document.write( "Question 648023: 8. The mean of a normal probability distribution is 60; the standard deviation is 5.
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document.write( "a. About what percent of the observations lie between 55 and 65?
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document.write( "b. About what percent of the observations lie between 50 and 70?
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document.write( "c. About what percent of the observations lie between 45 and 75? \r
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document.write( "PLEASE HELP!!! \n" );
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Algebra.Com's Answer #406530 by Theo(13342) You can put this solution on YOUR website! mean is 60. \n" ); document.write( "standard deviation is 5. \n" ); document.write( "calculate a z-score for each of the possibilities. \n" ); document.write( "z-score = actual score minus mean divided by standard deviation. \n" ); document.write( "for a: \n" ); document.write( "z-score = (55-60)/5 and (65-60)/5 which results in: \n" ); document.write( "z-score of -1 and 1. \n" ); document.write( "this is 1 standard deviation above and below the mean which is approximately 68% of the distribution. \n" ); document.write( "for b: \n" ); document.write( "z-score = (50-60)/5 and (70-60)/5 which results in: \n" ); document.write( "z-score of -2 and 2. \n" ); document.write( "this is 2 standard deviations above and belowthe mean which is approximately 95% of the distribution. \n" ); document.write( "for c: \n" ); document.write( "z-score = (45-60)/5 and (75-60)/5 which results in: \n" ); document.write( "z-score of -3 and 3. \n" ); document.write( "this is 3 standard deviations above and below the mean which is approximately 99.7% of the distribution. \n" ); document.write( "the actual figures are more detailed, but these are the generally recognized rounded figures that are normally used. \n" ); document.write( "check out this link for further information regarding normal distribution and standard deviations from the mean. \n" ); document.write( "http://www.mathsisfun.com/data/standard-normal-distribution.html \n" ); document.write( " |