Question

farm sizes the average farm in the united states in a certain year contained 441 acres. the standard deviation is 32 acres. use chebyshevs theorem to find the minimum percentage of data values that will fall in the range of 329 - 553 acres. round your answer to the nearest percentage. minimum percentage of data values that will fall in the range of 329 - 553 acres is %.

Explanation:

Step1: Calculate the number of standard - deviations away from the mean

The mean $\mu = 441$ and the standard deviation $\sigma=32$.
For the lower bound $x_1 = 329$, the number of standard - deviations $k_1$ is $k_1=\frac{\mu - x_1}{\sigma}=\frac{441 - 329}{32}=\frac{112}{32}=3.5$.
For the upper bound $x_2 = 553$, the number of standard - deviations $k_2$ is $k_2=\frac{x_2-\mu}{\sigma}=\frac{553 - 441}{32}=\frac{112}{32}=3.5$.

Step2: Apply Chebyshev's theorem

Chebyshev's theorem states that the proportion of data within $k$ standard - deviations of the mean is at least $1-\frac{1}{k^{2}}$.
Here $k = 3.5$, so the proportion $P$ is $P = 1-\frac{1}{k^{2}}=1-\frac{1}{3.5^{2}}=1-\frac{1}{12.25}\approx1 - 0.0816 = 0.9184$.

Step3: Convert the proportion to a percentage

To convert the proportion to a percentage, we multiply by 100. So the percentage is $0.9184\times100 = 91.84\%\approx92\%$.

Answer:

92%