Problems

Solution: \begin{align*} &&\kappa_{X-a} &= \frac{\mathbb{E}\left(\left(X-a-(\mu-a)\right)^4\right)}{\sigma_{X-a}^4}-3 \\ &&&= \frac{\mathbb{E}\left(\left(X-\mu\right)^4\right)}{\sigma_X^4}-3\\ &&&= \kappa_X \end{align*}

1. \(\,\) \begin{align*} && \kappa &= \frac{\mathbb{E}((X-\mu)^4)}{\sigma^4} - 3 \\ &&&= \frac{\mathbb{E}((\mu+\sigma Z-\mu)^4)}{\sigma^4} - 3 \\ &&&= \frac{\mathbb{E}((\sigma Z)^4)}{\sigma^4} - 3 \\ &&&= \mathbb{E}(Z^4)-3\\ &&&= \int_{-\infty}^{\infty} x^4\frac{1}{\sqrt{2\pi}} \exp \left ( - \frac12x^2 \right)\d x -3 \\ &&&= \left [\frac{1}{\sqrt{2\pi}}x^{3} \cdot \left ( -\exp \left ( - \frac12x^2 \right)\right) \right]_{-\infty}^{\infty} + \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty 3x^2 \exp \left ( - \frac12x^2 \right) \d x - 3 \\ &&&= 0 + 3 \textrm{Var}(Z) - 3 =0 \end{align*}
2. \(\,\) \begin{align*} && \mathbb{E}(T^4) &= \mathbb{E} \left [\left ( \sum\limits_{r=1}^n Y_r\right)^4\right] \\ &&&= \mathbb{E} \left [ \sum_{r=1}^n Y_r^4+\sum_{i\neq j} 4Y_iY_j^3+\sum_{i\neq j} 6Y_i^2Y_j^2+\sum_{i\neq j \neq k} 12Y_iY_jY_k^2 +\sum_{i\neq j\neq k \neq l}24 Y_iY_jY_kY_l\right] \\ &&&= \sum_{r=1}^n \mathbb{E} \left [ Y_r^4 \right]+\sum_{i\neq j} \mathbb{E} \left [ 4Y_iY_j^3\right]+\sum_{i\neq j} \mathbb{E} \left [ 6Y_i^2Y_j^2\right]+\sum_{i\neq j \neq k} \mathbb{E} \left [ 12Y_iY_jY_k^2\right] +\sum_{i\neq j\neq k \neq l} \mathbb{E} \left [ 24 Y_iY_jY_kY_l\right] \\ &&&= \sum_{r=1}^n \mathbb{E} \left [ Y_r^4 \right]+4\sum_{i\neq j} \mathbb{E} \left [ Y_i]\mathbb{E}[Y_j^3\right]+6\sum_{i\neq j} \mathbb{E} \left [ Y_i^2]\mathbb{E}[Y_j^2\right]+12\sum_{i\neq j \neq k} \mathbb{E} \left [ Y_i]\mathbb{E}[Y_j]\mathbb{E}[Y_k^2\right] +24\sum_{i\neq j\neq k \neq l} \mathbb{E} \left [ Y_i]\mathbb{E}[Y_j]\mathbb{E}[Y_k]\mathbb{E}[Y_l\right] \\ &&&= \sum_{r=1}^n \mathbb{E} \left [ Y_r^4 \right]+6\sum_{i\neq j} \mathbb{E} \left [ Y_i^2]\mathbb{E}[Y_j^2\right] \end{align*}
3. Without loss of generality, we may assume they all have mean zero. Therefore we can consider the sitatuion as in the previous case with \(T\) and \(Y_i\)s. Note that \(\mathbb{E}(Y_i^4) = \sigma^4(\kappa + 3)\) and \(\textrm{Var}(T) = n \sigma^2\) \begin{align*} && \kappa_T &= \frac{\mathbb{E}(T^4)}{(\textrm{Var}(T))^2} - 3 \\ &&&= \frac{\sum_{r=1}^n \mathbb{E} \left [ Y_r^4 \right]+6\sum_{i\neq j} \mathbb{E} \left [ Y_i^2\right]\mathbb{E}\left[Y_j^2\right]}{n^2\sigma^4}-3 \\ &&&= \frac{n\sigma^4(\kappa+3)+6\binom{n}{2}\sigma^4}{n^2\sigma^4} -3\\ &&&= \frac{\kappa}{n} + \frac{3n + \frac{6n(n-1)}{2}}{n^2} - 3 \\ &&&= \frac{\kappa}{n} + \frac{3n^2}{n^2}-3 \\ &&&= \frac{\kappa}{n} \end{align*}

