Step 1
Given;
Step 2
State Chebychev's theorem
Thus;
[tex]\begin{gathered} k=2 \\ 1-\frac{1}{2^2}=1-\frac{1}{4}=\frac{3}{4} \end{gathered}[/tex]
The empirical formula that applies to this is about 2 standard deviations of the mean
[tex]\begin{gathered} (\mu+2\sigma)\text{ and \lparen}\mu-2\sigma) \\ (88+2(6))\text{ and \lparen88-2\lparen6\rparen\rparen} \\ 100\text{ and 76} \end{gathered}[/tex]
Answer;
[tex]At\text{ least 75\% of the exam scores falls between 76 and 100}[/tex]
