document.write( "Question 1186632: Dawson's Repair Service orders parts from an electronic company, which advertises its parts to be no more than 11% defective. What is the probability that Bill Dawson finds 12 or more parts out of a sample of 100 to be defective? \n" ); document.write( "
Algebra.Com's Answer #817555 by Boreal(15235)\"\" \"About 
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Find the probability that 11, 10, 9,... are defective.
\n" ); document.write( "This can be done using the binomial distribution.
\n" ); document.write( "for 11 it is 100C11*.11^11*.89^89=0.1265
\n" ); document.write( "for 10 it is 100C10*.11^10*.89^90=0.1251
\n" ); document.write( "for 9 it is 0.1112
\n" ); document.write( "for 8 it is 0.0881
\n" ); document.write( "for 7, 0.0613
\n" ); document.write( "etc. down to 0
\n" ); document.write( "OR, can do binomcdf(100,.11,11) which will get everything from the left side to 11. That probability is 0.5794.
\n" ); document.write( "The answer is the complement or 1-0.5794=0.4206. (Also 1-binomcdf(100,0.11,11))
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\n" ); document.write( "can check with normal approximation
\n" ); document.write( "np=11
\n" ); document.write( "np(1-p)=V=11(.89)=9.79
\n" ); document.write( "sqrt(V)=sd=3.13
\n" ); document.write( "z > (11.5-11)/3.13, with continuity correction factor >0.5/3.13 or 0.16. That probability is 0.4364, which is a reasonable approximation of the actual answer.
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