Using Chebyshev's theorem, solve these problems for a distribution with a mean of 64 and a standard deviation of 17 . Round \( k \) to at least 2 decimal places and final answers to at least one decimal place if needed. Part 1 of 2 At least 88.9 \% of the values will fall between 13 and 115 . Part 2 of 2 At least \( \% \) of the values will fall between 18 and 110 .
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To find \( k \) using Chebyshev's theorem, we first need to calculate the number of standard deviations the values are from the mean. For Part 1, the mean is 64, and the values are between 13 and 115. The distance from 64 to 13 is \( 64 - 13 = 51 \), and from 64 to 115 is \( 115 - 64 = 51 \). Since both distances are the same, the total distance is \( 51 \). The standard deviation is 17, so \( k = \frac{51}{17} = 3.00 \). Chebyshev’s theorem states that at least \( \frac{1}{1 - \frac{1}{k^2}} \) of the values will fall within k standard deviations. Plugging in \( k = 3 \), we get \( \frac{1}{1 - \frac{1}{3^2}} = \frac{1}{1 - \frac{1}{9}} = \frac{1}{\frac{8}{9}} = \frac{9}{8} = 1.125 \), or at least 88.9%. For Part 2, we need to find how many standard deviations the values of 18 and 110 are from the mean. The distance from 64 to 18 is \( 64 - 18 = 46 \), and from 64 to 110 is \( 110 - 64 = 46 \). Again, both distances are the same, totaling \( 46 \). Thus, \( k = \frac{46}{17} \approx 2.71 \). Now we apply Chebyshev's theorem again: at least \( \frac{1}{1 - \frac{1}{(2.71)^2}} \). Calculating gives \( (2.71)^2 \approx 7.3441 \), meaning at least \( \frac{1}{1 - \frac{1}{7.3441}} \approx \frac{7.3441}{6.3441} \approx 1.157 \) or about 86.5%. So the final answers are: Part 1: At least 88.9% of the values will fall between 13 and 115, and Part 2: At least 86.5% of the values will fall between 18 and 110.
