document.write( "Question 1192857: In a normal distribution with mean 56 and standard deviation 21, how large a sample must be taken so that there will be at least a 90% percent chance that its mean is greater than
\n" ); document.write( "52? Recall that Z= x-mu/sigma/sqrt(n)
\n" ); document.write( "Can a tutor please assist me, what should I do to find n?
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Algebra.Com's Answer #824801 by math_tutor2020(3817)\"\" \"About 
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\n" ); document.write( "mu = population mean = 56
\n" ); document.write( "sigma = population standard deviation = 21\r
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\n" ); document.write( "\n" ); document.write( "We're wanting to find the value of n that will make
\n" ); document.write( "P(xbar > 52) = 0.90
\n" ); document.write( "to be possible.\r
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\n" ); document.write( "\n" ); document.write( "This is equivalent to finding a value of k such that
\n" ); document.write( "P(Z > k) = 0.90
\n" ); document.write( "and that in turn is the same as
\n" ); document.write( "P(Z < k) = 0.10\r
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\n" ); document.write( "\n" ); document.write( "I'll use this Z table
\n" ); document.write( "https://www.ztable.net/
\n" ); document.write( "to find that P(Z < -1.28) = 0.10027 which is the closest we can get to 0.10
\n" ); document.write( "So P(Z < -1.28) = 0.10 approximately
\n" ); document.write( "and
\n" ); document.write( "P(Z > -1.28) = 0.90 approximately\r
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\n" ); document.write( "\n" ); document.write( "The key takeaway is that z = -1.28 is the approximate z score. \r
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\n" ); document.write( "\n" ); document.write( "That z score is then used to solve for n like so
\n" ); document.write( "Z = (x-mu)/(sigma/sqrt(n))
\n" ); document.write( "-1.28 = (52-56)/(21/sqrt(n))
\n" ); document.write( "-1.28*(21/sqrt(n)) = -4
\n" ); document.write( "-26.88 = -4sqrt(n)
\n" ); document.write( "sqrt(n) = -26.88/(-4)
\n" ); document.write( "sqrt(n) = 6.72
\n" ); document.write( "n = (6.72)^2
\n" ); document.write( "n = 45.1584
\n" ); document.write( "n = 45\r
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\n" ); document.write( "\n" ); document.write( "Answer: 45
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