Solution:

1. \begin{align*} && \frac12 &= \mathbb{P}(X \leq x_m) \\ \Leftrightarrow && \frac12 &= \mathbb{P}(e^X \leq e^{x_m} = y_m) \end{align*} Therefore the median is \(y_m = e^{x_m}\)
2. \begin{align*} && \mathbb{P}(Y \leq y) &= \mathbb{P}(e^X \leq y) \\ &&&= \mathbb{P}(X \leq \ln y) \\ &&&= F(\ln y) \\ \Rightarrow && f_Y(y) &= f(\ln y)/y \\ \\ && f'_Y(y) &= \frac{f'(\ln y) - f(\ln y)}{y^2} \end{align*} Therefore since the mode satisfies \(f'_Y = 0\) we must have \(f'(\ln \lambda ) = f(\ln \lambda)\)
3. This is the integral of the pdf of \(N(\mu + \sigma^2, \sigma^2)\) and therefore is clearly \(1\). \begin{align*} && \E[Y] &= \int_{-\infty}^{\infty} e^x \cdot \frac{1}{\sqrt{2\pi \sigma^2}} e^{-(x-\mu)^2/(2\sigma^2)} \d x \\ &&&= \frac{1}{\sqrt{2\pi \sigma^2}} \int_{-\infty}^{\infty} \exp (x - (x-\mu)^2/(2\sigma^2)) \d x\\ &&&= \frac{1}{\sqrt{2\pi \sigma^2}} \int_{-\infty}^{\infty} \exp ((2x \sigma^2- (x-\mu)^2)/(2\sigma^2)) \d x\\ &&&= \frac{1}{\sqrt{2\pi \sigma^2}} \int_{-\infty}^{\infty} \exp (-(x-\mu-\sigma^2)^2+2\mu \sigma^2-\sigma^4)/(2\sigma^2)) \d x\\ &&&= \frac{1}{\sqrt{2\pi \sigma^2}} \int_{-\infty}^{\infty} \exp (-(x-\mu+\sigma^2)^2)/(2\sigma^2)+\mu +\frac12\sigma^2) \d x\\ &&&= \e^{\mu +\frac12\sigma^2}\frac{1}{\sqrt{2\pi \sigma^2}} \int_{-\infty}^{\infty} \exp (-(x-\mu-\sigma^2)^2)/(2\sigma^2)) \d x\\ &&&= \e^{\mu +\frac12\sigma^2} \end{align*}
4. Notice that \(y_m = e^\mu < e^{\mu + \tfrac12 \sigma^2} = \E[Y]\), so it suffices to prove that \(\lambda < e^{\mu}\) Notice that \(f'(x) - f(x) = f(x)[-(x-\mu)/\sigma^2 - 1]\) and therefore \(\ln y - \mu = -\sigma^2\) so \(\lambda = e^{\mu - \sigma^2}\) which is clearly less than \(e^{\mu}\) as required.

Solution:

