Question

using chebyshevs theorem, solve these problems for a distribution with a mean of 60 and a standard deviation of 23. 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 % of the values will fall between 14 and 106. part 2 of 2 at least % of the values will fall between 19 and 101.

Explanation:

Step1: Recall Chebyshev's Theorem

Chebyshev's Theorem states that for any dataset, the proportion of values within \( k \) standard deviations of the mean is at least \( 1 - \frac{1}{k^2} \), where \( k > 1 \). First, we need to find \( k \) for each interval.

Step2: Find \( k \) for Part 1 (14 to 106)

The mean \( \mu = 60 \), standard deviation \( \sigma = 23 \).
Lower bound: \( \mu - k\sigma = 14 \)
\( 60 - 23k = 14 \)
\( 23k = 60 - 14 = 46 \)
\( k = \frac{46}{23} = 2.00 \)

Step3: Calculate the proportion for Part 1

Using Chebyshev's formula: \( 1 - \frac{1}{k^2} = 1 - \frac{1}{2^2} = 1 - \frac{1}{4} = \frac{3}{4} = 0.75 \), so \( 75\% \).

Step4: Find \( k \) for Part 2 (19 to 101)

Lower bound: \( \mu - k\sigma = 19 \)
\( 60 - 23k = 19 \)
\( 23k = 60 - 19 = 41 \)
\( k = \frac{41}{23} \approx 1.78 \) (rounded to two decimal places)

Step5: Calculate the proportion for Part 2

Using Chebyshev's formula: \( 1 - \frac{1}{k^2} = 1 - \frac{1}{(1.78)^2} \approx 1 - \frac{1}{3.1684} \approx 1 - 0.316 = 0.684 \), so \( 68.4\% \) (rounded to one decimal place).

Answer:

Part 1: 75%
Part 2: 68.4%