(a) Normalisation gives \[\int_0^a k(ax-x^2)\,\mathrm dx=k\left[\frac{ax^2}{2}-\frac{x^3}{3}\right]_0^a=\frac{ka^3}{6}=1.\]

Therefore \(\boxed{k=6/a^3}\).

(b) The density is symmetric about \(x=a/2\), so without calculation \(\boxed{\mathrm E(X)=a/2}\).

(c)(i) Integrate up to \(a/4\): \[\begin{aligned}P(X<a/4)&=\frac6{a^3}\left[\frac{ax^2}{2}-\frac{x^3}{3}\right]_0^{a/4}\\&=\frac6{a^3}\left(\frac{a^3}{32}-\frac{a^3}{192}\right)\\&=\boxed{\frac5{32}}.\end{aligned}\]

The factor \(a^3\) cancels, so the probability is independent of \(a\).

(c)(ii) By symmetry, \(P(X>3a/4)=P(X<a/4)=5/32\). Subtract the two equal tail probabilities:

\[P(a/4<X<3a/4)=1-2\left(\frac5{32}\right)=\boxed{\frac{11}{16}}=0.6875.\]

The area under the density is a triangle of base 6 and height \(a\). Normalisation gives \[\frac12(6)a=1\quad\Longrightarrow\quad a=\frac13.\]

The line passes through \((0,1/3)\) and \((6,0)\), so \[f(x)=\frac13-\frac{x}{18},\qquad0\le x\le6.\]

Find the second moment by integration: \[\begin{aligned}\mathrm E(X^2)&=\int_0^6x^2\left(\frac13-\frac{x}{18}\right)\,\mathrm dx\\&=\left[\frac{x^3}{9}-\frac{x^4}{72}\right]_0^6\\&=24-18=6.\end{aligned}\]

Use the given mean \(\mathrm E(X)=2\): \[\operatorname{Var}(X)=\mathrm E(X^2)-[\mathrm E(X)]^2=6-2^2=\boxed{2}.\]

(a) The density is non-negative and non-zero on its support, so \(\lambda>0\). Its integral must equal 1:

\[\int_0^{\ln2}\frac87\lambda e^{-\lambda x}\,\mathrm dx=\frac87\left[-e^{-\lambda x}\right]_0^{\ln2}=\frac87(1-2^{-\lambda})=1.\]

\[1-2^{-\lambda}=\frac78\quad\Longrightarrow\quad2^{-\lambda}=\frac18=2^{-3}\quad\Longrightarrow\quad\boxed{\lambda=3}.\]

(b) With \(\lambda=3\), \[\mathrm E(X)=\int_0^{\ln2}\frac{24}{7}x e^{-3x}\,\mathrm dx.\]

Use integration by parts with \(u=x\), \(\mathrm dv=e^{-3x}\mathrm dx\), so \(\mathrm du=\mathrm dx\) and \(v=-\tfrac13e^{-3x}\):

\[\int x e^{-3x}\,\mathrm dx=-\frac{x}{3}e^{-3x}+\frac13\int e^{-3x}\,\mathrm dx=-\frac{x}{3}e^{-3x}-\frac19e^{-3x}.\]

Since \(e^{-3\ln2}=1/8\), \[\begin{aligned}\mathrm E(X)&=\frac{24}{7}\left[-\frac{x}{3}e^{-3x}-\frac19e^{-3x}\right]_0^{\ln2}\\&=\frac{24}{7}\left(-\frac{\ln2}{24}-\frac1{72}+\frac19\right)\\&=\boxed{\frac13-\frac{\ln2}{7}}=0.234312\ldots\approx\boxed{0.234}\quad\text{(3 s.f.)}.\end{aligned}\]

(a) A probability density has total integral 1, so

\[\int_0^a(a-x)\,\mathrm{d}x=\left[ax-\frac{x^2}{2}\right]_0^a=\frac{a^2}{2}=1.\]

The support must have positive length, so \(a>0\) and \(a=\sqrt2\). Now integrate \(xf(x)\):

\[\mathrm E(X)=\int_0^a x(a-x)\,\mathrm{d}x=\left[\frac{ax^2}{2}-\frac{x^3}{3}\right]_0^a=\frac{a^3}{6}=\frac{\sqrt2}{3}.\]

This proves the required form, with \(\boxed{k=1/3}\).

(b) Let the median be \(m\). Then \(P(X\le m)=1/2\), giving

\[\int_0^m(\sqrt2-x)\,\mathrm{d}x=\sqrt2m-\frac{m^2}{2}=\frac12\quad\Longrightarrow\quad m^2-2\sqrt2m+1=0.\]

\[m=\sqrt2\pm1.\]

The value \(\sqrt2+1\) is outside the support \([0,\sqrt2]\), so reject it. The exact median is \(\boxed{\sqrt2-1}\).

