document.write( "Question 1107340: : In the example in problem 18, assume that the mean is 300 and the standard deviation is 25. Using a normal curve table, what scores would be the top and bottom scores to find a)the middle 50% of architects b)the middle 90% of architects and c)the middle 99% of architects?\r
\n" ); document.write( "\n" ); document.write( "q 18 Suppose that the scores of architects on a particular creativity test are normally distributed. Using a normal curve table, what percentage of architects have Z scores a)above .10 b)below .10 c)above.20 d)below .20 e)above 1.10 f)below 1.10 g)above -.10 and h)below -.10 \n" ); document.write( "
Algebra.Com's Answer #722509 by Boreal(15235)\"\" \"About 
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The middle 50% need probability of .2500 on both sides.
\n" ); document.write( "This is z=+/-0.67
\n" ); document.write( "z*sigma=+/-16.75
\n" ); document.write( "scores are (283.25, 316.75)
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\n" ); document.write( "z for the middle 90% is +/-1.645
\n" ); document.write( "z*sigma=+/-41.1
\n" ); document.write( "scores are (258.9, 341.1)
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\n" ); document.write( "middle 99% z is +/- 2.576
\n" ); document.write( "z*sigma=+/-25*2.576=+/-64.4
\n" ); document.write( "scores are (235.6, 364.4)\r
\n" ); document.write( "\n" ); document.write( ">0.1=0.4602
\n" ); document.write( "<0.1=0.5398
\n" ); document.write( ">0.20=0.4207
\n" ); document.write( "<0.20=0.5793
\n" ); document.write( ">1,10=0.1357
\n" ); document.write( "<1.10=0.8643
\n" ); document.write( ">-0.10=0.5398
\n" ); document.write( "<-0.10=0.4602\r
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