document.write( "Question 675089: Hi, I just need help on doing the steps. I figured out it was 30, but do not remember how to do it. Thanks!\r
\n" ); document.write( "\n" ); document.write( "Assume that body temperatures of healthy adults are normally distributed with a mean of 98.2°F and a standard deviation of 0.62°F. Is P(98 < xbar < 98.4) greater for n =15 or
\n" ); document.write( "n = 30? \n" ); document.write( "
Algebra.Com's Answer #419532 by stanbon(75887)\"\" \"About 
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Assume that body temperatures of healthy adults are normally distributed with a mean of 98.2°F and a standard deviation of 0.62°F.
\n" ); document.write( "Is P(98 < xbar < 98.4) greater for n =15 or n = 30?\r
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\n" ); document.write( "Standard deviation for the distribution of x-bar is s/sqrt(n)
\n" ); document.write( "So std is smaller when n is larger.
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\n" ); document.write( "But z-value = (x-u)/[s/sqrt(n)]
\n" ); document.write( "so z-value and n are directly related;
\n" ); document.write( "so z-value is larger as n is larger
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\n" ); document.write( "Therefore the probability is greater when n = 30.\r
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\n" ); document.write( "Note:
\n" ); document.write( "z(98) = (98-98.2)/[0.62/sqrt(15)] = -1.2493
\n" ); document.write( "z(98) = (98-98.2)/[0.62/sqrt(30)] = -1.7668
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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