Chapter 6 - The Normal Distribution - 6-2 Applications of the Normal Distribution - Exercises 6-2 - Page 339: 19

1927.9 and 1692.1

We have $\mu=1810,\sigma=92$ the middle 80% means 10% below and 10% above which gives the probability of $P(z1)=0.1$ and $P(z2)=1-0.1=0.9$ Use Table E, we have $z1=-1.281,z2=1.281$ $\frac{X1-1810}{92}=-1.281, X1=1692.1$ $\frac{X2-1810}{92}=1.281, X2=1927.9$ The maximum and minimum sizes of the homes the contractor should build are 1927.9 and 1692.1