Answer: (a) \(a=-\frac{1}{5}\), \(b=\frac{7}{25}\).

(b) \(f(x)=\begin{cases}\frac{1}{5},&1\leqslant x\lt4,\\ \frac{x}{25},&4\leqslant x\leqslant6,\\ 0,&\text{otherwise.}\end{cases}\)

(c) \(\operatorname{Var}(X)=\frac{48659}{22500}\approx 2.16\).

(d) The 10th percentile is \(1.5\), and the 90th percentile is \(\sqrt{31}\approx5.57\).

(a) Since \(X\) is continuous, the cumulative distribution function must be continuous at the joining points.

At \(x=1\), the value must match the left-hand value \(0\), so

\(\frac{1}{5}(1)+a=0\), giving \(a=-\frac{1}{5}\).

At \(x=4\), the two formulae must agree. Using \(a=-\frac{1}{5}\),

\(\frac{1}{5}(4)-\frac{1}{5}=\frac{3}{5}\).

So

\(\frac{1}{50}(4^2)+b=\frac{3}{5}\).

Hence

\(\frac{16}{50}+b=\frac{3}{5}\), so \(\frac{8}{25}+b=\frac{15}{25}\), and therefore \(b=\frac{7}{25}\).

(b) The probability density function is found by differentiating the cumulative distribution function on each interval:

\(f(x)=F'(x)\).

Therefore

\(f(x)=\begin{cases}\frac{1}{5},&1\leqslant x\lt4,\\ \frac{x}{25},&4\leqslant x\leqslant6,\\ 0,&\text{otherwise.}\end{cases}\)

(c) We use \(\operatorname{Var}(X)=\mathrm{E}(X^2)-\{\mathrm{E}(X)\}^2\).

First find \(\mathrm{E}(X^2)\):

\(\mathrm{E}(X^2)=\int_{-\infty}^{\infty}x^2f(x)\,dx\)

\(=\frac{1}{5}\int_1^4 x^2\,dx+\frac{1}{25}\int_4^6 x^3\,dx\).

So

\(\mathrm{E}(X^2)=\left[\frac{x^3}{15}\right]_1^4+\left[\frac{x^4}{100}\right]_4^6\)

\(=\frac{64-1}{15}+\frac{1296-256}{100}\)

\(=\frac{63}{15}+\frac{1040}{100}=\frac{21}{5}+\frac{52}{5}=\frac{73}{5}\).

Given that \(\mathrm{E}(X)=\frac{529}{150}\),

\(\operatorname{Var}(X)=\frac{73}{5}-\left(\frac{529}{150}\right)^2\)

\(=\frac{48659}{22500}\approx2.16\).

(d) The 10th percentile is the value \(x\) such that \(F(x)=0.10\). Since \(F(4)=\frac{3}{5}=0.6\), this lies in the interval \(1\leqslant x\lt4\).

So

\(\frac{x}{5}-\frac{1}{5}=\frac{1}{10}\).

Multiplying by \(10\),

\(2x-2=1\), so \(x=\frac{3}{2}=1.5\).

The 90th percentile is the value \(x\) such that \(F(x)=0.90\). Since this is greater than \(F(4)=0.6\), it lies in the interval \(4\leqslant x\leqslant6\).

So

\(\frac{x^2}{50}+\frac{7}{25}=\frac{9}{10}\).

Now \(\frac{7}{25}=\frac{14}{50}\) and \(\frac{9}{10}=\frac{45}{50}\), so

\(\frac{x^2}{50}=\frac{31}{50}\).

Thus \(x^2=31\). Since \(x\) is in the interval \(4\leqslant x\leqslant6\),

\(x=\sqrt{31}\).