document.write( "Question 1194427: Suppose that the weight X of individual male patients registered at a certain diet clinic is
\n" ); document.write( "distributed normally with mean μ = 190 and variance σ2 = 100. Consider 𝑥̅𝑛=25 defined as the
\n" ); document.write( "sample mean of 25 individual observations of X. Find the values “a” and “b” such
\n" ); document.write( "that 𝑃(𝑎 ≤ 𝑥̅𝑛=25 ≤ 𝑏) = 0.8.
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Algebra.Com's Answer #826642 by Boreal(15235)\"\" \"About 
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variance is 100; sd is sqrt(100) or 10
\n" ); document.write( "sample sd is 10/sqrt(n)=2
\n" ); document.write( "to get a probability of 0.8 in the middle of the normal curve is z=-1.28 to z=+1.28
\n" ); document.write( "z*SE is the interval, so the interval is +/-1.28*2, or 2.56 with is added to and subtracted from 190 to get the interval.
\n" ); document.write( "the interval is (187.44, 192.56) whatever the weight units are, probably pounds.
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