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Answers without the blur. Sign up and see all textbooks for free! Q.5.24

Expert-verified Found in: Page 348 ### Introductory Statistics

Book edition OER 2018
Author(s) Barbara Illowsky, Susan Dean
Pages 902 pages
ISBN 9781938168208 # 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 $X~U\left(1.5,4.5\right)$.What is the ${90}^{th}$ percentile of square footage for homes?

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

See the step by step solution

## Step 1: Given information

Table, Sample mean = 2.50

Standard deviation = 0.8302

## Step 2: Solution

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

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

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

$=\frac{1}{3}$

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

$P\left(x

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

$\left(k-1.5\right)=0.90×3$

$k=2.7+1.5$

$k=4.2$ ### Want to see more solutions like these? 