Probability and Statistics (Quartiles) Q: Find the upper and lower quartiles of the random variabl

Probability and Statistics (Quartiles)
Q: Find the upper and lower quartiles of the random variable.
$f\left(x\right)=\frac{1}{x\mathrm{ln}\left(1.5\right)}$
for $4\le x\le 6$
I set $F\left(x\right)=0.25$ and $F\left(x\right)=0.75$ to denote $Q1$ and $Q3$
This is what I got for the answers:
For
$F\left(X\right)=0.25,x=9.86$
For
$F\left(X\right)=0.75,x=3.28$
However, these answers are wrong as compared with my textbook. It should be:
For
$F\left(X\right)=0.25,x=4.43$
For
$F\left(X\right)=0.75,x=5.42$
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Diya Bass
For $x$ between $4$ and $6$ we have
$F\left(x\right)=Pr\left(X\le x\right)={\int }_{4}^{x}\frac{1}{\mathrm{ln}\left(1.5\right)t}\phantom{\rule{thinmathspace}{0ex}}dt=\frac{1}{\mathrm{ln}\left(1.5\right)}\left(\mathrm{ln}x-\mathrm{ln}4\right).$
For the first quartile ${q}_{1}$, we want
$\frac{1}{\mathrm{ln}\left(1.5\right)}\left(\mathrm{ln}\left({q}_{1}\right)-\mathrm{ln}4\right)=0.25.$
To solve, note that the above equation is equivalent to
$\mathrm{ln}\left({q}_{1}/4\right)=\left(0.25\right)\mathrm{ln}\left(1.5\right).$
Equivalently,
$\frac{{q}_{1}}{4}=\mathrm{exp}\left(\left(0.25\right)\mathrm{ln}\left(1.5\right)\right)=\left(1.5{\right)}^{0.25}.$
We get ${q}_{1}\approx 4.427$. I am sure you can now take care of ${q}_{3}$