iyiswad9k

2022-05-02

Probability and Statistics (Quartiles)

Q: Find the upper and lower quartiles of the random variable.

$f(x)=\frac{1}{x\mathrm{ln}(1.5)}$

for $4\le x\le 6$

I set $F(x)=0.25$ and $F(x)=0.75$ to denote $Q1$ and $Q3$

This is what I got for the answers:

For

$F(X)=0.25,x=9.86$

For

$F(X)=0.75,x=3.28$

However, these answers are wrong as compared with my textbook. It should be:

For

$F(X)=0.25,x=4.43$

For

$F(X)=0.75,x=5.42$

Please help me. I don't know what I'm doing wrong.

Q: Find the upper and lower quartiles of the random variable.

$f(x)=\frac{1}{x\mathrm{ln}(1.5)}$

for $4\le x\le 6$

I set $F(x)=0.25$ and $F(x)=0.75$ to denote $Q1$ and $Q3$

This is what I got for the answers:

For

$F(X)=0.25,x=9.86$

For

$F(X)=0.75,x=3.28$

However, these answers are wrong as compared with my textbook. It should be:

For

$F(X)=0.25,x=4.43$

For

$F(X)=0.75,x=5.42$

Please help me. I don't know what I'm doing wrong.

Diya Bass

Beginner2022-05-03Added 20 answers

For $x$ between $4$ and $6$ we have

$F(x)=Pr(X\le x)={\int}_{4}^{x}\frac{1}{\mathrm{ln}(1.5)t}\phantom{\rule{thinmathspace}{0ex}}dt=\frac{1}{\mathrm{ln}(1.5)}(\mathrm{ln}x-\mathrm{ln}4).$

For the first quartile ${q}_{1}$, we want

$\frac{1}{\mathrm{ln}(1.5)}(\mathrm{ln}({q}_{1})-\mathrm{ln}4)=\mathrm{0.25.}$

To solve, note that the above equation is equivalent to

$\mathrm{ln}({q}_{1}/4)=(0.25)\mathrm{ln}(1.5).$

Equivalently,

$\frac{{q}_{1}}{4}=\mathrm{exp}((0.25)\mathrm{ln}(1.5))=(1.5{)}^{0.25}.$

We get ${q}_{1}\approx 4.427$. I am sure you can now take care of ${q}_{3}$

$F(x)=Pr(X\le x)={\int}_{4}^{x}\frac{1}{\mathrm{ln}(1.5)t}\phantom{\rule{thinmathspace}{0ex}}dt=\frac{1}{\mathrm{ln}(1.5)}(\mathrm{ln}x-\mathrm{ln}4).$

For the first quartile ${q}_{1}$, we want

$\frac{1}{\mathrm{ln}(1.5)}(\mathrm{ln}({q}_{1})-\mathrm{ln}4)=\mathrm{0.25.}$

To solve, note that the above equation is equivalent to

$\mathrm{ln}({q}_{1}/4)=(0.25)\mathrm{ln}(1.5).$

Equivalently,

$\frac{{q}_{1}}{4}=\mathrm{exp}((0.25)\mathrm{ln}(1.5))=(1.5{)}^{0.25}.$

We get ${q}_{1}\approx 4.427$. I am sure you can now take care of ${q}_{3}$

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