1. \(\,\) \begin{align*} && \mathbb{E}(Z| a < Z < b) &= \mathbb{E}(Z\mathbb{1}_{(a,b)}) /\mathbb{E}(\mathbb{1}_{(a,b)}) \\ &&&= \int_a^b z \phi(z) \d z \Big / (\Phi(b) - \Phi(a)) \\ &&&= \frac{\int_a^b \frac{1}{\sqrt{2 \pi}}z e^{-\frac12 z^2} \d z}{\Phi(b) - \Phi(a)} \\ &&&= \frac{\frac1{\sqrt{2\pi}} \left [-e^{-\frac12 z^2} \right]_a^b}{\Phi(b) - \Phi(a)} \\ &&&= \frac{\frac1{\sqrt{2\pi}} \left (e^{-\frac12 a^2}-e^{-\frac12 b^2} \right)}{\Phi(b) - \Phi(a)} \\ \end{align*}
2. \(\,\) \begin{align*} && \mathbb{E}(X |X > 0) &= \mathbb{E}(\mu + \sigma Z | \mu + \sigma Z > 0) \\ &&&= \mathbb{E}(\mu + \sigma Z | Z > -\tfrac{\mu}{\sigma}) \\ &&&= \mathbb{E}(\mu| Z > -\tfrac{\mu}{\sigma})+ \sigma \mathbb{E}(Z | Z > -\tfrac{\mu}{\sigma})\\ &&&= \mu+ \sigma \mathbb{E}(Z | Z > -\tfrac{\mu}{\sigma})\\ \end{align*} Hence \begin{align*} &&\mathbb{E}(|X|) &= \mathbb{E}(X | X > 0)\mathbb{P}(X > 0) - \mathbb{E}(X | X < 0)\mathbb{P}(X < 0) \\ &&&=\left ( \mu+ \sigma \mathbb{E}(Z | Z > -\mu /\sigma)\right)(1-\Phi(-\mu/\sigma)) - \left ( \mu+ \sigma \mathbb{E}(Z | Z < -\mu /\sigma)\right)\Phi(-\mu/\sigma) \\ &&&= \mu(1 - 2\Phi(-\mu/\sigma)) + \sigma \frac{e^{-\frac12\mu^2/\sigma^2}}{\sqrt{2\pi}(1-\Phi(-\mu/\sigma))}(1-\Phi(-\mu/\sigma)) + \sigma \frac{e^{-\frac12\mu^2/\sigma^2}}{\sqrt{2 \pi} \Phi(-\mu/\sigma)} \Phi(-\mu/\sigma) \\ &&&= \mu(1 - 2\Phi(-\mu/\sigma)) + \sigma \sqrt{\frac{2}{\pi}} \exp(-\tfrac12 \mu^2/\sigma^2) \end{align*} Finally, \begin{align*} && \textrm{Var}(|X|) &= \mathbb{E}(|X|^2) - [\mathbb{E}(|X|)]^2 \\ &&&= \mu^2 + \sigma^2 - m^2 \end{align*}

Solution:

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2. Let \(Y \sim N(0,\frac14)\), then: \begin{align*} &&\int_0^\infty \frac{1}{\sqrt{2 \pi \cdot \frac14}} e^{-2x^2} \, dx &= \frac12\\ \Rightarrow && \int_0^\infty e^{-2x^2} &= \frac{\sqrt{\pi}}{2 \sqrt{2}} \\ \Rightarrow && k &= \boxed{\frac{2\sqrt{2}}{\sqrt{\pi}}} \end{align*}
3. \begin{align*} \mathbb{E}[X] &= \int_0^\infty x f(x) \, dx \\ &= \frac{2\sqrt{2}}{\sqrt{\pi}}\int_0^\infty x e^{-2x^2}\, dx \\ &= \frac{2\sqrt{2}}{\sqrt{\pi}} \left [-\frac{1}{4}e^{-2x^2} \right]_0^\infty \\ &= \frac{1}{\sqrt{2\pi}} \\ \end{align*} In order to calculate \(\mathbb{E}(X^2)\) it is useful to consider the related computation \(\mathbb{E}(Y^2)\). In fact, by symmetry, these will be the same values. Therefore \(\mathbb{E}(X^2) = \mathbb{E}(Y^2) = \mathrm{Var}(Y) = \frac{1}{4}\) (since \(\mathbb{E}(Y) = 0\)). Therefore \(\mathrm{Var}(Y) = \mathbb{E}(Y^2) - \mathbb{E}(Y)^2 = \frac14 - \frac{1}{2\pi}\)
4. \begin{align*} && \mathbb{P}(X < x) &= \frac12 \\ \Leftrightarrow && 2\mathbb{P}(0 \leq Y < x) &= \frac12 \\ \Leftrightarrow && 2\l \mathbb{P}(Y < x) - \frac12 \r &= \frac12 \\ \Leftrightarrow && \mathbb{P}(Y < x)&= \frac34 \\ \Leftrightarrow && \mathbb{P}(\frac{Y-0}{1/2} < \frac{x}{1/2})&= \frac34 \\ \Leftrightarrow && \mathbb{P}(Z < \frac{x}{1/2})&= \frac34 \\ \Leftrightarrow && \Phi(2x)&= \frac34 \\ \Leftrightarrow && 2x &= 0.6744895\cdots \\ \Leftrightarrow && x &= 0.3372\cdots \\ \Leftrightarrow && &= 0.337 \, (3 \text{sf}) \\ \end{align*}

