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Q.5.24

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Introductory Statistics
Found in: Page 348
Introductory Statistics

Introductory Statistics

Book edition OER 2018
Author(s) Barbara Illowsky, Susan Dean
Pages 902 pages
ISBN 9781938168208

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Short Answer

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 90th percentile of square footage for homes?

The value of the 90thpercentile of square footage for homes is 4.2.

See the step by step solution

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(1.5,4.5),

f(x)= height

=14.51.5

=13

So, the 90th percentile will be,

P(x<k)= base × height

0.90=(k1.5)13

(k1.5)=0.90×3

k=2.7+1.5

k=4.2

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