(a) The triangular area under the density must be 1, so \(ab/2=1\), giving \(b=2/a\). The line through \((0,0)\) and \((a,b)\) has equation

\[f(x)=\frac ba x=\frac{2x}{a^2},\qquad0\le x\le a.\]

Let the median be \(m\). Then \[\int_0^m\frac{2x}{a^2}\,\mathrm dx=\left[\frac{x^2}{a^2}\right]_0^m=\frac{m^2}{a^2}=\frac12.\]

Hence \(m^2=a^2/2\). Since \(0<m<a\), take the positive root: \(\boxed{m=a/\sqrt2}\), as required.

(b) Integrate \(xf(x)\): \[\mathrm E(X)=\int_0^a\frac{2x^2}{a^2}\,\mathrm dx=\left[\frac{2x^3}{3a^2}\right]_0^a=\frac{2a}{3}.\]

The given mean is 4, so \(2a/3=4\), yielding \(\boxed{a=6}\).

(a) Expanding the density gives \(f(x)=-\frac34(x^2-8x+15)\), so \[\mathrm P(X>4.5)=-\frac34\int_{4.5}^{5}(x^2-8x+15)\,dx=\left[-\frac{x^3}{4}+3x^2-\frac{45x}{4}\right]_{4.5}^{5}=\frac5{32}.\]

(b) The density is symmetric about \(x=4\), so the median is \(4\) minutes.

(c) By symmetry, \(\mathrm P(X<3.5)=\mathrm P(X>4.5)=5/32\). The central probability is therefore \[\mathrm P(3.5<X<4.5)=1-2\left(\frac5{32}\right)=\frac{11}{16}=0.6875.\]

(a) A density integrates to 1. Hence \[1=k\int_0^2(2x^2-x^3)\,dx=k\left[\frac{2x^3}{3}-\frac{x^4}{4}\right]_0^2=k\left(\frac{16}{3}-4\right)=\frac{4k}{3},\] giving \(k=3/4\).

(b)(i) By the definition of the median for this continuous distribution, \(\mathrm P(X\le m)=0.5\).

(b)(ii) First find the mean: \[\mathrm E(X)=\frac34\int_0^2(2x^3-x^4)\,dx=\frac34\left[\frac{x^4}{2}-\frac{x^5}{5}\right]_0^2=\frac65.\] Then \[\mathrm P(X\le6/5)=\frac34\left[\frac{2x^3}{3}-\frac{x^4}{4}\right]_0^{6/5}=\frac{297}{625}=0.4752.\] This is less than 0.5, so the mean lies below the median. Therefore \[\mathrm P(\mathrm E(X)\le X\le m)=0.5-0.4752=0.0248=\frac{31}{1250}.\]

(a) The total area under a density is 1. The triangle has area \(ab/2\), so \[\frac{ab}{2}=1,\qquad b=\frac2a.\]

(b) The line is \(f(x)=a-(a/b)x\) for \(0\le x\le b\). Therefore \[\mathrm E(X)=\int_0^b x\left(a-\frac{a}{b}x\right)dx=\left[\frac{ax^2}{2}-\frac{ax^3}{3b}\right]_0^b=\frac{ab^2}{6}=\frac{b}{3}=\frac{2}{3a}.\] Equating this to \(4/9\) gives \(2/(3a)=4/9\), hence \(a=3/2\) and \(b=4/3\).

(c) The density is now \(f(x)=3/2-9x/8\) on \([0,4/3]\). Hence \[\int_0^k\left(\frac32-\frac98x\right)dx=\frac34\quad\Longrightarrow\quad\frac{3k}{2}-\frac{9k^2}{16}=\frac34.\] Rearranging, \[3k^2-8k+4=0\quad\Longrightarrow\quad(3k-2)(k-2)=0.\] The root \(k=2\) lies outside the support \([0,4/3]\), so reject it. The required value is \(k=2/3\).

