Finite Math and Applied Calculus (6th Edition)

Published by Brooks Cole
ISBN 10: 1133607705
ISBN 13: 978-1-13360-770-0

Chapter 8 - Section 8.4 - Measures of Dispersion - Exercises - Page 589: 12

Answer

Here, the variance is: $(2-4.5)^2*(.05)+(4-4.5)^2*(.75)+(6-4.5)^2*(.1)+(8-4.5)^2*(.1)=1.95$ The standard deviation is the square root of the variance: The standard deviation here equals to $\sqrt{1.95}=1.4$

Work Step by Step

In order to calculate the standard deviation, we have to get the expected value of the probability distribution. Here, we can calculate it as: $p_{1}\times X_1+p_2 \times X_2 +...$ Where $p$ is the probability, $X$ is the value for every possible value of the distribution. The expected value is: $(.05)*2+(.75)*4+(.1)*6+(.1)*8=4.5$ The variance of X can be calculated as: $(X_1-E(X))^2*p_1+(X_2-E(X))^2*p_2+...$ Here, the variance is: $(2-4.5)^2*(.05)+(4-4.5)^2*(.75)+(6-4.5)^2*(.1)+(8-4.5)^2*(.1)=1.95$ The standard deviation is the square root of the variance: The standard deviation here equals to $\sqrt{1.95}=1.4$
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