Suppose that prices of recently sold homes in Sherwood Forest have a mean of $226,492 with a standard deviation of $3,451. Assuming that no information is given about distribution of the prices of recently sold homes in Sherwood Forest, what is the minimum percentage of recently sold homes with prices between $218,156 and $234,828.
Solution :
Given that ,
mean = = $ 226,492
standard deviation = = $ 3,451
P( 218,156 < x < 234,828) = P[(218,156 - 226,492 ) / 3,451) < (x - ) / < ( 234,828 - 226,492) / 3,451) ]
= P(-2.42 < z < 2.42)
= P(z < 2.42 ) - P(z < -2.42)
Using z table,
= 0.9922 - 0.0078
= 0.9844
The percentage is = 98.44%
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