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: 25

Answer

$\approx2.14$

Work Step by Step

The mean of $n$ numbers is the sum of the numbers divided by $n$. Hence the mean: $\frac{9+5+9+5+9+5+9+5 }{8}=7$ 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: $\sqrt{\frac{4(9-7)^2+4(5-7)^2}{8-1}}\approx2.14$
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