document.write( "Question 332224: Can you please show me how to answer this? \r
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document.write( "For a particular sample of 50 scores on a psychology exam, the following results were obtained.
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document.write( "First quartile = 66 Third quartile = 92 Standard deviation = 9 Range = 49
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document.write( "Mean = 74 Median = 82 Mode = 83 Midrange = 73
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document.write( "Answer each of the following; show all work.
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document.write( "- What score was earned by more students than any other score? Why?
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document.write( "- What was the highest score earned on the exam?
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document.write( "- What was the lowest score earned on the exam?
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document.write( "- According to Chebyshev's Theorem, how many students scored between 56 and 92?
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document.write( "- Assume that the distribution is normal. Based on the Empirical Rule, how many students scored between 56 and 92?
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Algebra.Com's Answer #238108 by jrfrunner(365)![]() ![]() You can put this solution on YOUR website! - What score was earned by more students than any other score? Why? \n" ); document.write( "Mode= most frequently occurring score, 83 \n" ); document.write( "===== \n" ); document.write( "- What was the highest score earned on the exam? \n" ); document.write( "highest score= midrange + range/2 = 73+49/2=97.5 \n" ); document.write( "==== \n" ); document.write( "- What was the lowest score earned on the exam? \n" ); document.write( "Lowest score = midrange - range/2 = 73-49/2=48.5 \n" ); document.write( "=== \n" ); document.write( "- According to Chebyshev's Theorem, how many students scored between 56 and 92? \n" ); document.write( "largest-average=92-74=18 \n" ); document.write( "this represent k=18/9=2 standard deviations from the mean \n" ); document.write( "thus CT states that \"at least\" 1-1/k^2 =1-1/4=75% of the scores will surround the average within 2 standard deviations. \n" ); document.write( "75% of 50 = 37.5 or 38 scores \n" ); document.write( "=== \n" ); document.write( "- Assume that the distribution is normal. Based on the Empirical Rule, how many students scored between 56 and 92? \n" ); document.write( "if the data is normally distributed, approx 95% of the scores will surround the mean within 2 standard deviations \n" ); document.write( "so.. 95% of 50 =47.5 or approx 48 \n" ); document.write( " \n" ); document.write( " |