document.write( "Question 1169227: A normal distribution has a mean of 85.8 and a standard deviation of 4.88. Find the data value corresponding to the value of z given. (Enter your answer to four decimal places.)
\n" ); document.write( "z = 0.65
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Algebra.Com's Answer #793880 by Theo(13342)\"\" \"About 
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m = 85.8
\n" ); document.write( "sd = 4.88
\n" ); document.write( "z = .65\r
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\n" ); document.write( "\n" ); document.write( "z = (x - m)/sd
\n" ); document.write( "z = z-score
\n" ); document.write( "x = raw score
\n" ); document.write( "m = mean
\n" ); document.write( "sd = standard deviation.\r
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\n" ); document.write( "\n" ); document.write( "when m = 85.8 and sd = 4.88 and z = .65, you get:
\n" ); document.write( ".65 = (x - 85.8) / 4.88
\n" ); document.write( "solve for x to get:
\n" ); document.write( "x = .65 * 4.88 + 85.8 = 88.972\r
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\n" ); document.write( "\n" ); document.write( "if they're equivalent, they should have the same area under the normal distribution curve to the left of them.\r
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\n" ); document.write( "\n" ); document.write( "confirm using a normal distribution calculator.
\n" ); document.write( "the one i used is the ti-84 plus.\r
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\n" ); document.write( "\n" ); document.write( "z-score of .65 has an area of .7421539567 to the left of it.
\n" ); document.write( "raw score of 88.972 has an area of .7421539567 to the left of it.
\n" ); document.write( "you get the same area to the left of the z-score as the raw score, so the z-score and the raw score are comparable in their relationship to their position under the normal distribution curve.\r
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\n" ); document.write( "\n" ); document.write( "an online one that you can use is found at https://stattrek.com/online-calculator/normal.aspx\r
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\n" ); document.write( "\n" ); document.write( "here's a display of the z-score result and the raw score result.
\n" ); document.write( "the area to the left of each is the same.
\n" ); document.write( "to get the z-score result, make the mean = 0 and the standard deviation = 1.
\n" ); document.write( "to get the raw score result, make the mean = 85.8 and the standard deviation = 4.88.\r
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\n" ); document.write( "\n" ); document.write( "the online calculator rounds the result to 3 decimal places.
\n" ); document.write( "this is usually sufficient.\r
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