Solution: The the number of people being directed towards the buses (each cruise) is \(X \sim B(1024, \tfrac12) \approx N(512, 256) \approx 16Z + 512\). Therefore without an extra bus, the expected profit is \(\mathbb{E}[\min(X, 15 \times 32)]\). With the extra bus, the extra profit is \(\mathbb{E}[\min(X, 16 \times 32)]\), therefore the expected extra profit is: \(\mathbb{E}[\min(X, 16 \times 32)]-\mathbb{E}[\min(X, 15 \times 32)] = \mathbb{E}[\min(X, 16 \times 32)-\min(X, 15 \times 32)] \) \begin{align*} \text{Expected extra profit} &= \mathbb{E}[\min(X, 16 \times 32)-\min(X, 15 \times 32)] \\ &= \mathbb{E}[\min(16Z+512, 16 \times 32)-\min(16Z+512, 15 \times 32)] \\ &= 16\mathbb{E}[\min(Z+32, 32)-\min(Z+32, 30)] \\ &=16\int_{-\infty}^{\infty} \left (\min(Z+32, 32)-\min(Z+32, 30) \right)p_Z(z) \d z \\ &= 16 \left ( \int_{-2}^{0} (z+32-30) p_Z(z) \d z + \int_0^\infty (32-30)p_Z(z) \d z \right) \\ &= 16 \left ( \int_{-2}^{0} (z+2) p_Z(z) \d z + \int_0^\infty 2p_Z(z) \d z \right) \\ &= 16 \left ( \int_{-2}^{0} zp_Z(z) \d z + 2\int_{-2}^\infty p_Z(z) \d z \right) \\ &= 16 \left ( \int_{-2}^{0} z \frac{1}{\sqrt{2\pi}} e^{-\frac12 z^2} \d z + 2(1-\Phi(2)) \right) \\ &= 32(1-\Phi(2)) + \frac{16}{\sqrt{2\pi}} \left [ -e^{-\frac12z^2} \right]_{-2}^0 \\ &= 32(1-\Phi(2)) - \frac{16}{\sqrt{2\pi}} \left ( 1-e^{-2}\right) \end{align*} Across \(50\) different runs, this profit is \[ 1600(1-\Phi(2)) - \frac{800}{\sqrt{2\pi}} \left ( 1-e^{-2}\right) \]

Solution:

