Question

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.

Answer #1

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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