Essentials of Statistics (5th Edition)

Published by Pearson
ISBN 10: 0-32192-459-2
ISBN 13: 978-0-32192-459-9

Chapter 7 - Estimates and Sample Sizes - Review - Cumulative Review Exercises - Page 369: 2

Answer

between -2.1 and 13.1

Work Step by Step

The mean can be counted by summing all the data and dividing it by the number of data: $\frac{6+4+2+7+2+12}{6}=5.5$. Standard deviation=$\sqrt{\frac{\sum (x-\mu)^2}{n-1}}=\sqrt{\frac{(6−5.5)^2+(4−5.5)^2+(2−5.5)^2+(7−5.5)^2+(2−5.5)^2+(12−5.5)^2}{5}}=3.8.$ If a value is usual, then it is at most two standard deviations far from the mean. $Minimum \ usual \ value=mean-2\cdot(standard \ deviation)=5.5-2\cdot 3.8=-2.1.$ $Maximum \ usual \ value=mean+2\cdot(standard \ deviation)=5.5+2\cdot 3.8=13.1.$
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