(a) Integrating the density, \[\mathrm P(X<\tfrac12)=\int_0^{1/2}(1+\cos\pi x)\,dx=\left[x+\frac{\sin\pi x}{\pi}\right]_0^{1/2}=\frac12+\frac{\sin(\pi/2)}{\pi}=\frac12+\frac1\pi.\]

(b) The mean is \[\mathrm E(X)=\int_0^1x(1+\cos\pi x)\,dx.\] For \(\int x\cos(\pi x)\,dx\), use integration by parts with \(u=x\), \(dv=\cos(\pi x)\,dx\): \[\int x\cos(\pi x)\,dx=\frac{x\sin\pi x}{\pi}-\int\frac{\sin\pi x}{\pi}\,dx=\frac{x\sin\pi x}{\pi}+\frac{\cos\pi x}{\pi^2}.\] Consequently \[\mathrm E(X)=\left[\frac{x^2}{2}+\frac{x\sin\pi x}{\pi}+\frac{\cos\pi x}{\pi^2}\right]_0^1=\frac12+\frac{\cos\pi-\cos0}{\pi^2}=\frac12-\frac2{\pi^2}.\]

(a) Normalisation gives \[1=\int_0^a\frac{kx^2}{a^2}\,dx=\left[\frac{kx^3}{3a^2}\right]_0^a=\frac{ka}{3}.\] Hence \(k=3/a\).

(b) With this value of \(k\), \[\mathrm E(X)=\int_0^a\frac{3x^3}{a^3}\,dx=\left[\frac{3x^4}{4a^3}\right]_0^a=\frac{3a}{4}.\] Since the mean is 1, \(a=4/3\).

(c) Let the median be \(m\). The cumulative probability on \([0,a]\) is \(x^3/a^3\). Therefore \[\frac{m^3}{a^3}=\frac12,\qquad m=\frac{a}{\sqrt[3]{2}}=\sqrt[3]{\frac{32}{27}}=1.06\quad(3\text{ s.f.}).\]

(a) Normalisation requires \[1=\int_0^b ax\,dx=\left[\frac{ax^2}{2}\right]_0^b=\frac{ab^2}{2},\] hence \(a=2/b^2\).

(b) First compute the mean: \[\mathrm E(X)=\int_0^b\frac{2x^2}{b^2}\,dx=\left[\frac{2x^3}{3b^2}\right]_0^b=\frac{2b}{3}.\] Therefore \[\mathrm P(X<\mathrm E(X))=\int_0^{2b/3}\frac{2x}{b^2}\,dx=\left[\frac{x^2}{b^2}\right]_0^{2b/3}=\frac{(2b/3)^2}{b^2}=\frac49.\]

(a) The density is \(f(x)=kx\) on \([0,a]\). Using the median, \[\int_0^{\sqrt2}kx\,dx=\frac12\quad\Longrightarrow\quad\left[\frac{kx^2}{2}\right]_0^{\sqrt2}=\frac12\quad\Longrightarrow\quad k=\frac12.\]

(b) Normalisation gives \[1=\int_0^a\frac{x}{2}\,dx=\frac{a^2}{4}.\] Since \(a>0\), \(a=2\). Therefore \[\mathrm E(X)=\int_0^2x\frac{x}{2}\,dx=\left[\frac{x^3}{6}\right]_0^2=\frac43.\]

(a)(i) The area under the line is the area of a trapezium: \[1=\frac12(a+b)(2)=a+b.\] Therefore \(b=1-a\).

(a)(ii) The line is \(f(x)=a+(b-a)x/2=a+(1-2a)x/2\). Hence \[\mathrm E(X)=\int_0^2 x\left(a+\frac{1-2a}{2}x\right)dx=\left[\frac{ax^2}{2}+\frac{(1-2a)x^3}{6}\right]_0^2=2a+\frac43(1-2a)=\frac43-\frac{2a}{3}.\] Equating to 1.2 gives \(4/3-2a/3=1.2\), hence \(a=0.2\).

(b) By symmetry, \[\mathrm P(-c<T<c)=\int_{-c}^{c}\frac12\cos t\,dt=\left[\frac12\sin t\right]_{-c}^{c}=\sin c.\] Thus \(\sin c=1/2\). Within the support, \(0<c<\pi/2\), so the required value is \(c=\pi/6\).

(a) Integrating the density, \[1=\int_2^3\left(\frac a{x^2}-\frac{18}{x^3}\right)\,dx=\left[-\frac a x+\frac9{x^2}\right]_2^3=\frac a6-\frac54.\] Hence \(a/6=9/4\), so \(a=27/2\).

(b) \[E(X)=\int_2^3x f(x)\,dx=\left[\frac{27}{2}\ln x+\frac{18}{x}\right]_2^3=\frac{27}{2}(\ln3-\ln2)+6-9=\frac{27}{2}\ln\frac32-3.\]

(a) a and b are the minimum and maximum possible times to complete the challenge, in hours.

