Applied Statistics and Probability for Engineers, 6th Edition

Published by Wiley
ISBN 10: 1118539710
ISBN 13: 978-1-11853-971-2

Chapter 2 - Section 2-2 - Interpretations of Probability - Exercises - Page 33: 2-64

Answer

a) $\frac{1}{10}$ b) $\frac{1}{2}$ c) $\frac{1}{2}$

Work Step by Step

Note that $S =$ {$95, 96, 97, 98, 99, 100, 101, 102, 103, 104$} and each possible outcome has probability $\frac{1}{10}$ of occuring. a) We want to find $P(100) = \frac{1}{10}$ b) We want to find $P(less than 100) = P(95) + P(96) + P(97) + P(98) + P(99) = \frac{1}{10} + \frac{1}{10} + \frac{1}{10} + \frac{1}{10} + \frac{1}{10} = \frac{5}{10} = \frac{1}{2}$ c) We want to find $P(98\leq x\leq102) = P(98) + P(99) + P(100) + P(101) + P(102) = \frac{1}{10} + \frac{1}{10} + \frac{1}{10} + \frac{1}{10} + \frac{1}{10} = \frac{5}{10} = \frac{1}{2}$
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