If X is normally distributed, with mean \(\mu\) and variance \(\sigma^2\), find an upper bound for the following probabilities, using Chebyshev’s Inequality.
Note, from Chebyshev’s Inequality: \[ \begin{aligned} P(|X - \mu| \geq k) \leq \frac{\sigma^2}{k^2} \end{aligned} \]
\[ \begin{aligned} P(|X - \mu| \geq \sigma) & \leq \frac{\sigma^2}{\sigma^2} \\ &= 1 \\ \end{aligned} \]
\[ \begin{aligned} P(|X - \mu| \geq 2\sigma) & \leq \frac{\sigma^2}{4\sigma^2} \\ &= \frac{1}{4} \\ \end{aligned} \]
\[ \begin{aligned} P(|X - \mu| \geq 3\sigma) & \leq \frac{\sigma^2}{9\sigma^2} \\ &= \frac{1}{9} \\ \end{aligned} \]
\[ \begin{aligned} P(|X - \mu| \geq 4\sigma) & \leq \frac{\sigma^2}{16\sigma^2} \\ &= \frac{1}{16} \\ \end{aligned} \]