Solution. (a). A density must integrate to 1 over its range: \[\int _0^2 k(x+2)^2\,dx = k\left [\frac {(x+2)^3}{3}\right ]_0^2 = k\cdot \frac {64-8}{3} = \frac {56k}{3} = 1 \implies k = \frac {3}{56}.\]

(b). Expanding \((x+2)^2 = x^2+4x+4\), \begin {align*} E(X) &= \frac {3}{56}\int _0^2 (x^3+4x^2+4x)\,dx = \frac {3}{56}\left [\frac {x^4}{4}+\frac {4x^3}{3}+2x^2\right ]_0^2\\ &= \frac {3}{56}\left (4+\frac {32}{3}+8\right ) = \frac {3}{56}\cdot \frac {68}{3} = \frac {17}{14} \approx 1.214,\\ E(X^2) &= \frac {3}{56}\int _0^2 (x^4+4x^3+4x^2)\,dx = \frac {3}{56}\left [\frac {x^5}{5}+x^4+\frac {4x^3}{3}\right ]_0^2\\ &= \frac {3}{56}\cdot \frac {496}{15} = \frac {62}{35}. \end {align*}

Hence \[\operatorname {Var}(X) = \frac {62}{35}-\left (\frac {17}{14}\right )^{2} = \frac {1736-1445}{980} = \frac {291}{980} \approx 0.2969,\] and the standard deviation is \(\sqrt {291/980} \approx 0.545\).

(c). For \(0\leq x\leq 2\), \[F(x) = \int _0^x \frac {3}{56}(t+2)^2\,dt = \frac {(x+2)^3-8}{56},\] so \[F(x) = \begin {cases} 0 & x<0\\[2pt] \frac {(x+2)^3-8}{56} & 0\leq x\leq 2\\[6pt] 1 & x>2.\end {cases}\] As a check, \(F(2) = (64-8)/56 = 1\).

(d). \[P(X\geq 1) = 1-F(1) = 1-\frac {27-8}{56} = 1-\frac {19}{56} = \frac {37}{56}.\]

(e). The two constants \(-8\) cancel when the c.d.f. is differenced: \begin {align*} P\left (\tfrac 12<X<\tfrac 32\right ) &= F\left (\tfrac 32\right )-F\left (\tfrac 12\right ) = \frac {1}{56}\left [\left (\tfrac {7}{2}\right )^{3}-\left (\tfrac {5}{2}\right )^{3}\right ]\\ &= \frac {1}{56}\cdot \frac {343-125}{8} = \frac {218}{448} = \frac {109}{224}. \end {align*}

Solution. (a). Integrate piece by piece and set the total to 1. The graph is a trapezium – a straight rise, a flat top, a straight fall – so the areas can also be read off geometrically: \[\underbrace {\int _0^1 kx\,dx}_{k/2} + \underbrace {\int _1^3 k\,dx}_{2k} + \underbrace {\int _3^4 k(4-x)\,dx}_{k/2} = 3k = 1 \implies k = \frac {1}{3}.\]

(b). Before integrating, notice that the density is symmetric about \(x=2\): the rising piece on \([0,1)\) mirrors the falling piece on \([3,4)\), and the flat piece is centred at 2. So \(E(X) = 2\) without any calculation. Confirming it, \[E(X) = \frac 13\left [\int _0^1 x^2\,dx + \int _1^3 x\,dx + \int _3^4 x(4-x)\,dx\right ] = \frac 13\left [\frac 13+4+\frac {5}{3}\right ] = 2.\] For the variance, \[E(X^2) = \frac 13\left [\int _0^1 x^3\,dx + \int _1^3 x^2\,dx + \int _3^4 x^2(4-x)\,dx\right ] = \frac {29}{6},\] so \[\operatorname {Var}(X) = \frac {29}{6}-2^2 = \frac {5}{6} \approx 0.833.\]

(c). Accumulate the pieces, carrying forward the area already covered: \[F(x) = \begin {cases} 0 & x<0\\[2pt] \frac {x^2}{6} & 0\leq x<1\\[6pt] \frac {x}{3}-\frac {1}{6} & 1\leq x<3\\[6pt] -\frac {x^2}{6}+\frac {4x}{3}-\frac {5}{3} & 3\leq x<4\\[6pt] 1 & x\geq 4. \end {cases}\] The pieces agree at the joins – \(F(1)=\tfrac 16\), \(F(3)=\tfrac 56\) – and \(F(4)=1\).

(d). \[P(X>2) = 1-F(2) = 1-\left (\frac 23-\frac 16\right ) = 1-\frac 12 = \frac 12,\] which is what the symmetry about \(x=2\) already told us.

(e). \[P\left (\tfrac 12<X\leq \tfrac 72\right ) = F\left (\tfrac 72\right )-F\left (\tfrac 12\right ) = \frac {23}{24}-\frac {1}{24} = \frac {11}{12}.\]

Solution. (a). \[\int _{-1}^{0}Kx^{2}\,dx = K\left [\frac {x^{3}}{3}\right ]_{-1}^{0} = K\left (0+\frac {1}{3}\right ) = \frac {K}{3} = 1 \implies K = 3.\] Note that \(K\) is positive even though the range is negative, because \(x^{2}\) is positive throughout – a density must never be negative.

(b). \[E(X) = \int _{-1}^{0}3x^{3}\,dx = 3\left [\frac {x^{4}}{4}\right ]_{-1}^{0} = 3\left (0-\frac {1}{4}\right ) = -\frac {3}{4},\] \[E(X^{2}) = \int _{-1}^{0}3x^{4}\,dx = 3\left [\frac {x^{5}}{5}\right ]_{-1}^{0} = 3\left (0+\frac {1}{5}\right ) = \frac {3}{5},\] \[\operatorname {Var}(X) = \frac {3}{5}-\left (-\frac {3}{4}\right )^{2} = \frac {48-45}{80} = \frac {3}{80} = 0.0375.\] A negative mean is perfectly sensible here: all the probability sits on \([-1,0]\), so \(X\) is negative with certainty. The density is heaviest near \(-1\), which pulls the mean below the midpoint \(-0.5\).

Solution. (a). The density must integrate to 1 over its range: \[\int _0^{a}2(1-x)\,dx = \left [2x-x^{2}\right ]_0^{a} = 2a-a^{2} = 1.\] So \(a^{2}-2a+1 = 0\), that is \((a-1)^{2} = 0\), giving \(a = 1\) as a repeated root. This is also the only sensible answer for another reason: \(2(1-x)\) is negative for \(x>1\), and a density cannot be negative, so the range could not have extended past 1 in any case.

(b). The upper limit 3 lies outside the range, where the density is zero, so the interval effectively stops at 1: \[P\left (\tfrac 12<X<3\right ) = P\left (\tfrac 12<X<1\right ) = \int _{1/2}^{1}2(1-x)\,dx = \left [2x-x^{2}\right ]_{1/2}^{1} = 1-\frac {3}{4} = \frac {1}{4}.\] Integrating all the way to 3 without noticing this would add a negative contribution and give a wrong answer – always check the stated range before substituting limits.