Statistics: Informed Decisions Using Data (4th Edition)

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

Chapter 6 - Section 6.2 - Assess Your Understanding - Vocabulary and Skill Building - Page 344: 34a

Answer

See the picture.

Work Step by Step

$P(x)={}_nC_{x}~p^x~(1-p)^{n-x}$ n = 9, p = 0.8 and 1 - p = 0.2 $P(0)={}_{9}C_{0}\times0.8^0\times0.2^9=\frac{9!}{0!\times9!}\times1\times0.2^9=1\times1\times0.2^9=0.000000512$ $P(1)={}_{9}C_{1}\times0.8^1\times0.2^8=\frac{9!}{1!\times8!}\times0.8^1\times0.2^8=9\times0.8\times0.2^8=0.00001843$ $P(2)={}_{9}C_{2}\times0.8^2\times0.2^7=\frac{9!}{2!\times7!}\times0.8^2\times0.2^7=36\times0.8^2\times0.2^7=0.0002949$ $P(3)={}_{9}C_{3}\times0.8^3\times0.2^6=\frac{9!}{3!\times6!}\times0.8^3\times0.2^6=84\times0.8^3\times0.2^6=0.002753$ $P(4)={}_{9}C_{4}\times0.8^4\times0.2^5=\frac{9!}{4!\times5!}\times0.8^4\times0.2^5=126\times0.8^4\times0.2^5=0.01652$ $P(5)={}_{9}C_{5}\times0.8^5\times0.2^4=\frac{9!}{5!\times4!}\times0.8^5\times0.2^4=126\times0.8^5\times0.2^4=0.06606$ $P(6)={}_{9}C_{6}\times0.8^6\times0.2^3=\frac{9!}{6!\times3!}\times0.8^6\times0.2^3=84\times0.8^6\times0.2^3=0.1762$ $P(7)={}_{9}C_{7}\times0.8^7\times0.2^2=\frac{9!}{7!\times2!}\times0.8^7\times0.2^2=36\times0.8^7\times0.2^2=0.3020$ $P(8)={}_{9}C_{8}\times0.8^8\times0.2^1=\frac{9!}{8!\times1!}\times0.8^8\times0.2=9\times0.8^8\times0.2=0.3020$ $P(9)={}_{9}C_{9}\times0.8^9\times0.2^0=\frac{9!}{9!\times0!}\times0.8^9\times1=1\times0.8^9\times1=0.1342$
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