document.write( "Question 974249: Assume that the heights of women are normally distributed with a mean of 63.5 inches and a standard deviation of 2.8 inches.
\n" ); document.write( "A. If 1 woman is randomly selected, find the probability that her height is less than 64 in.
\n" ); document.write( "B. If 44 women are randomly selected, find the probability that they have a mean height less than 64 in.
\n" ); document.write( "So far I got part A.
\n" ); document.write( " 64-63.5/2.8
\n" ); document.write( " =0.18
\n" ); document.write( " The probability is approximately 0.5714 (0.1, .08) on the \"Positive z Scores\"
\n" ); document.write( "\"Standard Normal (z) Distribution: Cumulative Area from the Left\"
\n" ); document.write( "For part B I am stuck on
\n" ); document.write( "In this case, the desired probability is for the mean of a sample of 44 women, therefore use the central limit theorem.
\n" ); document.write( "The probability is approx.?
\n" ); document.write( "2.8/sqrt44
\n" ); document.write( "Please help me and show the work so that I understand…I am not just looking for an answer! Thank you
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Algebra.Com's Answer #596214 by stanbon(75887)\"\" \"About 
You can put this solution on YOUR website!
Assume that the heights of women are normally distributed with a mean of 63.5 inches and a standard deviation of 2.8 inches.
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\n" ); document.write( "A. If 1 woman is randomly selected, find the probability that her height is less than 64 in.\r
\n" ); document.write( "\n" ); document.write( "So far I got part A.
\n" ); document.write( "(64-63.5)/2.8 = 0.18
\n" ); document.write( "P(x < 64) = P(z< 0.18) = is approximately 0.5714
\n" ); document.write( "You are correct
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\n" ); document.write( "B. If 44 women are randomly selected, find the probability that they have a mean height less than 64 in.
\n" ); document.write( "For part B I am stuck on
\n" ); document.write( "In this case, the desired probability is for the mean of a sample of 44 women, therefore use the central limit theorem.
\n" ); document.write( "The standard deviation of the sample means is 2.8/sqrt44 = 0.42
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\n" ); document.write( "Since n > 30, use the z-distribution.
\n" ); document.write( "z(64) = (64-63.5)/0.42 = 1.19
\n" ); document.write( "Then P(x-bar < 64) = P(z < 1.19) = 0.88
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\n" ); document.write( "Hope that helps.
\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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