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 193: 5.22

Answer

Mean =μ = $∑xP(x)$ = 3.05 units Thus, it is expected that 3.05 units of remote starting systems are installed on a given day. Standard Deviation = $\sqrt∑x^{2}P(x) -μ^{2}$ = $\sqrt 10.75-3.05^{2}$ =1.2031 units of remote starting system If k = 2, at least 75% of the number of remote starting systems installed on a given day lie between $μ-2σ$ and $μ+2σ$. μ =3.05, σ = 1.2031 $μ-2σ$ = 3.05-2(1.2031) = 0.6438 $μ+2σ$ = 3.05+2(1.2031) = 5.4562 Using Chebyshev's theorem, we can state that at least 75% of the remote starting systems are expected to install 0.6438 to 5.4562 units on a given day.
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