Introductory Statistics 9th Edition

Published by Wiley
ISBN 10: 1-11905-571-7
ISBN 13: 978-1-11905-571-6

Chapter 5 - Section 5.3 - Mean and Standard Deviation of a Discrete Random Variable - Exercises - Page 192: 5.21

Answer

Mean =μ = $∑xP(x)$ = 2.5546 units of defective tires Thus, it is expected that 2.5546 units of defective tires are on this fleet of limos, with a standard deviation of 1.3155 units of defective tires. Standard Deviation = $\sqrt∑x^{2}P(x) -μ^{2}$ = $\sqrt 8.2564-2.5546^{2}$ =1.3155 units of defective tires If k = 2, at least 75% of the number of defective tires on a limo lie between $μ-2σ$ and $μ+2σ$. μ =2.5546, σ = 1.3155 $μ-2σ$ = 2.5546-2(1.3155) = -0.0764 $μ+2σ$ = 2.5546+2(1.3155) = 5.1856 Using Chebyshev's theorem, we can state that at least 75% of the number of tires are expected to contain -0.0764 to 5.1856 defective tires on a limo.
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