1. A fair coin is tossed 100 times. The expected number of heads is 50, and the standard deviation for the number of heads is (100 · 1/2 · 1/2)1/2 = 5. What does Chebyshev’s Inequality tell you about the probability that the number of heads that turn up deviates from the expected number 50 by three or more standard deviations (i.e., by at least 15)?

Solution:

Chebyshev’s inequality is given as:

\[P(|X - \mu| \ge k\sigma ) \le \frac{\sigma^2}{k^2 \sigma^2}=\frac{1}{k^2}, \\ \space where \space E(x) = \mu, \space and \space V(X) = \sigma^2 \space and \space \mu= k\sigma = k \]

\[P(|X - 50| \ge 15) = \frac{1}{k^2}\]

Calculate k:

\[k \sigma = 15\] \[k.5 = 15 \] \[\begin{align*} \therefore k& = \frac{15}{5} \\ &= 3 \end{align*}\]

\[ \begin{align*} P(|X - 50| \ge 15) &= \frac{1}{k^2} \\ &= \frac{1}{ \frac{1}{3^2}} \\ &= \frac{1}{9} \end{align*} \]

There is a \(\frac{1}{9}\) probability that the number of heads deviates from the expected number of 50 by three or more standard deviations.