Chebyshev’s Inequality. Page 322, #10

A student’s score on a particular calculus final is a random variable with values of [0, 100], mean 70, and variance 25.


(a) Find a lower bound for the probability that the student’s score will fall between 65 and 75.

Given data:

  • Mean (\(\mu\)) = 70
  • Variance (\(\sigma^2\)) = 25
  • Standard Deviation (\(\sigma\)) = 5


Step 1 - Calculate \(k\)

\[ k = \frac{|\text{Range boundary} - \mu|}{\sigma} \]

\[ k = \frac{70 - 65}{5} = 1 \]

Step 2 - Chebyshev’s inequality:

\[ P(|X - 70| < 5) \geq 1 - \frac{1}{1^2} = 0 \]

Chebyshev’s inequality does not provide a useful boundary for small values of \(k\).


(b) If 100 students take the final, find a lower bound for the probability that the class average will fall between 65 and 75.

Step 1 - Calculate score variance for 100 students

\[ \sigma_{\text{avg}}^2 = \frac{\sigma^2}{n} = \frac{25}{100} = 0.25 \]

Step 2 - Calculate Standard deviation \[ \sigma_{\text{avg}} = 0.5 \]

Step 3 - Calculate \(k\) for scores between 65 and 75: \[ k = \frac{70 - 65}{0.5} = 10 \]

Chebyshev’s inequality

\[ P(|\bar{X} - 70| < 5) \geq 1 - \frac{1}{10^2} = 0.99 \]

There is 99% probability that the class average will fall between 65 and 75.