document.write( "Question 1121493: Estimate the indicated probability by using the normal distribution as an approximation of the binomial distribution. A multiple choice exam consists of 60 questions. Each question has 4 possible answers of which one is correct. If all the answers are random guesses, calculate the probability of obtaining at least 20% of correct answers. \n" ); document.write( "
Algebra.Com's Answer #737388 by rothauserc(4718)\"\" \"About 
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20% = 20/100 = 1/5
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\n" ); document.write( "1/5 * 60 = 12 questions
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\n" ); document.write( "Probability (P) ( guessing at least 12 questions correctly out of 60 ) = 1 - P (guessing less than 12 questions correctly)
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\n" ); document.write( "using normal distribution, we calculate P (X < 12) including the correction factor of -0.5
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\n" ); document.write( "P of guessing one answer correctly is 1/4 and 60 * 1/4 = 15(this is our mean)
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\n" ); document.write( "standard deviation = sqrt(1/4 * 3/4 * 60) = 3.3541
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\n" ); document.write( "apply 0.5 correction factor to 12, 12 - 0.5 = 11.5
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\n" ); document.write( "Z(11.5) = (11.5 - 15) / 3.3541 = -1.0435 approximately -1.04
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\n" ); document.write( "next we look for the associate probability in the table of z-values, it is 0.1492
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\n" ); document.write( "1 - 0.1492 = 0.8508 approximately 0.85
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\n" ); document.write( "Probability (P) ( guessing at least 12 questions correctly out of 60 ) = 0.85
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\n" ); document.write( "Note the binomial probability for guessing at least 12 questions correctly out of 60 is 0.852
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