# Solved (Free): A contractor decides to build homes that will include the middle 80% of the market. If the average size of homes built is 1810 square feet

#### ByDr. Raju Chaudhari

Apr 12, 2021

A contractor decides to build homes that will include the middle 80% of the market. If the average size of homes built is 1810 square feet, find the maximum and minimum sizes of the homes the contractor should build. Assume the standard deviation is 92 square feet and the variable is normally distributed.

#### Solution

Given that mean $\mu=1810$ and standard deviation $=\sigma = 92$.

Let the maximum size be "a" and the minimum size be "b" for the middle 80% of the home.

$P(a\leq X\leq b) =1-\alpha=0.8$ means

$P(X < a)= (1-\alpha)/2 = 0.1$

and $P(X > b) = (1-\alpha)/2 = 0.1$.

 \begin{aligned} & P(X < a) =0.1\\ &\Rightarrow P\big(\frac{X-\mu}{\sigma} < \frac{a-1810}{92}\big)=0.1\\ &\Rightarrow P(Z < \frac{a-1810}{92}\big)=0.1\\ &\Rightarrow \frac{a-1810}{92}= -1.282\\ &\quad\quad\text{(From normal table)}\\ &\Rightarrow a = 1810 +( -1.282)* 92\\ &\Rightarrow a = 1692.056 \end{aligned}

And

 \begin{aligned} & P(X > b) = 0.1\\ &\Rightarrow P(X < b) =0.9\\ &\Rightarrow P\big(\frac{X-\mu}{\sigma} < \frac{b-1810}{92}\big)=0.9\\ &\Rightarrow P(Z < \frac{b-1810}{92}\big)=0.9\\ &\Rightarrow \frac{b-1810}{92}= 1.282\\ &\quad\quad\text{(From normal table)}\\ &\Rightarrow b = 1810 + 1.282* 92\\ &\Rightarrow b = 1927.944 \end{aligned}

The maximum size of the home the contractor should build will be 1692.056 square feet and the minimum size of the home the contractor should build will be 1927.944 square feet.