document.write( "Question 834023: Exam scores are normally distributed with a mean of 81 and a standard deviation of 9.\r
\n" ); document.write( "\n" ); document.write( "a) What is the minimum score one must have to be in the top 4% of the students taking the exam?\r
\n" ); document.write( "\n" ); document.write( "b) What percentage of scores are below 90?\r
\n" ); document.write( "\n" ); document.write( "c) What percentage of scores are above 45?\r
\n" ); document.write( "\n" ); document.write( "d) If 30 people take this exam, how many would you expect to get above a 93%?\r
\n" ); document.write( "\n" ); document.write( "e) If you have a test score in the 45th percentile, what is its z-score?\r
\n" ); document.write( "\n" ); document.write( "f) What is the z-score and percentile for an exam score 1.2 standard deviations below the mean?
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Algebra.Com's Answer #502797 by stanbon(75887)\"\" \"About 
You can put this solution on YOUR website!
Exam scores are normally distributed with a mean of 81
\n" ); document.write( "and a standard deviation of 9.
\n" ); document.write( "a) What is the minimum score one must have to be in the top 4% of the students taking the exam?
\n" ); document.write( "Find the z-score with a right tail of 4%:
\n" ); document.write( "invNorm(0.96) = 1.7507
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\n" ); document.write( "Find the corresponding score: 81+1.7507*9 = 96.76
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\n" ); document.write( "b) What percentage of scores are below 90?
\n" ); document.write( "z(90) = (90-81)/9 = 1
\n" ); document.write( "P(x < 90) = P(z < 1) = 0.8413
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\n" ); document.write( "c) What percentage of scores are above 45?
\n" ); document.write( "Find z(45) then find the Probability z is above that z value.
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\n" ); document.write( "d) If 30 people take this exam, how many would you expect to get above a 93%?
\n" ); document.write( "Ans:0.07*30 = 2.1
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\n" ); document.write( "e) If you have a test score in the 45th percentile, what is its z-score?
\n" ); document.write( "invNorm(0.45) = -0.1257
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\n" ); document.write( "f) What is the z-score and percentile for an exam score 1.2 standard deviations below the mean?
\n" ); document.write( "z = -1.2
\n" ); document.write( "score = 81-1.2*9 = 70.2
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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