Question

The data that follow are the square footage (in 1,000 feet squared) of 28 homes.

The sample mean = 2.50 and the sample standard deviation = 0.8302. The distribution can be written as $X~U(1.5,4.5)$.

What is the ${90}^{th}$percentile of square footage for homes?

Short answer

The value of the ${90}^{th}$percentile of square footage for homes is $4.2$.

Step-by-step solution

Given information

Table,

Sample mean = 2.50

Standard deviation = 0.8302

Solution

Here, we need to find the probability density function or height of the $X~U(1.5,4.5)$,

$f\left(x\right)=\text{height}$

$=\frac{1}{4.5-1.5}$

$=\frac{1}{3}$

So, the ${90}^{\text{th}}$percentile will be,

$P(x<k)=\text{base}\times \text{height}$

$0.90=(k-1.5)\left(\frac{1}{3}\right)$

$(k-1.5)=0.90\times 3$

$k=2.7+1.5$

$k=4.2$