1. \(\,\) \begin{align*} && \mathbb{P}\left (\mu - \tfrac12 \sigma \leq X \right) &= \mathbb{P}\left (X \leq \mu + \tfrac12 \sigma \right) \tag{by symmetry} \\ &&&= b \\ \Rightarrow && \mathbb{P} \left (\mu - \tfrac12 \sigma \leq X \leq \mu + \sigma \right) &= a - (1-b) = a+b - 1\\ \\ && \mathbb{P} \left ( X \le \mu +\tfrac12 \sigma \vert X \ge \mu - \tfrac12 \sigma \right ) &= \frac{ \mathbb{P} \left (\mu - \tfrac12 \sigma \leq X \leq \mu + \tfrac12 \sigma \right)}{\mathbb{P} \left ( X \ge \mu - \tfrac12 \sigma \right )} \\ &&&= \frac{b-(1-b)}{1-(1-b)} \\ &&&= \frac{2b-1}{b} \end{align*}
2. 1. Let \(Y\) be the volume of milk in the carton I bring home, we are interested in: \begin{align*} && \mathbb{P}(Y \geq 500 | Y \leq 505) &= \frac{\mathbb{P}(500 \leq Y \leq 505)}{\mathbb{P}(Y \leq 505)} \\ &&&=\frac{\mathbb{P}(500 \leq Y \leq 505|\text{skimmed})\mathbb{P}(\text{skimmed})+\mathbb{P}(500 \leq Y \leq 505|\text{full fat})\mathbb{P}(\text{full fat})}{\mathbb{P}(Y \leq 505|\text{skimmed})\mathbb{P}(\text{skimmed})+\mathbb{P}(Y \leq 505|\text{full fat})\mathbb{P}(\text{full fat})} \\ &&&= \frac{\frac35 \cdot \mathbb{P}(\mu \leq X \leq \mu + \tfrac12 \sigma) + \frac25 \cdot \mathbb{P}(\mu+\tfrac12 \sigma \leq X \leq \mu +\sigma)}{\frac35 \cdot \mathbb{P}(X \leq \mu + \tfrac12 \sigma) + \frac25 \cdot \mathbb{P}(X \leq \mu +\sigma)} \\ &&&= \frac{\frac35 \cdot(b-\tfrac12) + \frac25 \cdot (a-b)}{\frac35 \cdot b + \frac25 \cdot a} \\ &&&= \frac{b+2a-\frac32}{3b+2a} \\ &&&= \frac{4a+2b-3}{4a+6b} \end{align*}
   2. \(70\%\) of cartons have contained at most 505 ml, so: \begin{align*} && \tfrac7{10} &= \mathbb{P}(Y \leq 505) \\ &&&= \mathbb{P}(Y \leq 505 | \text{ skimmed}) \mathbb{P}(\text{skimmed}) + \mathbb{P}(Y \leq 505 | \text{ full fat}) \mathbb{P}(\text{full fat}) \\ &&&= \mathbb{P}(X \leq \mu + \tfrac12 \sigma) \cdot \tfrac35 + \mathbb{P}(X\leq \mu + \sigma ) \cdot \tfrac25 \\ \Rightarrow && 7 &= 6b+ 4a \end{align*} \(\tfrac13\) of cartons containing 495 ml contained full fat milk: \begin{align*} && \tfrac13 &= \mathbb{P}(\text{full fat} | Y \geq 495) \\ &&&= \frac{\mathbb{P}(\text{full fat and} Y \geq 495) }{\mathbb{P}(Y \geq 495)} \\ &&&= \frac{\mathbb{P}(X \geq \mu)\frac25}{\mathbb{P}(X \geq \mu)\cdot \frac25+\mathbb{P}(X \geq \mu-\tfrac12 \sigma)\cdot \frac35} \\ &&&= \frac{\frac15}{\frac12 \cdot \frac25 + b\frac35}\\ &&&= \frac{1}{1+ 3b }\\ \Rightarrow && 3b+1 &= 3 \\ \Rightarrow && b &= \frac23 \\ && a &= \frac34 \end{align*}

Solution: \begin{align*} && 1 &= \int_{-\infty}^{\infty} f(x) \d x \\ &&&= k[1 + \lambda] \\ \Rightarrow && k &= \frac{1}{1+\lambda} \\ \\ && \mu &= \int_{-\infty}^\infty x f(x) \d x \\ &&&= k \int_{-\infty}^\infty x \phi(x) \d x + k \lambda \int_{-\infty}^{\infty} x g(x) \d x \\ &&&= k \cdot 0 + k \lambda \cdot \frac{\lambda}{2} \\ &&&= \frac{\lambda^2}{2(1+\lambda)} \\ \\ && \E[X^2] &= \int_{-\infty}^\infty x^2 f(x) \d x \\ &&&= k \int_{-\infty}^\infty x^2 \phi(x) \d x + k \lambda \int_{-\infty}^{\infty} x^2 g(x) \d x \\ &&&= k \cdot 1 + k \lambda \int_0^{\lambda} \frac{x^2}{\lambda} \d \lambda \\ &&&= k + \frac{k \lambda^3}{3} \\ &&&= \frac{3+\lambda^3}{3(1+\lambda)} \\ && \var[X] &= \frac{3+\lambda^3}{3(1+\lambda)} - \frac{\lambda^4}{4(1+\lambda)^2} \\ &&& = \frac{(3+\lambda^3)4(1+\lambda) - 3\lambda^4}{12(1+\lambda)^2} \\ &&&= \frac{\lambda^4+4\lambda^3+12\lambda + 12}{12(1+\lambda)^2} \end{align*}

