Answer:
(a) Lower quartile: \(1\). Upper quartile: \(8-\sqrt{19}\). So the interquartile range is \(7-\sqrt{19}\approx 2.64\).
(b) \(\mathrm{E}(X^3)=34.56\) \(\left(=\frac{864}{25}\right)\).
(c) The probability density function of \(Y\) is
\(g(y)=\begin{cases}\frac{1}{60}(32y-4y^3),&0\le y\le \sqrt6,\\0,&\text{otherwise.}\end{cases}\)
For \(0\le x\le 6\), the cumulative distribution function is
\(F(x)=\frac{1}{60}(16x-x^2).\)
Since \(X\) is continuous, quartiles are found by solving \(F(x)=0.25\) and \(F(x)=0.75\).
(a) For the lower quartile \(Q_1\),
\(\frac{1}{60}(16x-x^2)=\frac14\).
So
\(16x-x^2=15\)
\(x^2-16x+15=0\)
\((x-1)(x-15)=0.\)
The value must lie in \([0,6]\), so \(Q_1=1\).
For the upper quartile \(Q_3\),
\(\frac{1}{60}(16x-x^2)=\frac34\).
So
\(16x-x^2=45\)
\(x^2-16x+45=0.\)
Using the quadratic formula,
\(x=\frac{16\pm\sqrt{16^2-4\cdot 45}}{2}=\frac{16\pm\sqrt{76}}{2}=8\pm\sqrt{19}.\)
Again the value must lie in \([0,6]\), so \(Q_3=8-\sqrt{19}\).
Hence
\(\text{IQR}=Q_3-Q_1=(8-\sqrt{19})-1=7-\sqrt{19}.\)
Therefore the interquartile range is \(\boxed{7-\sqrt{19}}\), approximately \(2.64\).
(b) Differentiate the cdf to get the pdf:
\(f(x)=F'(x)=\frac{1}{60}(16-2x)=\frac{1}{30}(8-x),\qquad 0\le x\le 6.\)
So
\(\mathrm{E}(X^3)=\int_0^6 x^3 f(x)\,dx=\frac{1}{30}\int_0^6 x^3(8-x)\,dx.\)
Expand the integrand:
\(\mathrm{E}(X^3)=\frac{1}{30}\int_0^6 (8x^3-x^4)\,dx.\)
Integrate:
\(\mathrm{E}(X^3)=\frac{1}{30}\left[2x^4-\frac{x^5}{5}\right]_0^6.\)
Substitute the limits:
\(\mathrm{E}(X^3)=\frac{1}{30}\left(2\cdot 6^4-\frac{6^5}{5}\right)=\frac{1}{30}\left(2592-\frac{7776}{5}\right).\)
\(2592=\frac{12960}{5}\), so
\(\mathrm{E}(X^3)=\frac{1}{30}\cdot \frac{5184}{5}=\frac{5184}{150}=\frac{864}{25}=34.56.\)
Thus \(\boxed{\mathrm{E}(X^3)=34.56}\).
(c) We are given \(Y=\sqrt{X}\), so \(X=Y^2\).
Since \(0\le X\le 6\), the range of \(Y\) is
\(0\le Y\le \sqrt6.\)
First find the cdf of \(Y\):
\(G(y)=P(Y\le y)=P(\sqrt{X}\le y).\)
For \(y\ge 0\), this is equivalent to
\(G(y)=P(X\le y^2)=F(y^2).\)
Hence, for \(0\le y\le \sqrt6\),
\(G(y)=\frac{1}{60}(16y^2-y^4).\)
Differentiate to obtain the pdf:
\(g(y)=G'(y)=\frac{1}{60}(32y-4y^3).\)
So the probability density function of \(Y\) is
\(g(y)=\begin{cases}\frac{1}{60}(32y-4y^3),&0\le y\le \sqrt6,\\0,&\text{otherwise.}\end{cases}\)
This can also be written as
\(g(y)=\frac{1}{15}(8y-y^3),\qquad 0\le y\le \sqrt6.\)
