Question

sale price of homes the average sale price of new one - family houses in the united states for a recent year was $242,700. find the range of values in which at least 93.75% of the sale prices will lie if the standard deviation is $47,900. round your k to the nearest whole number.
the range of values is between $\square$ and $\square$.

Explanation:

Step1: Recall Chebyshev's Theorem

Chebyshev's Theorem states that for any dataset, the proportion of data within \( k \) standard deviations of the mean is at least \( 1 - \frac{1}{k^2} \), where \( k > 1 \). We know that at least \( 93.75\% = 0.9375 \) of the data lies within some range. So we set up the equation:

$$ 1 - \frac{1}{k^2} = 0.9375 $$

Step2: Solve for \( k \)

Subtract \( 0.9375 \) from both sides:

$$ \frac{1}{k^2} = 1 - 0.9375 = 0.0625 $$

Take the reciprocal of both sides:

$$ k^2 = \frac{1}{0.0625} = 16 $$

Take the square root of both sides:

$$ k = \sqrt{16} = 4 $$

Step3: Calculate the range

The mean \( \mu = 242700 \) and the standard deviation \( \sigma = 47900 \). The range within \( k = 4 \) standard deviations of the mean is \( \mu - k\sigma \) to \( \mu + k\sigma \).

First, calculate the lower bound:

$$ \mu - k\sigma = 242700 - 4 \times 47900 = 242700 - 191600 = 51100 $$

Then, calculate the upper bound:

$$ \mu + k\sigma = 242700 + 4 \times 47900 = 242700 + 191600 = 434300 $$

Answer:

The range of values is between $51100 and $434300.