document.write( "Question 1106396: Assume that women's heights are normally distributed with a mean given by mu equals 64.3 in μ=64.3 in, and a standard deviation given by sigma equals 3.1 in σ=3.1 in.
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document.write( "(a) If 1 woman is randomly selected, find the probability that her height is less than 65 in.
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document.write( "(b) If 50 women are randomly selected, find the probability that they have a mean height less than 6565 in. \n" );
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Algebra.Com's Answer #721357 by Boreal(15235) You can put this solution on YOUR website! z=(x-mean)/sd \n" ); document.write( "z< (65-64.3)/3.1=z < +0.23 \n" ); document.write( "that probability is 0.5893\r \n" ); document.write( "\n" ); document.write( "The second one is divided by the standard error, which is 3.1/sqrt(50) \n" ); document.write( "That quotient is 0.438 \n" ); document.write( "z < 0.7/0.4380 <1.60 \n" ); document.write( "That probability is 0.9450; much less likely to find an average >65. \n" ); document.write( " |