Statistics: Informed Decisions Using Data (4th Edition)

Published by Pearson
ISBN 10: 0321757270
ISBN 13: 978-0-32175-727-2

Chapter 6 - Section 6.3 - Assess Your Understanding - Applying the Concepts - Page 352: 19a

Answer

See the picture.

Work Step by Step

$P(x)=\frac{(λt)^x}{x!}e^{-λt}$ λ = 0.2 and t = 30 $P(0)=\frac{(0.2\times30)^0}{0!}e^{-0.2\times30}=0.002479$ $P(1)=\frac{(0.2\times30)^1}{1!}e^{-0.2\times30}=0.01487$ $P(2)=\frac{(0.2\times30)^2}{2!}e^{-0.2\times30}=0.04462$ $P(3)=\frac{(0.2\times30)^3}{3!}e^{-0.2\times30}=0.08924$ $P(4)=\frac{(0.2\times30)^4}{4!}e^{-0.2\times30}=0.1339$ $P(5)=\frac{(0.2\times30)^5}{5!}e^{-0.2\times30}=0.1606$ $P(6)=\frac{(0.2\times30)^6}{6!}e^{-0.2\times30}=0.1606$ $P(7)=\frac{(0.2\times30)^7}{7!}e^{-0.2\times30}=0.1377$ $P(8)=\frac{(0.2\times30)^8}{8!}e^{-0.2\times30}=0.1033$ $P(9)=\frac{(0.2\times30)^9}{9!}e^{-0.2\times30}=0.06884$ $P(10)=\frac{(0.2\times30)^{10}}{10!}e^{-0.2\times30}=0.04130$ $P(11)=\frac{(0.2\times30)^{11}}{11!}e^{-0.2\times30}=0.02253$ $P(12)=\frac{(0.2\times30)^{12}}{12!}e^{-0.2\times30}=0.01126$ $P(13)=\frac{(0.2\times30)^{13}}{13!}e^{-0.2\times30}=0.005199$ $P(14)=\frac{(0.2\times30)^{14}}{14!}e^{-0.2\times30}=0.002228$ $P(15)=\frac{(0.2\times30)^{15}}{15!}e^{-0.2\times30}=0.0008913$ $P(16)=\frac{(0.2\times30)^{16}}{16!}e^{-0.2\times30}=0.0003342$
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