document.write( "Question 931785: A normal population has a mean of 66 and a standard deviation of 4. You select a sample of 48.
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\n" ); document.write( "Compute the probability the sample mean is: (Round z values to 2 decimal places and final answers to 4 decimal places.)
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\n" ); document.write( "(a) Less than 65.\r
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\n" ); document.write( "\n" ); document.write( "(b) Between 65 and 67.\r
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\n" ); document.write( "\n" ); document.write( "(c) Between 67 and 68.\r
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\n" ); document.write( "\n" ); document.write( "(d) Greater than 68.
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Algebra.Com's Answer #565816 by ewatrrr(24785)\"\" \"About 
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mean of 66 and a standard deviation of 4.
\n" ); document.write( "n = 48, \"z+=blue+%28x+-+68%29%2Fblue%284%2Fsqrt%2848%29%29+=blue+%28x+-+68%29%2Fblue%28.58%29+\"
\n" ); document.write( "P(x < 65) = P(z <-1/.58) = normalcdf(-100, -1.72)
\n" ); document.write( "P(65 < x <67) = P(-1.72 < z < 1.72) = normalcdf(-1.72, 1.72)
\n" ); document.write( "P( 67< x < 68) = P(1.72 < z < 3.45)=normalcdf(1.72, 3.45)
\n" ); document.write( "P(x > 68) = P(z > 3.45) = normalcdf(3.45, 100)
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