9}{b-a} = 0.25[/tex]
5 - a = 0.25 (b - a) ==> 0.75a + 0.25b = 5 ==> 3a + b = 20
b - 9 = 0.25 (b - a) ==> 0.25a + 0.75b = 9 ==> a + 3b = 36
Eliminate a :
(3a + b) - 3 (a + 3b) = 20 - 3×36
-8b = -88
==> b = 11 ==> a = 3
and so P(X = x) = 1/(11 - 3) = 1/8 for all 3 < x < 11.
(b)
[tex]\displaystyle P(6<X<7) = \int_6^7f_X(x)\,\mathrm dx = \int_6^7\frac{\mathrm dx}8 = \boxed{\frac18}[/tex]
(c) The distribution function is then
[tex]\displaystyle F_X(x) = \int_{-\infty}^x f_X(t)\,\mathrm dt = \begin{cases}0&\text{if }x\le3 \\ \dfrac x8 &\text{if }3<x<11 \\ 1&\text{if }x\ge11\end{cases}[/tex]
