Question

Assume that the prices paid for housing within a neighborhood have a normal distribution, with mean $100,000, and a standard deviation of $35,000.

1. What percentages of houses in the neighborhood have prices between $90,000 and $130,000?

2. What price of housing is such that only 12% of all houses in that neighborhood have lower prices?

Answer 1

solution

given that

(A)P(90000< x <130000 ) = P[(90000 -100000) / 35000< (x - ) / < (130000-100000) / 35000 )]

= P( -0.29< Z <0.86 )

= P(Z <0.86 ) - P(Z <-0.29 )

Using z table   

= 0.8051-0.3859

= 0.4192

answer=41.92%

(B)

Using standard normal table,

P(Z < z) = 12%

= P(Z < z) = 0.12  

= P(Z <- 1.18 ) = 0.12

z = - 1.18 Using standard normal z table,

Using z-score formula  

x= z * +

x= -1.18*35000+100000

x= 58700