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2. \(\,\) \begin{align*} && \mathbb{P}(X \leq x) &= \int_{-\infty}^x f(x) \d x \\ &&&= \begin{cases} \frac13 \Phi(x) & \text{if } x < 0 \\ \frac13\Phi(x) + \frac13x & \text{if } 0 \leq x \leq 2 \\ \frac13 \Phi(x) + \frac23 & \text{if } 2 < x \end{cases} \end{align*} When \(\lambda = 2\), \(\mu = \frac{4}{6} = \frac23\), \(\sigma^2 = \frac{16+32+24+12}{12 \cdot 9} = \frac{7}{9}\), so \(\mu + 2 \sigma = \frac23 + \frac{2\sqrt7}{3}>2\). Therefore \begin{align*} && \P(0 < X < \mu + 2\sigma) &= \frac13 \Phi\left (\frac{2+2\sqrt{7}}{3} \right) + \frac23 - \Phi(0) \\ &&&= \tfrac13 \cdot 0.9921 +\tfrac23 - \tfrac12 \\ &&&= 0.4974 \end{align*}

Solution: If \(\sigma\) is smaller we should expect our sample to have a mean closer to the true mean. Therefore we should use a two sided test which accepts \(\mathrm{H}_0\) if the mean is very close to the true mean. Suppose \(\textrm{H}_0\) is true, ie \(\sigma = \sigma_0\), then note that \(X \sim N(\mu, \frac{\sigma_0^2}{n})\) \begin{align*} && 1-\alpha &= \mathbb{P}(\mu - c < X < \mu + c) \\ &&&= \mathbb{P}(\mu - c < \frac{\sigma_0}{\sqrt{n}} Z + \mu < \mu + c) \\ &&&= \mathbb{P}(- \frac{c\sqrt{n}}{\sigma_0} < Z<\frac{\sqrt{n}c}{\sigma_0}) \\ &&&= \mathbb{P}(Z<\frac{\sqrt{n}c}{\sigma_0}) -\mathbb{P}( Z<-\frac{\sqrt{n}c}{\sigma_0}) \\ &&&= \mathbb{P}(Z<\frac{\sqrt{n}c}{\sigma_0}) -(1-\mathbb{P}( Z<\frac{\sqrt{n}c}{\sigma_0})) \\ &&&= 2\mathbb{P}(Z<\frac{\sqrt{n}c}{\sigma_0})-1 \\ \Rightarrow && \Phi(\frac{\sqrt{n}c}{\sigma_0})&=1 - \tfrac12 \alpha \\ \Rightarrow && \frac{\sqrt{n}c}{\sigma_0} &= z_{\alpha} \\ && c &= \frac{\sigma_0 z_{\alpha}}{\sqrt{n}} \end{align*} Under \(\mathrm{H}_1\), \(\sigma = \sigma_1\) so \begin{align*} && \beta &= \mathbb{P}(\mu - c < X < \mu + c) \\ &&&= \mathbb{P}(-\frac{c\sqrt{n}}{\sigma_1} < Z < \frac{\sqrt{n}c}{\sigma_1}) \\ &&&= \mathbb{P}(-\frac{\sigma_0}{\sigma_1} z_{\alpha}< Z < \frac{\sigma_0}{\sigma_1} z_{\alpha}) \\ &&&= 2\Phi(\frac{\sigma_0}{\sigma_1} z_{\alpha})-1 \end{align*} which does not depend on \(n\). Suppose both \(\alpha<0.05\) and \(\beta<0.05\), then \(z_{\alpha} > 1.96\) and \(\Phi(\frac{\sigma_0}{\sigma_1}1.96)<0.525 \Rightarrow \frac{\sigma_0}{\sigma_1}1.96 < 0.063 \Rightarrow \frac{\sigma_1}{\sigma_0} > \frac{1.96}{0.063} = 31.1 \) so the ratio of variances needs to be larger than \(31.1\).