(b) \[1=\int_a^b x^{-2}\,dx=\frac1a-\frac1b\Rightarrow b-a=ab\Rightarrow a=\frac b{b+1}.\]

(c) \[E(X)=\int_a^b x^{-1}\,dx=\ln\frac ba=\ln(b+1)=\ln3.\] Thus \(b+1=3\), so \(b=2\) and \(a=2/3\).

(d) If the median is \(m\), \[\int_{2/3}^m x^{-2}\,dx=\frac32-\frac1m=\frac12,\] so \(m=1\) hour.

(a) The area under the density is 1: \(\pi a^2/4=1\). Hence \(a=2/\sqrt\pi\), taking the positive root.

(b) The circle equation is \(x^2+y^2=a^2=4/\pi\). In the first quadrant, \(y=\sqrt{4/\pi-x^2}\), so \[f(x)=\sqrt{4/\pi-x^2},\qquad0\le x\le2/\sqrt\pi.\]

(c) \[E(X)=\int_0^{2/\sqrt\pi}x\sqrt{4/\pi-x^2}\,dx=\left[-\frac13(4/\pi-x^2)^{3/2}\right]_0^{2/\sqrt\pi}=\frac{8}{3\pi^{3/2}}.\]

(a) \[1=k\int_0^\pi(1+\cos x)\,dx=k[x+\sin x]_0^\pi=k\pi,\] so \(k=1/\pi\).

(b) \(F(x)=(x+\sin x)/\pi\) on the domain. \[F(0.83)=0.499088<0.5,\quad F(0.84)=0.504408>0.5.\] Since \(F\) is continuous and increasing here, \(0.83<m<0.84\).

(c) Integrating \(x\cos x\) by parts, \[E(X)=\frac1\pi\int_0^\pi(x+x\cos x)\,dx=\frac1\pi\left[\frac{x^2}{2}+x\sin x+\cos x\right]_0^\pi=\frac\pi2-\frac2\pi.\]

(a) \[1=\int_0^{\sqrt2}(ax-x^3)\,dx=\left[\frac{ax^2}{2}-\frac{x^4}{4}\right]_0^{\sqrt2}=a-1,\] so \(a=2\).

(b) \[\int_0^m(2x-x^3)\,dx=m^2-\frac{m^4}{4}=\frac12.\] Thus \(m^4-4m^2+2=0\), so \(m^2=2\pm\sqrt2\). Only \(m^2=2-\sqrt2\) lies in the domain, giving \(m=\sqrt{2-\sqrt2}=0.765\).

(c) \[E(X)=\int_0^{\sqrt2}(2x^2-x^4)\,dx=\left[\frac{2x^3}{3}-\frac{x^5}{5}\right]_0^{\sqrt2}=\frac{8\sqrt2}{15}.\]

(a) By symmetry, \(P(-1<X<2)=117/256\), and \(P(X<2)=1/2\). Hence \[P(X\le-1)=\frac12-\frac{117}{256}=\frac{11}{256},\quad P(X>-1)=1-\frac{11}{256}=\frac{245}{256}.\]

(b) \[k\int_2^5(12+4x-x^2)\,dx=k\left[12x+2x^2-\frac{x^3}{3}\right]_2^5=39k=\frac{117}{256},\] giving \(k=3/256\).

(c) \(2+x-x^2=(2-x)(x+1)\ge0\) gives the domain \([-1,2]\). Symmetry about \(x=1/2\) gives \(E(X)=1/2\). \[E(X^2)=\frac29\int_{-1}^2(2x^2+x^3-x^4)\,dx=\frac29\left[\frac{2x^3}{3}+\frac{x^4}{4}-\frac{x^5}{5}\right]_{-1}^2=\frac7{10}.\] Thus \(\operatorname{Var}(X)=7/10-1/4=9/20=0.45\).

(a) Symmetry about 3 gives \(P(X<3)=1/2\). Hence \[p+\frac{13}{10}p\le\frac12,\qquad \frac{23}{10}p\le\frac12,\qquad p\le\frac5{23}.\]

(b) Reflection about 3 gives \(P(3<X<6-a)=P(a<X<3)=p+\frac{13}{10}p\). Thus \[P(b<X<6-a)=\frac{13}{10}p+p+\frac{13}{10}p=\frac{18}{5}p.\]

(c) \[\int_a^2\frac{6x-x^2}{36}\,dx=\frac5{27},\qquad \left[\frac{x^2}{12}-\frac{x^3}{108}\right]_a^2=\frac5{27}.\] This reduces to \[a^3-9a^2+8=0=(a-1)(a^2-8a-8).\] The roots are \(1,4-2\sqrt6,4+2\sqrt6\). Only \(1\) belongs to \(0\le a<2\), so \(a=1\).

