Suppose that prices of recently sold homes in one neighborhood have a mean of $270,000 with a standard deviation of $8400. Using Chebyshev's Theorem, state the range in which at least 88.9% of the data will reside. Please do not round your answers.

The range in which at least 88.9% of the data will reside, ($244,800, $295,200).

Step-by-step explanation:

The Chebyshev's theorem states that, if X is a random variable with mean µ and standard deviation σ then for any positive number k, we have  

Here .

Then the value of k is:

Then we know that,

|X - µ| ≥ kσ

⇒ µ - kσ ≤ X ≤ µ + kσ.

Here it is given that mean (µ) = $270,000 and standard deviation (σ) = $8400.

Then, the price range is given by,

Thus, the range in which at least 88.9% of the data will reside, ($244,800, $295,200).

what's that

the 8 in 8,420 is in the thousands place so it's value is 8 thousand.

the 8 in 6,870 is in the hundreds place so it's value is 8 hundred.

divide 8000 by 800 to find the answer:

8000 / 800 = 10

the 8 in 8,420 is 10 times greater than the 8 in 6,870