A student’s score on a particular calculus final is a random variable with values of [0,100], mean 70, and variance 25.
\(P(65 \leq X \leq 75)\)
We can subtract the average (\(\mu=70\)) from each element of the inequality.
\(P(65-70 \leq X-70 \leq 75-70)\)
\(P(-5 \leq X-70 \leq 5)\)
This can also be rewritten as
\(1-P(|X-70|\geq 5)\)
In this form, as per Chebyshev’s inequality, we know this probability will solve out to \(\frac{\sigma^{2}}{\in^{2}}\).
\(1-\frac{25}{5^2}\)
\(1-\frac{25}{25}\)
\(1-1\)
\(0\)
The lower bound for the probability that the student’s score will fall between 65 and 75 is 0.
The \(\mu\) remains the same, but the \(\sigma^{2}\) changes to \(\frac{25}{100}\).
Again, \(P(65 \leq X \leq 75)\) reduces to \(1-P(|X-70|\geq 5)\). Using Chebyshev’s inequality, we know this probability will solve out to \(\frac{\sigma^{2}}{\in^{2}}\).
\(1-\frac{\frac{25}{100}}{25}\)
\(1-\frac{1}{100}\)
\(\frac{99}{100}\)
The lower bound for the probability that the class average will fall between 65 and 75 if 100 students take the final is 0.99.