document.write( "Question 933350: the mean is 50, and standard deviation is 6. what can you say about the fraction of numbers that lie between 32-68?\r
\n" ); document.write( "\n" ); document.write( "find the least possible percentage of numbers in a data set lying within 3/2 standard deviation of the mean?\r
\n" ); document.write( "\n" ); document.write( "I need to fingure these our for a final This is not the fianl but if i can figure some of these out I can plug the numbers in my instruct said. I have many more im struggling so bad
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Algebra.Com's Answer #568641 by Theo(13342)\"\" \"About 
You can put this solution on YOUR website!
first question:\r
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\n" ); document.write( "\n" ); document.write( "mean is 50
\n" ); document.write( "standard deviation is 6.
\n" ); document.write( "fraction that lies between 32 and 68 is calculated as follows:\r
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\n" ); document.write( "\n" ); document.write( "first you have to find the z-score, then you find the percent using the z-score table.\r
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\n" ); document.write( "\n" ); document.write( "the formuls for z-score is z = (x-m) / s, where z = the z-score and x is the raw score and m is the mean and s is the stnadard deviation.\r
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\n" ); document.write( "\n" ); document.write( "you want the z-score for 32 and 68.\r
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\n" ); document.write( "\n" ); document.write( "formula for raw score of 32 is z = (32 - 50) / 6 which gets you z = -3\r
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\n" ); document.write( "\n" ); document.write( "formula for raw score of 68 is z = (68 - 50) / 6 which gets you z = 3\r
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\n" ); document.write( "\n" ); document.write( "you are looking for the area under the normal distribution curve that is between a z-score of -3 and a z-score of 3.\r
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\n" ); document.write( "\n" ); document.write( "look in the z-score table to find the percent of the area to the left of the z-score.\r
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\n" ); document.write( "\n" ); document.write( "from the table, you will find that the ratio to the left of -3 is equal to .0013 and the ratio0 to the left of 3 is equal to .9987.\r
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\n" ); document.write( "\n" ); document.write( "subtract the smaller ratio from the larger ratio to get the ratio that is between -3 and 3 is equal to .9974.\r
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\n" ); document.write( "\n" ); document.write( "a ratio of .9974 means that 99.74% of the area under the normal distribution curve is between a z-score of -3 and 3.\r
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\n" ); document.write( "\n" ); document.write( "this means that the fraction of numbers between 32 and 68 is equal to .9987 which means that 99.87 of all the numbers in the data set that gave you a mean of 50 with a standard deviation of 6 can be found between 32 and 68.\r
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\n" ); document.write( "\n" ); document.write( "if you grabbed a number at random from the data set, you would find that the number was between 32 and 68 99.74% of the number of times that you grabbed, and would be less than 32 and greater than 68 only .26% of the number of times that you grabbed.\r
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\n" ); document.write( "\n" ); document.write( "put another way, if you grabbed a number at random 10,000 times, you would get a number between 32 and 68 for 9974 of those times and get a number less than 32 or greater than 68 only 26 of those times.\r
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\n" ); document.write( "\n" ); document.write( "second question:\r
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\n" ); document.write( "\n" ); document.write( "if the number lies with 3/2 standard deviations from the mean, then you would calculate the percent as follows.\r
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\n" ); document.write( "\n" ); document.write( "a z-score tells you the number of standard deviations you are from the mean.\r
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\n" ); document.write( "\n" ); document.write( "a z-score of 3 tells you that you are 3 standard deviations above the mean.
\n" ); document.write( "a z-score of -3 tells you that you are 3 standard deviations below the mean.\r
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\n" ); document.write( "\n" ); document.write( "we solved for that in the first problem.\r
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\n" ); document.write( "\n" ); document.write( "in this problem, you are looking for a z-swcore of between -3/2 and 3/2 because:\r
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\n" ); document.write( "\n" ); document.write( "a z-score of 3/2 means you are 3/2 standard deviations above the mean.
\n" ); document.write( "a z-score of -3/2 means you are 3/2 standard deviations below the mean.\r
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\n" ); document.write( "\n" ); document.write( "same drill.
\n" ); document.write( "look for a z-score of 3/2 in the z-score table and find the ratio to the left of that z-score.
\n" ); document.write( "look for a z-score of -3/2 in the z-score table and find the ratio to the left of that z-score.
\n" ); document.write( "subtract the smaler ratio from the larger ratio and you have the ratio between those z-scores.
\n" ); document.write( "multiply by 100 and you have the percent of the area under the normal distribution curve that is between the z-scores of -3/2 and 3/2.\r
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\n" ); document.write( "\n" ); document.write( "a z-score of 3/2 is equal to a z-score of 1.5.
\n" ); document.write( "a z-score of -3/2 is equal to a z-score of -1.5.\r
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\n" ); document.write( "\n" ); document.write( "area to the left of a z-score of 1.5 is equal to .9332
\n" ); document.write( "area to the left of a z-score of -1.5 is equal to .0668\r
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\n" ); document.write( "\n" ); document.write( ".9332 - .0668 = .8664\r
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\n" ); document.write( "\n" ); document.write( "this is the ratio or proportion of the area under the normal distribution curve that is between a z-score of -1.5 and 1.5.\r
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\n" ); document.write( "\n" ); document.write( "the percent is equal to 100 * .8664 which is equal to 86.64\r
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\n" ); document.write( "\n" ); document.write( "the z-score table i am using can be found here:\\r
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\n" ); document.write( "\n" ); document.write( "http://www.stat.ufl.edu/~athienit/Tables/Ztable.pdf\r
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\n" ); document.write( "\n" ); document.write( "the area shown is the area that is to the left of the z-score.\r
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\n" ); document.write( "\n" ); document.write( "that area is the ratio of the area to the left of the z-score to the total area under the normal distribution curve.\r
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\n" ); document.write( "\n" ); document.write( "the total area under the normal distribution curve is always equal to 1.\r
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\n" ); document.write( "\n" ); document.write( "the ratio multiplied by 100 is equal to the percent.\r
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