Using Chebyshev's Theorem, it is found that:
At least 75% of the exam scores fall between 76 and 92.
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[tex]P = 100(1 - \frac{1}{2^{2}})[/tex]
[tex]P = 100(1 - \frac{1}{4})[/tex]
[tex]P = 100(\frac{3}{4})[/tex]
[tex]P = 75[/tex]
84 - 2(4) = 76
84 + 2(4) = 92
A similar problem is given at https://brainly.com/question/23612895