document.write( "Question 1066859: Find the standard score (z) such that the area above the mean and below z is about 34% of the area under the normal curve.\r
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Algebra.Com's Answer #682069 by stanbon(75887)\"\" \"About 
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Find the standard score (z) such that the area above the mean and below z is about 34% of the area under the normal curve.
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\n" ); document.write( "Sketch a normal curve with horizontal axis being z-values.
\n" ); document.write( "Make the mean as z = 0 in the middle of the z-axis.
\n" ); document.write( "Shade the area to the left of z = 0 and the some area to the
\n" ); document.write( "right of z = 0 approximating 0.34.
\n" ); document.write( "The total shaded area is 0.50+0.34 = 0.84
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\n" ); document.write( "Use your z-chart to find the z-value corresponding to
\n" ); document.write( "0.34 area to the right of z = 0. That chart value is 1.
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\n" ); document.write( "OR use a TI-84 calculator to find the answer as follows:
\n" ); document.write( "That z-value has a left tail of 0.50+0.34 = 0.84
\n" ); document.write( "Ans: z = InvNorm(0.84) = 1
\n" ); document.write( "Note:: The calculator gives an answer of 0.994457..
\n" ); document.write( "I rounded that off to z = 1.
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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