Solution: If Henry catches the first train, his journey time is \(N(55+30,25+144) = N(85,13^2)\). He is on time if the journey takes less than \(105\) minutes, \(\frac{20}{13}\) std above the mean. If he catches the second train, his journey times is \(N(65+25, 16+9) = N(90, 5^2)\). He is on time if his journey takes less than \(80\) minutes, ie \(\frac{10}{5} = 2\) standard deviations above the mean. This is more likely than from the first train. \(A = 1 - \Phi(20/13)\) is the probability he is late from the first train. \(B = 1 - \Phi(2)\) is the probability he is late from the second train. The expected number of lates is \(L = M \cdot p \cdot A + M \cdot (1-p) \cdot B\), since \(B \leq A\) we must have \(BM \leq L \leq AM\) \begin{align*} && \frac35 &= \frac{pA}{pA + (1-p)B} \\ \Rightarrow && 3(1-p)B &= 2pA \\ \Rightarrow && p(2A+3B) &= 3B \\ \Rightarrow && p &= \frac{3B}{2A+3B} \end{align*}

Solution:

1. \(\,\) \begin{align*} && \sqrt{p(1-p)} &= \sqrt{p-p^2} \\ &&&= \sqrt{\tfrac14-(\tfrac12-p)^2} \\ &&&\leq \sqrt{\tfrac14} = \tfrac12 \end{align*} Therefore the maximum is \(\tfrac12\) when \(p=\frac12\)
2. Notice that our estimate \(\hat{p}\) will (for large \(n\)) be follow a normal distribution \(N(p, pq/n)\) by either the normal approximation to the binomial or central limit theorem. We would like \(0.01 > \mathbb{P}\left ( |\hat{p}-p| < 0.01 \right)\) or in other words \begin{align*} && 0.01 &> \mathbb{P}\left ( |\hat{p}-p| > 0.01 \right) \\ &&&=\mathbb{P}\left ( |\sqrt{\frac{pq}{n}}Z+p-p| > 0.01 \right) \\ &&&= \mathbb{P} \left (|Z|>\frac{0.01\sqrt{n}}{\sqrt{pq}}\right) \end{align*} therefore we need \(\frac{0.01\sqrt{n}}{\sqrt{pq}}> 2.58 \Rightarrow n > 258^2 pq \approx 2^{14} \approx 16\,000\), where we are using \(pq = \frac14\) as the worst case possibility and \(258 \approx 256 = 2^8\)
3. If we were looking at when we are looking at left handed people (maybe ~\(10\%\), we would be looking at \(pq = \frac{9}{100}\) so we need a smaller sample). If we are looking at millionaires (an even smaller again percentage), we would need an even smaller sample. This is surprising since you would expect you would need a larger sample to accurately gauge smaller proportions. However, this surprise can be resolved by considering that this is an absolute error. For smaller values the relative error is larger, but the absolute error is smaller.

Solution: Let \(Z \sim N(0,1)\), with a pdf of \(f(x) = \frac{1}{\sqrt{2\pi}} \exp(-x^2/2)\) \begin{align*} && 1 &= \int_{-\infty}^\infty Ax^2 \exp(-x^2/2) \d x \\ &&&= A\sqrt{2\pi} \int_{-\infty}^\infty x^2 \frac{1}{\sqrt{2\pi}} \exp(-x^2/2) \d x \\ &&&= A\sqrt{2\pi} \E[Z^2] = A\sqrt{2\pi} \\ \Rightarrow && A &= \frac{1}{\sqrt{2\pi}} \end{align*}

1. The probability of seeing a result as extreme as \(87.3\) is \begin{align*} \mathbb{P}(X > 87.3) &= \frac{1}{\sqrt{2\pi}}\int_{87.3}^{\infty} x^2 \exp(-x^2/2) \d x \\ &= \left [ -\frac{1}{\sqrt{2\pi}}x \exp(-x^2/2)\right]_{87.3}^{\infty}+\int_{87.3}^{\infty}\frac{1}{\sqrt{2\pi}} \exp(-x^2/2) \d x \\ &\approx 0 +(1- \Phi(87.3)) \\ &\approx 0 \end{align*} It is very unlikely this data point has come from our distribution rather than one with a higher mean, therefore our faith is very shaken.
2. If there are 1000 trials of this, we would expect the sample mean to be distributed according to the CLT. Each sample has mean \(0\) and variance \(\E[X^2] = \int_{-\infty}^\infty x^4 \frac{1}{\sqrt{2\pi}} \exp(-x^2/2) \d x = \E[Z^4] = 3\), therefore the sample mean is \(N(0, 3/1000)\). Therefore the probability of being \(0.23\) away is \begin{align*} && \mathbb{P}(S > 0.23) &= \mathbb{P}\left (Z > \frac{0.23}{\sqrt{3/1000}} \right) \\ &&&= \mathbb{P}\left (Z > \frac{0.23}{\sqrt{30}/100} \right) \\ &&&\approx \mathbb{P}\left (Z > \frac{0.23}{0.055} \right) \\ &&& \approx 0 \end{align*} again our faith should be shaken

