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: 11

Answer

See below.

Work Step by Step

a) The mean of $n$ numbers is the sum of the numbers divided by $n$. Hence the mean: $\frac{85+95+90+85+100 }{5}=91$ b) The deviation is the difference of the data item and the mean. Hence here the deviations: $85-91=-6,95-91=4,90-91=-1,85-91=-6,100-91=9$ c) The sums of the deviations: $-6+4-1-6+9=0$
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