Let $D_1$ and $D_2$ represent two draws at random with replacement from a population with $E(D_1) = 10$ and $SD(D_1) = 2$. ....

Question

Let $D_1$ and $D_2$ represent two draws at random with replacement from a population with $E(D_1) = 10$ and $SD(D_1) = 2$. Find a number $c$ so that $P(|D_1-D_2|<c) \ge 99\%$.

Since both draws are from the same population with replacement,

$$E(D_2) = 10$$

$$SD(D_2) = 2$$

About $99\%$ of $D_1$ falls within $3σ$ of $E(D_1) = E(D_2)$.

About $99\%$ of $D_2$ falls within $3σ$ of $E(D_1) = E(D_2)$.

Thus, about $99\%$ of $D_1 - D_2$ will always be within $3σ$ of $E(D_1) = E(D_2)$

$$P(|D_1-D_2|<c) \ge 99\%$$

$$c = (E(D_1) + 3σ) - (E(D_1) - 3σ) = (10 + 6) - (10 - 6) = 12$$

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But the textbook answer is:

$10 \sqrt{8}$

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Edit 1:

Given equation: $P(|D_1-D_2|<c) \ge 0.99$

By Chebychev's Inequality, we know:

$P(|X-E(X)| \ge kSD(X)) \le \frac{1}{k^2}$

This looks similar to the given equation. If we could somehow alter this to the form of the equation given, then we have found $c$.

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First, we fix the inequalities:

$- P(|X-E(X)| \ge kSD(X)) \ge -\frac{1}{k^2}$

$1 - P(|X-E(X)| \ge kSD(X)) \ge 1 - \frac{1}{k^2}$

$P(|X-E(X)| \lt kSD(X)) \ge 1 - \frac{1}{k^2}$, the inequalities now match the given equation.

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$1 - \frac{1}{k^2} = 0.99$ when $k=10$

$P(|X-E(X)| \lt 10SD(X)) \ge 1 - \frac{1}{10^2}$

$P(|X-E(X)| \lt 10SD(X)) \ge 0.99$

$D_2$ in $D_1-D_2$ is not an expected value of $D_1$ so we can't choose $X = D_1$ and $E(X) = E(D_1) = D_2$. But we can choose $X = D_1-D_2$ and hope that $E(X) = E(D_1-D_2) = 0$.

$P(|(D_1-D_2)-E(D_1-D_2)| \lt 10SD(D_1-D_2)) \ge 0.99$

$P(|(D_1-D_2)-(E(D_1)-E(D_2))| \lt 10SD(D_1-D_2)) \ge 0.99$

$P(|(D_1-D_2)-(10-10)| \lt 10SD(D_1-D_2)) \ge 0.99$

$P(|D_1-D_2| \lt 10SD(D_1-D_2)) \ge 0.99$

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Thus,

$c$

$= 10SD(D_1-D_2)$

$= 10 \sqrt{Var(D_1-D_2)}$

$= 10 \sqrt{Var(D_1 + (-D_2))}$, $D_1$ and $D_2$ are independent from each other.

$= 10 \sqrt{Var(D_1) + Var(-D_2)}$

$= 10 \sqrt{Var(D_1) + Var(D_2)}$

$= 10 \sqrt{(SD(D_1))^2 + (SD(D_2))^2}$

$= 10 \sqrt{2^2 + 2^2}$

$= 10 \sqrt{8}$

Answer 1

This is Chebyshev's inequality:

Chebyshev's Inequality: Let $X$ be a random variable with finite mean $\mu_X$ and finite non-zero variance $\sigma_X^2$. Then for any real number $k>0$, $$P(|X-\mu_X|\geq k\sigma_X) \leq\frac{1}{k^2}$$

We can use this to find that: $$1 - P(|X-\mu_X|\geq k\sigma_X) \geq 1 - \frac{1}{k^2} \Longrightarrow P(|X-\mu_X|<k\sigma_X) \geq 1 - \frac{1}{k^2}$$

You are interested in the case $1 - \frac{1}{k^2} = 0.99$ or $k=10$. Your random variable is $X = D_1 - D_2$ which has mean $\mu_X = \Bbb E[D_1 - D_2] = 0$. The variance of $X$ is: $$\Bbb V[X] = \Bbb V[D_1 - D_2] = \Bbb V[D_1] + \Bbb V[D_2] = 4+4 = 8$$ So its standard deviation is $\sigma_X=\sqrt{8}$. Putting these together, we get: $$P(|D_1 - D_2|<10\sqrt{8})\geq0.99$$