document.write( "Question 123192: THis is homework can you help?\r
\n" ); document.write( "\n" ); document.write( "Suppose that the heights of adult women in the United States are normally distributed with a mean of 64.5 inches and a standard deviation of 2.4 inches. Jennifer is taller than 80% of the population of U.S. women. How tall (in inches) is Jennifer? Carry your intermediate computations to at least four decimal places. Round your answer to at least one decimal place.
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Algebra.Com's Answer #90411 by Edwin McCravy(20055)\"\" \"About 
You can put this solution on YOUR website!
THis is homework can you help?
\n" ); document.write( "Suppose that the heights of adult women in the United States are normally distributed with a mean of 64.5 inches and a standard deviation of 2.4 inches. Jennifer is taller than 80% of the population of U.S. women. How tall (in inches) is Jennifer? Carry your intermediate computations to at least four decimal places. Round your answer to at least one decimal place.
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document.write( "There are two different kinds of normal tables that are used, \r\n" );
document.write( "and some people don't use tables at all, but either the TI-84 \r\n" );
document.write( "calculator or the computer program Excel. \r\n" );
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document.write( "Using tables:\r\n" );
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document.write( "If you have the kind of table that has both negative and positive z \r\n" );
document.write( "values in it, then you'll find the values .7995 and .8023, because 80%\r\n" );
document.write( "or .8000 is between those two numbers. That is, the z-value entry which \r\n" );
document.write( "is closest to having 80% of the area to the left of it. It's a matter of\r\n" );
document.write( "looking through the body of the table until you find the entries which\r\n" );
document.write( ".8 is between:   Those listings correspond to z-values 0.84 and 0.85.  \r\n" );
document.write( "Since .7995 is closer to .8000 than .8023, we choose z-value 0.84.\r\n" );
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document.write( "If you have the kind of table that has only positive z-values, then you\r\n" );
document.write( "subtract .8 - .5, getting .3. then you'll find the values .2995 and .3023,\r\n" );
document.write( "because .3 (or .3000) is between those two numbers. That is, the z-value \r\n" );
document.write( "entry which is closest to having 30% of the area to the right of the\r\n" );
document.write( "middle. the left of it. It's a matter of looking through the body of the \r\n" );
document.write( "table until you find the listings correspond to z-values 0.84 and 0.85.  \r\n" );
document.write( "Since .2995 is closer to .3000 than .3023, we choose z-value 0.84.\r\n" );
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document.write( "From either table, we get z = 0.84\r\n" );
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document.write( "Then we use the formula\r\n" );
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document.write( "x = m + sz \r\n" );
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document.write( "x = 64.5 + (2.4)(0.84) = 66.516\r\n" );
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document.write( "If you use a TI-83 or 84 calculator,\r\n" );
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document.write( "Press \r\n" );
document.write( "2nd \r\n" );
document.write( "VARS\r\n" );
document.write( "3\r\n" );
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document.write( "Then type\r\n" );
document.write( ".8\r\n" );
document.write( ",  the comma key\r\n" );
document.write( "64.5\r\n" );
document.write( ",  the comma key again\r\n" );
document.write( "2.4\r\n" );
document.write( ")  the close parenthesis key\r\n" );
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document.write( "(The screen should now read invNorm(.8,64.5,2.4)\r\n" );
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document.write( "press\r\n" );
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document.write( "ENTER\r\n" );
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document.write( "Read 66.51989096\r\n" );
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document.write( "Edwin
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