Question

Question 3 of 3 (2 points) 6.2 Section Exercise 20 If the average price of a new one family home is $246.300 with a standard deviation of $15,000, find the minimum and maximum prices of the houses that a contractor will build to satisfy the middle 40% of the market. Assume that the variable is normally distributed. Round-value calculations to 2 decimal places and final answers to the nearest dollar. Source: New York Times Almanac. Minimum price : S Maximum price : 5 TE

Answer 1

Solution

$\small \mathbf{Minimum\;price:}$	$\small \$\mathbf{{\color{Red} 238500}}$
$\small \mathbf{Maximum\;price:}$	$\small \mathbf{\$\mathbf{{\color{Red} 254100}}}$

$\small \mathbf{given}:\;\mu=246300,\;\sigma= 15000$

formula : Z = *-*

$\small \mathbf{Middle\;40\%}\Rightarrow P(-z\leq Z\leq z)=0.40$

NOTE: Pla <Z<b) =PZ<b) - P(Z <a

$\small \small \Rightarrow P(Z\leq z)-P(Z\leq -z)=0.40$

= P(Z <=) - [1 - P(Z <=)) = 0.40 [:P(Z <->) = 1 - P(Z <:)

$\small \small \Rightarrow 2P(Z\leq z)-1=0.40$

$\small \small \Rightarrow P(Z\leq z)=\frac{1+0.40}{2}$

$\small \small \Rightarrow P(Z\leq z)=0.70$

Refer Standard normal table/Z-table, Lookup for z-score corresponding to area 0.70 to the left of the normal curve or use excel formula "=NORM.S.INV(0.70)" to find the z-score.

$\small \small \Rightarrow P(Z\leq \mathbf{0.52})=0.70$

$\small \small \mathbf{\therefore z=0.52}$

$\small \small \mathbf{\Rightarrow P(-0.52\leq Z\leq 0.52)=0.40}$

$\small {\color{blue} \mathbf{Minimum\;price\;(z=-0.52)}}$

$\small \small -0.52=\frac{X_{min}-246300 }{15000}$

$\small X_{min}=246300 +\left (-0.52*15000 \right )$

$\small X_{min}=246300 -7800$

$\small \mathbf{X_{min}=\$\mathbf{238500}}$

$\small {\color{blue} \mathbf{Maximum\;price\;(z=-0.52)}}$

$\small 0.52=\frac{X_{max}-246300 }{15000}$

$\small X_{max}=246300 +\left (0.52*15000 \right )$

$\small X_{max}=246300+7800$

$\small \mathbf{X_{max}=\$\mathbf{254100}}$