Thinking Mathematically (6th Edition)

Published by Pearson
ISBN 10: 0321867327
ISBN 13: 978-0-32186-732-2

Chapter 12 - Statistics - 12.3 Measures of Dispersion - Exercise Set 12.3 - Page 800: 19

Answer

$\approx3.46$

Work Step by Step

The mean of $n$ numbers is the sum of the numbers divided by $n$. Hence the mean: $\frac{7+9+9+15 }{4}=10$ The standard deviation of $x_1,x_2,...,x_n$ is (where $\overline{x}$ is the mean of the data values): $\sqrt{\frac{(x_1-\overline{x})^2+(x_2-\overline{x})^2+...+(x_n-\overline{x})^2}{n-1}}$. Hence here the the standard deviation is: $\sqrt{\frac{(7-10)^2+(9-10)^2+(9-10)^2+(15-10)^2}{4-1}}\approx3.46$
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