(a) \[E(X)=\int_a^b x\frac1{x^2}\,dx=[\ln x]_a^b=\ln\frac ba=\ln2.\] Thus \(b/a=2\), so \(b=2a\).

(b) The density integrates to 1: \[\int_a^b x^{-2}\,dx=\left[-\frac1x\right]_a^b=\frac1a-\frac1b=1.\] With \(b=2a\), this becomes \(1/(2a)=1\), so \(a=1/2\) and \(b=1\).

(c) Let the median be \(m\). Then \[\int_{1/2}^{m}x^{-2}\,dx=\frac12,\qquad 2-\frac1m=\frac12,\qquad m=\frac23.\]

(a) The function is nonnegative. The total area under its graph is \(\frac12(2)(1)=1\), so it could be a density.

(b) A density has total area 1, so \[\frac12\pi r^2=1,\qquad r=\sqrt{\frac2\pi}\approx0.798.\]

(c)(i) The curve is decreasing across 10 to 20. The area to the left of 15 is greater than half the total area, so the median is less than 15.

(c)(ii) \[E(X)=\int_{10}^{20}x\left(\frac{40}{x^2}-\frac1{10}\right)\,dx=\left[40\ln x-\frac{x^2}{20}\right]_{10}^{20}=40\ln2-15\approx12.7\text{ minutes}.\]

(a)(i) The line passes through the origin, so \(f(x)=kx\). Normalisation gives \[\int_0^4kx\,dx=8k=1,\qquad k=\frac18.\]

(a)(ii) \[E(X)=\int_0^4x\frac{x}{8}\,dx=\left[\frac{x^3}{24}\right]_0^4=\frac83.\]

(b) A decreasing triangular density on \([0,a]\) has height \(2/a\) at 0 and equation \(g(w)=2(a-w)/a^2\). Since the median is 1, \(a>1\), and \[\int_0^1\frac{2(a-w)}{a^2}\,dw=\frac12,\qquad \frac{2a-1}{a^2}=\frac12.\] Thus \(a^2-4a+2=0\), giving \(a=2\pm\sqrt2\). The smaller root is below the median, so \[a=2+\sqrt2\approx3.41.\]

\[E(X)=\int_0^1x\frac32(1-x^2)\,dx=\frac32\left[\frac{x^2}{2}-\frac{x^4}{4}\right]_0^1=\frac38.\]

(a) The function is nonnegative, and its total area is \[\frac12\pi\left(\sqrt{\frac2\pi}\right)^2=1.\] Therefore it is a valid density.

(b) In right triangle \(OAB\), \[\cos\angle AOB=\frac{OA}{OB}=\frac{\sqrt{1/\pi}}{\sqrt{2/\pi}}=\frac1{\sqrt2},\] so \(\angle AOB=\pi/4\). The sector area is \[\frac12OB^2\frac{\pi}{4}=\frac12\frac2\pi\frac\pi4=\frac14.\] Also \(AB=\sqrt{OB^2-OA^2}=\sqrt{1/\pi}\), so the triangle area is \(\frac12 OA\cdot AB=1/(2\pi)\). The desired area to the right of \(AB\) is the sector minus the triangle: \[P\!\left(X>\sqrt{\frac1\pi}\right)=\frac14-\frac1{2\pi}\approx0.0908.\]

(a) By symmetry, \(P(X>1.8)=P(X<1.2)=a+b\). Remove this and the area below 0.6: \[P(0.6<X<1.8)=1-a-(a+b)=1-2a-b.\]

(b)(i) Normalisation gives \[1=k\int_0^3(9x^2-6x^3+x^4)\,dx=k\left[3x^3-\frac32x^4+\frac{x^5}{5}\right]_0^3=k\frac{81}{10}.\] Hence \(k=10/81\).

(b)(ii) Symmetry gives \(E(X)=3/2\). Also \[E(X^2)=\frac{10}{81}\int_0^3(9x^4-6x^5+x^6)\,dx\] \[=\frac{10}{81}\left[\frac{9x^5}{5}-x^6+\frac{x^7}{7}\right]_0^3=\frac{18}{7}.\] Therefore \[\operatorname{Var}(X)=\frac{18}{7}-\left(\frac32\right)^2=\frac9{28}\approx0.321.\]

(a) Symmetry gives \(E(X)=1\). Translating by 1 produces \(E(S)=2\) with the same variance: the parabola runs from 0 to 4, with highest point \((2,3/8)\).

