document.write( "Question 1098608: Given that z is a standard normal random variable, what is the value of z if the area to the left of z is 0.0559 \n" ); document.write( "
Algebra.Com's Answer #713217 by Theo(13342)\"\" \"About 
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if you have a calculator that does statistics, like the texas instruments TI-84 plus, than you would take the inverse norm of .559 and the calculator would tell you that the z-score that has .559 area under the distribution curve to the left of it is .1484343318 which you would probably round to 2 decimal places and you would get .15\r
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\n" ); document.write( "\n" ); document.write( "if you use a standard z-score table, you would look for an area to the left of a z-score that is closest to .559.\r
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\n" ); document.write( "\n" ); document.write( "you would probably find that .5557 = .14 and .5596 = .15
\n" ); document.write( "since .5596 is closer to .559 than .5557, you would go with z = .15\r
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\n" ); document.write( "\n" ); document.write( "there are online calculator that would get you the inverse.\r
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\n" ); document.write( "\n" ); document.write( "one such calculator can be found at http://stattrek.com/online-calculator/normal.aspx\r
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\n" ); document.write( "\n" ); document.write( "give it the area to the left of the z-score and it will tell you the z-score.\r
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\n" ); document.write( "\n" ); document.write( "the table i used can be found at http://www.stat.ufl.edu/~athienit/Tables/Ztable.pdf\r
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\n" ); document.write( "\n" ); document.write( "here's a picture of what i found.\r
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\n" ); document.write( "\n" ); document.write( "the areas to the left of the z-score are .5557 and .5596.\r
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\n" ); document.write( "\n" ); document.write( "the row that they are on shows a z-score of .1
\n" ); document.write( "the columns that they are on show z-scores of .04 and .05
\n" ); document.write( "you add .1 to .04 and you get .14
\n" ); document.write( "you add .1 to .05 and you get .15\r
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\n" ); document.write( "\n" ); document.write( ".15 has the area to the left of the z-score that is closest to .559, therefore choose .15.\r
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\n" ); document.write( "\n" ); document.write( "if you want better accuracy, you can interpolate manually or use a calculator.\r
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\n" ); document.write( "\n" ); document.write( "most times rounding to the nearest 2 decimal digits on the z-score is good enough.\r
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\n" ); document.write( "\n" ); document.write( "a manual interpolation is not quite as accurate as a calculator generated result, but it gets you closer than you would have been otherwise.\r
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\n" ); document.write( "\n" ); document.write( "a manual interpolation of what this table shows would get you the following.\r
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\n" ); document.write( "\n" ); document.write( ".5557 to .559 is equal to .0033
\n" ); document.write( ".5557 to .5596 is equal to .0039
\n" ); document.write( ".0033 / .0039 = .8461538462
\n" ); document.write( ".14 to .15 = .01
\n" ); document.write( ".8461538462 * .01 = .008415385
\n" ); document.write( ".14 + .008415385 = .148415385 rounded to 3 decimal places = .148\r
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\n" ); document.write( "\n" ); document.write( "your manual approximation would be .148 which is pretty close to what the calculator would have given you.\r
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\n" ); document.write( "\n" ); document.write( "so, you would either go with .15 or with .148, depending on the accuracy required.\r
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\n" ); document.write( "\n" ); document.write( "visually, your area to the left of the z-score of .148 would look like this:\r
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\n" ); document.write( "\n" ); document.write( "this is through the use of another calculator by david m. lane that gets your results and gives you a visual of what it looks like.\r
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\n" ); document.write( "\n" ); document.write( "that calculator can be found at http://davidmlane.com/hyperstat/z_table.html\r
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