Question

Suppose that prices of recently sold homes in one neighborhood have a mean of $225,000 with a standard deviation of $6700. Using Chebyshev's Theorem, what is the minimum percentage of recently sold homes with prices between $211,600 and $238,400

Answer 1

Answer:

$211600 = 225000 -k*6700$

$k = \frac{225000-211600}{6700}= 2$

$238400 = 225000 +k*6700$

$k = \frac{238400-225000}{6700}= 2$

So then the % expected would be:

$1- \frac{1}{2^2}= 1- 0.25 =0.75$

So then the answer would be 75%

Step-by-step explanation:

For this case we have the following info given:

$\mu = 225000$ represent the true mean

$\sigma =6700$ represent the true deviation

And for this case we want to find the minimum percentage of sold homes between $211,600 and $238,400.

From the chebysev theorem we know that we have $1 -\frac{1}{k^2}$ % of values within $\mu \pm k\sigma$ if we use this formula and the limit given we have:

$211600 = 225000 -k*6700$

$k = \frac{225000-211600}{6700}= 2$

$238400 = 225000 +k*6700$

$k = \frac{238400-225000}{6700}= 2$

So then the % expected would be:

$1- \frac{1}{2^2}= 1- 0.25 =0.75$

So then the answer would be 75%