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2005 Paper 1 Q13
D: 1500.0 B: 1516.0
probability conditional Bayes' theorem normal distribution symmetric distribution mean and standard deviation law of total probability

The random variable \(X\) has mean \(\mu\) and standard deviation \(\sigma\). The distribution of \(X\) is symmetrical about \(\mu\) and satisfies: \[\P \l X \le \mu + \sigma \r = a \mbox{ and } \P \l X \le \mu + \tfrac{1}{ 2}\sigma \r = b\,,\] where \(a\) and \(b\) are fixed numbers. Do not assume that \(X\) is Normally distributed.

  1. Determine expressions (in terms of \(a\) and \(b\)) for \[ \P \l \mu-\tfrac12 \sigma \le X \le \mu + \sigma \r \mbox{ and } \P \l X \le \mu +\tfrac12 \sigma \; \vert \; X \ge \mu - \tfrac12 \sigma \r.\]
  2. My local supermarket sells cartons of skimmed milk and cartons of full-fat milk: \(60\%\) of the cartons it sells contain skimmed milk, and the rest contain full-fat milk. The volume of skimmed milk in a carton is modelled by \(X\) ml, with \(\mu = 500\) and \(\sigma =10\,\). The volume of full-fat milk in a carton is modelled by \(X\) ml, with \(\mu = 495\) and \(\sigma = 10\,\).
    1. Today, I bought one carton of milk, chosen at random, from this supermarket. When I get home, I find that it contains less than 505 ml. Determine an expression (in terms of \(a\) and \(b\)) for the probability that this carton of milk contains more than 500 ml.
    2. Over the years, I have bought a very large number of cartons of milk, all chosen at random, from this supermarket. \(70\%\) of the cartons I have bought have contained at most 505 ml of milk. Of all the cartons that have contained at least 495 ml of milk, one third of them have contained full-fat milk. Use this information to estimate the values of \(a\) and \(b\).

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*}
    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*}

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