Respuesta :
Answer:
[tex] 211600 = 225000 -k*6700[/tex]
[tex] k = \frac{225000-211600}{6700}= 2[/tex]
[tex] 238400 = 225000 +k*6700[/tex]
[tex] k = \frac{238400-225000}{6700}= 2[/tex]
So then the % expected would be:
[tex] 1- \frac{1}{2^2}= 1- 0.25 =0.75[/tex]
So then the answer would be 75%
Step-by-step explanation:
For this case we have the following info given:
[tex] \mu = 225000[/tex] represent the true mean
[tex]\sigma =6700[/tex] 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 [tex]1 -\frac{1}{k^2}[/tex] % of values within [tex]\mu \pm k\sigma[/tex] if we use this formula and the limit given we have:
[tex] 211600 = 225000 -k*6700[/tex]
[tex] k = \frac{225000-211600}{6700}= 2[/tex]
[tex] 238400 = 225000 +k*6700[/tex]
[tex] k = \frac{238400-225000}{6700}= 2[/tex]
So then the % expected would be:
[tex] 1- \frac{1}{2^2}= 1- 0.25 =0.75[/tex]
So then the answer would be 75%
