document.write( "Question 1021892: The mean and standard deviation of exam score are 49% and 9% respectively.Assuming that examintion are normally distributed,calculate the probability of score:
\n" ); document.write( "A.less than 50%
\n" ); document.write( "B.more than 45%
\n" ); document.write( "C.between 52% and 63%
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Algebra.Com's Answer #637601 by Fombitz(32388)\"\" \"About 
You can put this solution on YOUR website!
Use the z score,
\n" ); document.write( "A.\"z=%28x-mu%29%2Fsigma=%2850-49%29%2F9=0.1111\"
\n" ); document.write( "\"P%28z%29=0.5442\"
\n" ); document.write( "B.\"z=-0.4444\"
\n" ); document.write( "\"P=1-P%28z%29=1-0.3284=0.6716\"
\n" ); document.write( "C.\"z%5B1%5D=%2852-50%29%2F9=0.3333\"
\n" ); document.write( "\"z%5B2%5D=1.5556\"
\n" ); document.write( "\"P=P%282%29-P%281%29=0.9401-+0.6306=0.3095\"
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