(b) Taking \(T=(X+1)/2\) keeps the mean at 1 and quarters the variance. Its density is twice as high: the parabola runs from 0 to 2, with highest point \((1,3/4)\). Both densities are zero outside their stated intervals.

(c) Write \(u=x-1\), so \(3+2x-x^2=4-u^2\). Then \[\frac3{32}\int_{-a}^{a}(4-u^2)\,du=\frac12.\] Thus \[\frac3{32}\left(8a-\frac{2a^3}{3}\right)=\frac12,\qquad12a-a^3=8,\] giving \(a^3-12a+8=0\).

(d) For \(F(a)=a^3-12a+8\), \[F(0.69)=0.048509>0,\qquad F(0.70)=-0.057<0.\] The polynomial is continuous, and \(dF/da=3a^2-12<0\) on this interval. Its relevant root therefore satisfies \(0.69<a<0.70\).

(a) Symmetry gives \(P(X<-1)=P(X>0)=1-p\). Therefore \[P(-1<X<0)=p-(1-p)=2p-1.\]

(b)(i) Normalisation gives \[\int_{-3}^{2}[a-b(x^2+x)]\,dx=1.\] Evaluating the antiderivative, \[\left[ax-b\left(\frac{x^3}{3}+\frac{x^2}{2}\right)\right]_{-3}^{2}=5a-\frac{55b}{6}=1,\] so \(30a-55b=6\).

(b)(ii) The graph meets the axis at \(x=2\), hence \(f(2)=a-6b=0\). Substituting \(a=6b\) gives \(125b=6\), so \[b=\frac6{125}=0.048,\qquad a=\frac{36}{125}=0.288.\]

The variable is uniform on \([0,a]\), with density \(1/a\). (a) The median is \(a/2\). (b) \[P(X>3a/4)=\frac{a-3a/4}{a}=\frac14.\]

(c) \[E(X)=\int_0^a\frac{x}{a}\,dx=\frac a2,\qquad E(X^2)=\int_0^a\frac{x^2}{a}\,dx=\frac{a^2}{3}.\] Therefore \[\operatorname{Var}(X)=\frac{a^2}{3}-\left(\frac a2\right)^2=\frac{a^2}{12}.\]

(d) \(p=b/a\). Each omitted end interval has length \(b/3\) and probability \(p/3\), so \[P(b/3<X<a-b/3)=1-\frac{2p}{3}.\]

(a) \[E(X)=\frac3{16}\int_2^4(4x^2-x^3)\,dx=\frac3{16}\left[\frac{4x^3}{3}-\frac{x^4}{4}\right]_2^4=\frac{11}{4}.\]

(b) \[E(X^2)=\frac3{16}\int_2^4(4x^3-x^4)\,dx=\frac3{16}\left[x^4-\frac{x^5}{5}\right]_2^4=\frac{39}{5}.\] Hence \[\operatorname{Var}(X)=\frac{39}{5}-\left(\frac{11}{4}\right)^2=\frac{19}{80}=0.2375.\]

(c) \[P(X<3)=\frac3{16}\left[2x^2-\frac{x^3}{3}\right]_2^3=\frac{11}{16}.\] This exceeds \(1/2\), so \(m<3\). Since \(P(X<m)=1/2\), \[P(m<X<3)=\frac{11}{16}-\frac12=\frac3{16}=0.1875.\]

(a)(i) \(P(X\ge0)=1\). (a)(ii) Symmetry gives \(P(W\ge0)=1/2\). (a)(iii) The support of \(W\) has two equal halves with the same shape as that of \(X\). Hence \(q=p/2\).

(b) Normalisation gives \[1=p\int_0^a(a^2-x^2)\,dx=p\left[a^2x-\frac{x^3}{3}\right]_0^a=\frac{2pa^3}{3}.\] Therefore \(p=3/(2a^3)\). The mean condition gives \[3=p\int_0^a x(a^2-x^2)\,dx=p\left[\frac{a^2x^2}{2}-\frac{x^4}{4}\right]_0^a=\frac{pa^4}{4}=\frac{3a}{8}.\] Thus \(a=8\).

(a) The quadratic density is symmetrical about \(x=2\), so its mean is 2.

(b) The density integrates to 1: \[1=-k\int_1^3(x^2-4x+3)\,dx=-k\left[\frac{x^3}{3}-2x^2+3x\right]_1^3=\frac{4k}{3}.\] Hence \(k=3/4\).

