1. Problem 4: Given the probability density function (pdf) of lifespan $f(x) = k \cos\left(\frac{\pi x}{1000}\right)$ for $0 \leq x \leq 500$, and 0 otherwise. (a) Show that $k = \frac{\pi}{1000}$. - The total probability must be 1, so integrate $f(x)$ over $[0,500]$: $$\int_0^{500} k \cos\left(\frac{\pi x}{1000}\right) dx = 1$$ - Compute the integral: $$k \left[ \frac{1000}{\pi} \sin\left(\frac{\pi x}{1000}\right) \right]_0^{500} = 1$$ - Evaluate sine terms: $$k \frac{1000}{\pi} \left( \sin\left(\frac{\pi \times 500}{1000}\right) - \sin(0) \right) = 1$$ - Since $\sin(\frac{\pi}{2}) = 1$, this becomes: $$k \frac{1000}{\pi} (1 - 0) = 1 \implies k = \frac{\pi}{1000}$$ (b) Find $E(X)$: - Use definition: $$E(X) = \int_0^{500} x f(x) dx = \int_0^{500} x \frac{\pi}{1000} \cos\left(\frac{\pi x}{1000}\right) dx$$ - Let $a = \frac{\pi}{1000}$, then: $$E(X) = a \int_0^{500} x \cos(a x) dx$$ - Integration by parts: Let $u = x$, $dv = \cos(a x) dx$, then $du = dx$, $v = \frac{\sin(a x)}{a}$. - So: $$\int x \cos(a x) dx = x \frac{\sin(a x)}{a} - \int \frac{\sin(a x)}{a} dx = \frac{x \sin(a x)}{a} + \frac{\cos(a x)}{a^2} + C$$ - Evaluate from 0 to 500: $$\left[ \frac{x \sin(a x)}{a} + \frac{\cos(a x)}{a^2} \right]_0^{500} = \frac{500 \sin(a \times 500)}{a} + \frac{\cos(a \times 500)}{a^2} - \frac{\cos(0)}{a^2}$$ - Substitute $a = \frac{\pi}{1000}$ and $a \times 500 = \frac{\pi}{2}$: $$= \frac{500 \times 1}{a} + \frac{0}{a^2} - \frac{1}{a^2} = \frac{500}{a} - \frac{1}{a^2}$$ - Multiply by $a$: $$E(X) = a \left( \frac{500}{a} - \frac{1}{a^2} \right) = 500 - \frac{1}{a} = 500 - \frac{1000}{\pi} \approx 181.69$$ (c) Find $P(X > 400 | X > 300)$: - Conditional probability: $$P(X > 400 | X > 300) = \frac{P(400 < X \leq 500)}{P(300 < X \leq 500)}$$ - Compute numerator: $$P(400 < X \leq 500) = \int_{400}^{500} f(x) dx = k \int_{400}^{500} \cos\left(\frac{\pi x}{1000}\right) dx$$ - Compute denominator: $$P(300 < X \leq 500) = k \int_{300}^{500} \cos\left(\frac{\pi x}{1000}\right) dx$$ - Integrate: $$\int \cos(a x) dx = \frac{\sin(a x)}{a}$$ - Numerator: $$k \left[ \frac{\sin(a x)}{a} \right]_{400}^{500} = \frac{k}{a} (\sin(a \times 500) - \sin(a \times 400))$$ - Denominator: $$\frac{k}{a} (\sin(a \times 500) - \sin(a \times 300))$$ - Substitute $k = a = \frac{\pi}{1000}$ and values: $$\sin(\frac{\pi}{2}) = 1, \sin(\frac{2\pi}{5}) \approx 0.9511, \sin(\frac{3\pi}{10}) \approx 0.8090$$ - So: $$P = \frac{1 - 0.9511}{1 - 0.8090} = \frac{0.0489}{0.191} \approx 0.256$$ --- 2. Problem 5: Sales $X$ with pdf $f(x) = 6x(1-x)$ for $0 \leq x \leq 1$. (a) Find mean and variance. - Mean: $$E(X) = \int_0^1 x \cdot 6x(1-x) dx = 6 \int_0^1 x^2 (1-x) dx = 6 \int_0^1 (x^2 - x^3) dx$$ - Compute integral: $$6 \left[ \frac{x^3}{3} - \frac{x^4}{4} \right]_0^1 = 6 \left( \frac{1}{3} - \frac{1}{4} \right) = 6 \times \frac{1}{12} = 0.5$$ - Variance: $$E(X^2) = \int_0^1 x^2 \cdot 6x(1-x) dx = 6 \int_0^1 x^3 (1-x) dx = 6 \int_0^1 (x^3 - x^4) dx$$ - Compute integral: $$6 \left[ \frac{x^4}{4} - \frac{x^5}{5} \right]_0^1 = 6 \left( \frac{1}{4} - \frac{1}{5} \right) = 6 \times \frac{1}{20} = 0.3$$ - Variance: $$Var(X) = E(X^2) - (E(X))^2 = 0.3 - 0.5^2 = 0.3 - 0.25 = 0.05$$ (b) Probability town runs out of petrol is 25%, so: $$P(X > s) = 0.25 \implies P(X \leq s) = 0.75$$ - Find $s$ such that: $$F(s) = \int_0^s 6x(1-x) dx = 0.75$$ - Compute CDF: $$F(s) = 6 \int_0^s (x - x^2) dx = 6 \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_0^s = 6 \left( \frac{s^2}{2} - \frac{s^3}{3} \right) = 3 s^2 - 2 s^3$$ - Solve: $$3 s^2 - 2 s^3 = 0.75$$ - Rearranged: $$2 s^3 - 3 s^2 + 0.75 = 0$$ - Numerically solve for $s$ in $[0,1]$ gives $s \approx 0.67$ to $0.68$. - Since $X$ is in tens of thousands litres, amount supplied is between 6700 and 6800 litres. --- 3. Problem 6: $X$ with pdf $f(x) = \frac{k}{x^2}$ for $2 \leq x \leq 4$. (a) Find $k$: $$\int_2^4 \frac{k}{x^2} dx = 1 \implies k \int_2^4 x^{-2} dx = 1$$ - Integral: $$\int_2^4 x^{-2} dx = \left[-\frac{1}{x}\right]_2^4 = -\frac{1}{4} + \frac{1}{2} = \frac{1}{4}$$ - So: $$k \times \frac{1}{4} = 1 \implies k = 4$$ (b) Find $E(X)$ and $Var(X)$: - $E(X)$: $$\int_2^4 x \cdot \frac{4}{x^2} dx = 4 \int_2^4 x^{-1} dx = 4 [\ln x]_2^4 = 4 \ln 2 \approx 2.7726$$ - $E(X^2)$: $$\int_2^4 x^2 \cdot \frac{4}{x^2} dx = 4 \int_2^4 1 dx = 4 (4 - 2) = 8$$ - Variance: $$Var(X) = E(X^2) - (E(X))^2 = 8 - (4 \ln 2)^2 \approx 8 - 7.686 = 0.314$$ (c) Interquartile range (IQR): - Find $Q_1$ and $Q_3$ such that: $$F(x) = \int_2^x \frac{4}{t^2} dt = 1 - \frac{2}{x}$$ - Solve for $x$ with $F(x) = 0.25$ and $0.75$: $$0.25 = 1 - \frac{2}{Q_1} \implies \frac{2}{Q_1} = 0.75 \implies Q_1 = \frac{2}{0.75} = \frac{8}{3} \approx 2.6667$$ $$0.75 = 1 - \frac{2}{Q_3} \implies \frac{2}{Q_3} = 0.25 \implies Q_3 = 8$$ - IQR: $$Q_3 - Q_1 = 8 - \frac{8}{3} = \frac{16}{3} \approx 5.3333$$