document.write( "Question 1134479: A study of the impact of executive networking on firm performance measured firm performance as annual return on equity​ (ROE), recorded as a percentage. The mean ROE for the firms studied was 14.93​% and the standard deviation was 21.74​%. Assume that these values represent mu and sigma for the population ROE distribution and that this distribution is normal. What value of ROE will be exceeded by 78​% of the​ firms?
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Algebra.Com's Answer #751866 by Theo(13342)\"\" \"About 
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the mean is 14.93%
\n" ); document.write( "the standard deviation is 21.74%\r
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\n" ); document.write( "\n" ); document.write( "you want to know what value of ROE will be exceeded by 78% of the firms.\r
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\n" ); document.write( "\n" ); document.write( "in a normal distributuion, 78% of the area under the nomral distribution curve to the right of the z-score would generate a z-score equal to -.7721932195.\r
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\n" ); document.write( "\n" ); document.write( "the z-score formula is z = (x - m) / s\r
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\n" ); document.write( "\n" ); document.write( "z is the z-score
\n" ); document.write( "x is the raw score
\n" ); document.write( "m is the raw score mean
\n" ); document.write( "s is the standard deviation\r
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\n" ); document.write( "\n" ); document.write( "this formula becomes -.7721932195 = (x - 14.93) / 21.74.\r
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\n" ); document.write( "\n" ); document.write( "solve for x to get x = 21.74 * -.7721932195 + 14.93 = -1.857480591 percent.\r
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\n" ); document.write( "\n" ); document.write( "this can be seen visually in the following graph.\r
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\n" ); document.write( "\n" ); document.write( "78% of the firms will provide a ROE greater than -1.857480591 percent.\r
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