Solution: Let \(X \sim N(0,1)\), and $\displaystyle X_{M}=\begin{cases} X & \text{if }X < M,\\ M & \text{if }X\geqslant M. \end{cases} $. Then we can calculate: \begin{align*} \mathbb{E}[X_M] &= \int_{-\infty}^M xf_X(x)\,dx + M\mathbb{P}(X \geq M) \\ &= \int_{-\infty}^M x \frac1{\sqrt{2\pi}}e^{-\frac12x^2}\,dx + M\mathbb{P}(X \geq M) \\ &= \left [ -\frac{1}{\sqrt{2\pi}}e^{-\frac12x^2} \right ]_{-\infty}^M + M (1-\mathbb{P}(X < M)) \\ &= -\phi(M) + M(1-\Phi(M)) \end{align*} Let \(B \sim B\left (1024, \frac12 \right)\) be the number of potential bus passengers. Then \(B \approx N(512, 256) = N(512, 16^2)\) which is a good approximation since both \(np\) and \(nq\) are large. The question is asking us, how much additional profit would the bus company get if they ran an additional bus. Currently each week they is (on average) \(512\) passengers worth of demand, but they can only supply \(496\) seats, so we should expect that there is demand for another bus. The question is how much that demand is worth. Using the first part of the question, we can see that their profit is something like a `capped normal', \(X_M\), except we are scaled and with a different cap. So we are interested in $\displaystyle Y_{M}=\begin{cases} B & \mbox{if }B< M,\\ M & \mbox{if }B\geqslant M. \end{cases}\(, but since \)B \approx N\left (512,16^2\right)$ this is similar to \begin{align*} Y_{M}&=\begin{cases} 16X+512 & \mbox{if }16X+512< M,\\ M & \mbox{if }16X+512\geq M. \end{cases} \\ &= \begin{cases} 16X+512 & \mbox{if }X< \frac{M-512}{16},\\ M & \mbox{if }X \geq \frac{M-512}{16}. \end{cases} \\ &= 16X_{\frac{M-512}{16}} + 512\end{align*} We are interested in \(\mathbb{E}[Y_{16\times31}]\) and \(\mathbb{E}[Y_{17\times31}]\), which are \(16\mathbb{E}[X_{-1}]+512\) and \(16\mathbb{E}[Y_{\frac{15}{16}}]+512\) Since \(\frac{15}{16} \approx 1\), lets look at \(16(\mathbb{E}[X_1] - \mathbb{E}[X_{-1}])\) \begin{align*} \mathbb{E}[X_1] - \mathbb{E}[X_{-1}] &= \left ( -\phi(1) + 1-\Phi(1)\right) - \left ( - \phi(-1) -(1 - \Phi(-1)) \right ) \\ &= -\phi(1) + \phi(-1) + 1-\Phi(1) + 1 - \Phi(-1) \\ &= 1 - \Phi(1) + \Phi(1) \\ &= 1 \end{align*} Therefore the extra \(31\) will fill roughly \(16\) of them. (This is a slight overestimate, which is worth bearing in mind). A better approximation might be that \(\mathbb{E}[X_t] - \mathbb{E}[X_{-1}] = \frac{t +1}{2}\) for \(t \approx 1\), (since we want something increasing). This would give us an approximation of \(15.5\), which is very close to the `true' answer. Therefore, over \(50\) bus runs, we should earn roughly \(800\) vloska extra from an additional bus. (Again an overestimate, and with an uncertain pay-off, they should consider offering maybe \(600\)). Since this is the future, we can quite easily calculate the exact values using the binomial distribution on a computer. This gives the true value as \(15.833\), and so they should pay up to \(791\)