(c) \[E(X^2)=-\frac34\int_1^3(x^4-4x^3+3x^2)\,dx=-\frac34\left[\frac{x^5}{5}-x^4+x^3\right]_1^3=\frac{21}{5}.\] Thus \[\operatorname{Var}(X)=\frac{21}{5}-2^2=\frac15=0.2.\]

(d) For one roll, \[P(X>2.5)=-\frac34\left[\frac{x^3}{3}-2x^2+3x\right]_{2.5}^{3}=\frac5{32}.\] A point is scored unless all three independent rolls are at most 2.5: \[P(\text{point})=1-\left(1-\frac5{32}\right)^3=1-\left(\frac{27}{32}\right)^3\approx0.399.\]

(a) \[P(X<1.2)=\frac1{18}\int_0^{1.2}(9-x^2)\,dx=\frac1{18}\left[9x-\frac{x^3}{3}\right]_0^{1.2}=\frac{71}{125}=0.568.\]

(b) \[E(X)=\frac1{18}\int_0^3(9x-x^3)\,dx=\frac1{18}\left[\frac{9x^2}{2}-\frac{x^4}{4}\right]_0^3=\frac98=1.125.\]

(c) By the definition of the median, \[\frac1{18}\int_0^m(9-x^2)\,dx=\frac12,\qquad\frac1{18}\left(9m-\frac{m^3}{3}\right)=\frac12.\] Multiplying by 54 and rearranging gives \(m^3-27m+27=0\).

(a)(i) \[1=k\int_0^2(4x-x^2)\,dx=k\left[2x^2-\frac{x^3}{3}\right]_0^2=\frac{16k}{3},\] so \(k=3/16\).

(a)(ii) \[E(X)=\frac3{16}\int_0^2(4x^2-x^3)\,dx=\frac3{16}\left[\frac{4x^3}{3}-\frac{x^4}{4}\right]_0^2=\frac54.\]

(b) Symmetry about \(y=2.5\) gives \(P(Y>5-a)=P(Y<a)=0.2\), while \(P(Y>2.5)=0.5\). Therefore \[P(2.5<Y<5-a)=0.5-0.2=0.3.\]

Symmetry gives \(P(X<3)=P(X>5)=1-20/27=7/27\). Therefore \[P(3<X<5)=\frac{20}{27}-\frac7{27}=\frac{13}{27}.\]

The density is symmetric about \(x=3\), so \(E(X)=3\). Normalisation gives \[1=k\int_0^6(6x-x^2)\,dx=k\left[3x^2-\frac{x^3}{3}\right]_0^6=36k,\] hence \(k=1/36\). Then \[E(X^2)=\frac1{36}\int_0^6(6x^3-x^4)\,dx=\frac1{36}\left[\frac{3x^4}{2}-\frac{x^5}{5}\right]_0^6=\frac{54}{5}.\] Therefore \(\operatorname{Var}(X)=54/5-3^2=9/5\).

(a) The triangular area is 1, so \[\frac12(p-1)p=1\quad\Longrightarrow\quad p^2-p-2=(p-2)(p+1)=0.\] Since \(p>1\), \(p=2\).

(b) The line joins \((1,0)\) to \((2,2)\), so \(f(x)=2x-2\) for \(1\le x\le2\). Hence \[E(X)=\int_1^2x(2x-2)\,dx=\left[\frac{2x^3}{3}-x^2\right]_1^2=\frac53.\]

(a) For one day, \[P(X>10)=\frac3{8000}\int_{10}^{20}(x-20)^2\,dx=\frac1{8000}\left[(x-20)^3\right]_{10}^{20}=\frac18.\] For two independent days, the probability is \((1/8)^2=1/64\approx0.0156\).

(b) \[E(X)=\frac3{8000}\int_0^{20}(x^3-40x^2+400x)\,dx=\frac3{8000}\left[\frac{x^4}{4}-\frac{40x^3}{3}+200x^2\right]_0^{20}=5.\]

(c) \[\frac3{8000}\int_0^m(x-20)^2\,dx=\frac12\quad\Longrightarrow\quad\frac{(m-20)^3+8000}{8000}=\frac12.\] Therefore \((m-20)^3=-4000\), and \[m=20-\sqrt[3]{4000}\approx4.13\text{ minutes}.\]

(d) The model assigns zero probability to a train being more than 20 minutes late, although this could occur in practice.

(a) The area under the density is 1: \[\frac12(k)\left(\frac k2\right)=1,\qquad k^2=4.\] Since \(k>0\), \(k=2\).

