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Q.5.24

Expert-verifiedFound in: Page 348

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 $X~U(1.5,4.5)$.

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

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

Table,

Sample mean = 2.50

Standard deviation = 0.8302

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,

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

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

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

$k=2.7+1.5$

$k=4.2$

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