document.write( "Question 1167442: Scores on a standardized exam are known to follow a normal distribution with standard deviation σ = 14. A researcher randomly selects 85 students and computes their average score. He reports that the mean score is 73.5, with a margin of error of 1.640.\r
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document.write( "How confident are you that the mean score for all students taking the exam is in the interval (71.86, 75.14)?\r
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document.write( "Confidence =
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document.write( " %
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Algebra.Com's Answer #792038 by Boreal(15235) You can put this solution on YOUR website! The half-interval is 1.64 \n" ); document.write( "this equals z*sigma/sqrt(n) or z*14/sqrt(85) \n" ); document.write( "so z=1.64*sqrt(85)/14=1.08 which is at the 86th percentile for z \n" ); document.write( "this would be 28% counting both sides or a CI of 72%.\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |