Question

the mean score on a statistics exam is 80 points, with a standard deviation of 6 points. apply chebychevs theorem to the data using ( k = 2 ). interpret the results. at least (%) of the exam scores fall between and (simplify your answers.)

Explanation:

Step1: Calculate the percentage using Chebyshev's formula

Chebyshev's Theorem states that for any number \(k>0\), at least \((1 - \frac{1}{k^{2}})\times100\%\) of the data lies within \(k\) standard deviations of the mean.
Given \(k = 2\), then \((1-\frac{1}{2^{2}})\times100\%=(1 - \frac{1}{4})\times100\%=\frac{3}{4}\times100\% = 75\%\)

Step2: Calculate the lower and upper bounds

The mean \(\mu=80\) and the standard deviation \(\sigma = 6\)
The lower bound is \(\mu - k\sigma=80-2\times6=80 - 12=68\)
The upper bound is \(\mu + k\sigma=80+2\times6=80 + 12 = 92\)

Answer:

At least \(75\%\) of the exam scores fall between \(68\) and \(92\).