(b) The straight line joins \((0,1)\) to \((2,0)\), so \(f(x)=1-x/2\) for \(0\le x\le2\). Hence \[E(X)=\int_0^2x\left(1-\frac x2\right)\,dx=\left[\frac{x^2}{2}-\frac{x^3}{6}\right]_0^2=\frac23.\]

(c) \[\int_p^1\left(1-\frac x2\right)\,dx=\left[x-\frac{x^2}{4}\right]_p^1=\frac14.\] Thus \(3/4-p+p^2/4=1/4\), or \(p^2-4p+2=0\). The root satisfying \(0\le p<1\) is \[p=2-\sqrt2\approx0.586.\]

(a) The total area is 1, so \(ka=1\) and \(k=1/a\).

(b) \[\mathrm E(X)=\frac1a\int_0^a x\,dx=\frac a2,\qquad \mathrm E(X^2)=\frac1a\int_0^a x^2\,dx=\frac{a^2}{3}.\] Therefore \[\operatorname{Var}(X)=\frac{a^2}{3}-\left(\frac a2\right)^2=\frac{a^2}{12}=3.\] Since \(a>0\), \(a=6\).

(a) The triangular area equals 1: \(\tfrac12(3)c=1\), so \(c=2/3\).

(b) The smaller triangle has base one third of the whole triangle, so its area is \((1/3)^2=1/9\). Thus \(\mathrm P(X>2)=1/9\).

(c) The line has equation \(f(x)=\frac23-\frac29x\) for \(0\le x\le3\). Therefore \[\mathrm E(X)=\int_0^3x\left(\frac23-\frac29x\right)\,dx=\left[\frac{x^2}{3}-\frac{2x^3}{27}\right]_0^3=1.\]

(a) A downward-opening parabola from (0, 0) to (20, 0), with maximum at (10, 0.075); the density is zero elsewhere.

(b) The density is symmetric about \(t=10\), so its mean is 10.

(c) \[\mathrm E(T^2)=\frac3{4000}\int_0^{20}(20t^3-t^4)\,dt=\frac3{4000}\left[5t^4-\frac{t^5}{5}\right]_0^{20}=120.\] Hence \(\operatorname{Var}(T)=120-10^2=20\).

(d) By symmetry, \(\mathrm P(T\le10-a)=\mathrm P(T\ge10+a)=1-p\). Thus the probability between the two limits is \(p-(1-p)=2p-1\).

(e) \[\mathrm P(8<T<12)=\frac3{4000}\int_8^{12}(20t-t^2)\,dt=\frac3{4000}\left[10t^2-\frac{t^3}{3}\right]_8^{12}=\frac{37}{125}=0.296.\]

(f) The model allows no waiting time above 20 minutes, although longer waits can occur.

(a) Since the total density is 1, \[1=\int_1^a\frac{k}{x^2}\,dx=k\left[-\frac1x\right]_1^a=k\left(1-\frac1a\right).\] Therefore \(k=a/(a-1)\), with \(a>1\).

(b) \[\mathrm E(X)=\int_1^a x\frac{k}{x^2}\,dx=k[\ln x]_1^a=\frac{a\ln a}{a-1}.\]

(c) Let the 60th percentile be \(m\). Then \[\frac{a}{a-1}\left(1-\frac1m\right)=\frac35.\] Thus \(1/m=(2a+3)/(5a)\), giving \(m=5a/(2a+3)\).

(a) The density is symmetric about \(x=7.5\), so \(\mathrm E(X)=7.5\).

(b) \[\mathrm E(X^2)=\frac6{125}\int_5^{10}(-x^4+15x^3-50x^2)\,dx=57.5.\] Therefore \(\operatorname{Var}(X)=57.5-7.5^2=1.25\).

(c) First, \[p_0=\mathrm P(X<6)=\frac6{125}\left[-\frac{x^3}{3}+\frac{15x^2}{2}-50x\right]_5^6=0.104.\] For two independent selections, \[\mathrm P(\text{exactly one below 6})=2p_0(1-p_0)=2(0.104)(0.896)=0.186368\approx0.186.\]

(a) \[\mathrm P(X>5)=\frac3{4000}\int_5^{10}(100-x^2)\,dx=\frac3{4000}\left[100x-\frac{x^3}{3}\right]_5^{10}=\frac5{32}=0.15625.\]

(b) \[\frac3{4000}\left[100x-\frac{x^3}{3}\right]_p^{10}=\frac14\quad\Longrightarrow\quad\frac3{4000}\left(\frac{2000}{3}-100p+\frac{p^3}{3}\right)=\frac14.\] Multiplying by 4000 and simplifying gives \(p^3-300p+1000=0\).

(c) The density is symmetric about \(x=0\). Since the upper tail beyond \(p\) has probability 0.25, the lower tail below \(-p\) also has probability 0.25. The central probability is therefore 0.50, and \(q=p=3.47\).