document.write( "Question 414068: The probability of a defective pen in a large office building is 0.08. If a random sample of 30 telephones is selected, find the following probabilities, assuming this experiment is a binomial experiment.\r
\n" ); document.write( "\n" ); document.write( "A. P[exactly 3 defective pens]
\n" ); document.write( "B. P[at most 3 defective pens]
\n" ); document.write( "C. P[no defective pens]
\n" ); document.write( "D. Determine mean and variance\r
\n" ); document.write( "\n" ); document.write( "For P[exactly 3] I came up with 0.219 and for P[at most 3] I came up with 0.784. Are these correct? Need help figuring out C and D. Thanks! \n" ); document.write( "

Algebra.Com's Answer #290425 by ewatrrr(24785) $\"\"$ $\"About$

You can put this solution on YOUR website!


\n" );
document.write( "Hi
\n" );
document.write( "Note: The probability of x successes in n trials is: 
\n" );
document.write( "P = nCx*  where p and q are the probabilities of success and failure respectively. 
\n" );
document.write( "In this case p= .08 and q = .92
\n" );
document.write( "nCx = 
\n" );
document.write( "A. P[exactly 3 defective pens] = .2188    |Good work
\n" );
document.write( "B. P[at most 3 defective pens] = .7842    |Good work
\n" );
document.write( "C. P[no defective pens] = (.92)^30 = .082
\n" );
document.write( "D. Determine mean and variance 
\n" );
document.write( "    mean = np = 30*.08 = 2.4
\n" );
document.write( " variance = npq = 30*.08*.92 = 2.208 \n" );
document.write( "

\n" );