$$ \begin{array}{l} ext { A7. (a) State and prove the Chebyshev's inequality. } \ \left.\begin{array}{l} ext { (b) The number of marriage licences issued in Harare per month is known to be a } \ ext { random variable with mean } 124 ext { and standard deviation } 7.5 ext {. Use Chebyshev's } \ ext { inequality to estimate a bound for the probability that is between } 64 ext { and } 184 \ ext { marriage licences will be issued next month. } \ ext { (c) State the properties of a cumulative density function. }\end{array} ight][8] \ {[4] }\end{array} $$

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$$ \begin{array}{l} 	ext { A7. (a) State and prove the Chebyshev's inequality. } \ \left.\begin{array}{l}	ext { (b) The number of marriage licences issued in Harare per month is known to be a } \ 	ext { random variable with mean } 124 	ext { and standard deviation } 7.5 	ext {. Use Chebyshev's } \ 	ext { inequality to estimate a bound for the probability that is between } 64 	ext { and } 184 \ 	ext { marriage licences will be issued next month. } \ 	ext { (c) State the properties of a cumulative density function. }\end{array}ight][8] \ {[4] }\end{array} $$

UpStudy AI · Accepted Answer

**(a) Chebyshev's Inequality** Let $$X$$ be a random variable with finite mean $$\mu$$ and finite variance $$\sigma^2$$. Chebyshev's inequality states that for any $$k > 0$$, $$ P(|X-\mu| \ge k\sigma) \le \frac{1}{k^2}. $$ An equivalent statement is that $$ P(|X-\mu| < k\sigma) \ge 1 - \frac{1}{k^2}. $$ *Proof:* 1. Define the non-negative random variable $$(X-\mu)^2$$. 2. By Markov's inequality, for any $$a > 0$$, $$ P((X-\mu)^2 \ge a) \le \frac{\mathbb{E}[(X-\mu)^2]}{a}. $$ 3. Choose $$a = k^2\sigma^2$$. Since $$\mathbb{E}[(X-\mu)^2] = \sigma^2$$, we have $$ P\big((X-\mu)^2 \ge k^2\sigma^2\big) \le \frac{\sigma^2}{k^2\sigma^2} = \frac{1}{k^2}. $$ 4. Notice that the event $$\{(X-\mu)^2 \ge k^2\sigma^2\}$$ is equivalent to $$\{|X-\mu| \ge k\sigma\}$$. Thus, $$ P(|X-\mu| \ge k\sigma) \le \frac{1}{k^2}. $$ This completes the proof of Chebyshev's inequality. --- **(b) Application to Marriage Licences in Harare** Let $$X$$ denote the number of marriage licences issued in Harare per month with mean $$\mu = 124$$ and standard deviation $$\sigma = 7.5$$. We are required to bound the probability that next month the number of licences $$X$$ will be between 64 and 184. This can be expressed as $$ P(64 \le X \le 184) = P(|X - 124| \le 60). $$ 1. Compute $$k$$ such that $$k\sigma = 60$$: $$ k = \frac{60}{7.5} = 8. $$ 2. Applying Chebyshev's inequality, $$ P(|X-\mu| < k\sigma) \ge 1 - \frac{1}{k^2} = 1 - \frac{1}{8^2} = 1 - \frac{1}{64} = \frac{63}{64}. $$ Thus, the probability that between 64 and 184 marriage licences will be issued is at least $$ \frac{63}{64} \approx 98.44\%. $$ --- **(c) Properties of a Cumulative Density Function** Let $$F(x)$$ denote the cumulative distribution function (CDF) of a random variable $$X$$. The properties of $$F(x)$$ are as follows: 1. **Non-decreasing:** $$F(x)$$ is a non-decreasing function; that is, if $$x_1 \le x_2$$, then $$ F(x_1) \le F(x_2). $$ 2. **Right-continuous:** $$F(x)$$ is right-continuous: $$ \lim_{y \to x^+} F(y) = F(x). $$ 3. **Limits at Infinity:** $$ \lim_{x \to -\infty} F(x) = 0 \quad \text{and} \quad \lim_{x \to \infty} F(x) = 1. $$ These properties ensure that $$F(x)$$ properly describes the probability distribution of the random variable $$X$$. **(a) Chebyshev's Inequality** Chebyshev's inequality states that for any random variable with mean $$\mu$$ and standard deviation $$\sigma$$, the probability that the variable deviates from the mean by at least $$k\sigma$$ is at most $$\frac{1}{k^2}$$. **(b) Marriage Licences in Harare** Given the mean is 124 and standard deviation is 7.5, the probability that next month's marriage licences will be between 64 and 184 is at least 98.44%. **(c) Properties of a Cumulative Density Function** A cumulative distribution function $$F(x)$$ is: 1. **Non-decreasing:** If $$x_1 \le x_2$$, then $$F(x_1) \le F(x_2)$$. 2. **Right-continuous:** $$\lim_{y \to x^+} F(y) = F(x)$$. 3. **Limits at Infinity:** $$\lim_{x \to -\infty} F(x) = 0$$ and $$\lim_{x \to \infty} F(x) = 1$$.

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