document.write( "Question 1187601: Suppose approximately 56% of all students study after midnight.\r
\n" ); document.write( "\n" ); document.write( "For random sample of n=3 students, let X be the number of students that study after midnight.\r
\n" ); document.write( "\n" ); document.write( "Complete the following table for the binomial distribution of X.
\n" ); document.write( "Keep at least 4 decimal places.\r
\n" ); document.write( "\n" ); document.write( "X P(X)
\n" ); document.write( "0
\n" ); document.write( "1
\n" ); document.write( "2
\n" ); document.write( "3 \r
\n" ); document.write( "\n" ); document.write( "Determine the mean and standard deviation of X.
\n" ); document.write( "Keep at least 4 decimal places.\r
\n" ); document.write( "\n" ); document.write( "The expected value of X=
\n" ); document.write( "
\n" ); document.write( "The standard deviation of X= \n" ); document.write( "

Algebra.Com's Answer #819211 by Boreal(15235) $\"\"$ $\"About$

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X P(X)
\n" ); document.write( "0 .44^3=0.0852
\n" ); document.write( "1 3C1*.56^1*.44^2=0.3252
\n" ); document.write( "2 3C2*0.56^2*0.44=0.4140
\n" ); document.write( "3 0.56^3=0.1756
\n" ); document.write( "They add to 1.0, as they must.
\n" ); document.write( "E(x)=np=3*0.56=1.68
\n" ); document.write( "V(x)=np(1-p)=1.68*0.44=0.7392
\n" ); document.write( "sd= sqrt (V)=0.8598 \n" ); document.write( "

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