Questions: Consider the sample of exam scores to the right, arranged in increasing order. The sample mean and sample standard deviation of these exam scores are, respectively, 83.4 and 16.9. Chebychev's rule states that for any data set and any real number k>1, at least 100(1-1 / k^2) % of the observations lie within k standard deviations to either side of the mean. Complete parts (a) and (b) below. 26 53 53 61 67 67 78 81 81 82 85 85 88 88 88 90 90 92 92 92 92 94 94 95 97 97 97 98 99 100 a. Use Chebychev's rule to obtain a lower bound on the percentage of observations that lie within two standard deviations to either side of the mean. % (Round to one decimal place as needed.)

Solution Steps

To solve this problem, we will use Chebychev's rule, which states that for any data set and any real number \( k > 1 \), at least \( 100 \left(1 - \frac{1}{k^2}\right) \% \) of the observations lie within \( k \) standard deviations to either side of the mean. For part (a), we need to find the percentage of observations that lie within two standard deviations of the mean.

1. Identify the value of \( k \) (which is 2 in this case).
2. Apply Chebychev's rule formula: \( 100 \left(1 - \frac{1}{k^2}\right) \% \).
3. Calculate the percentage.

We are given that \( k = 2 \).

Chebychev's rule states that for any data set and any real number \( k > 1 \), at least \( 100 \left(1 - \frac{1}{k^2}\right) \% \) of the observations lie within \( k \) standard deviations to either side of the mean.

Substitute \( k = 2 \) into the formula: \[ 100 \left(1 - \frac{1}{2^2}\right) \% = 100 \left(1 - \frac{1}{4}\right) \% = 100 \left(1 - 0.25\right) \% = 100 \times 0.75 \% = 75.0 \% \]

Final Answer