document.write( "Question 476009: I'm having problems with this one any help would be appreciated\r
\n" ); document.write( "\n" ); document.write( "In a standard normal distribution, what z value corresponds to 17% of the data between the mean and the z value? \n" ); document.write( "

Algebra.Com's Answer #326434 by MathLover1(20850) $\"\"$ $\"About$

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\n" ); document.write( "I believe your question is to find a range going from the mean to a $\"z-value\"$ on the standard normal distribution that corresponds to $\"17\"$ % of the area. A normal distribution goes from values of minus infinity to positive infinity. A standard normal distribution has a $\"mean\"$ of $\"0\"$ and an $\"standard\"$ $\"deviation\"$ of $\"1\"$ .\r
\n" ); document.write( "\n" ); document.write( "It is usually best if you draw a diagram, in this case a bell shape curve with $\"mean+=+0\"$ . The area to the left of the mean is $\"50\"$ % of the total area. We find a z value that corresponds to $\"67\"$ % (50% + 17%) of the area to the left of this value. This can be done either with a lookup table or a spreadsheet program. I prefer excel, + $\"norminv%280.67%29+=+0.44\"$ .\r
\n" ); document.write( "\n" ); document.write( "The problem could also be worded to find the area going from a z-value to the mean. In this case, we must find a z-value that corresponds to $\"33\"$ % (50%-17%). Using Excel, I calculate + $\"norminv%280.33%29+=+-0.44\"$ . \n" ); document.write( "

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