document.write( "Question 1197391: The line width of a tool used for semiconductor manufacturing is assumed to be normally distributed with a mean of 0.5 micrometer and a standard deviation of 0.05 micrometer.
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Algebra.Com's Answer #830646 by ewatrrr(24785)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "Hi  \r\n" );
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document.write( "Normal Distribution: µ = .5 and σ = .05\r\n" );
document.write( "Using TI or similarly an inexpensive calculator like an Casio fx-115 ES plus\r\n" );
document.write( "normalcdf(smaller value, larger value, µ, σ).\r\n" );
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document.write( "P(x >.62 ) = normalcdf( .62, 9999, .5, .05) = .0082\r\n" );
document.write( "P(.47 ≤ x ≤ .63) = normalcdf( .47, .63, .5, .05) = .7211\r\n" );
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document.write( "The line width of 90% of samples is below what value\r\n" );
document.write( "  z= Invnorm(.90) = 1.28\r\n" );
document.write( "\"blue+%28x%29+=+blue%28sigma%29%2Az+%2B+mu+\" =  .05(1.28) + .5  \r\n" );
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document.write( "Recommend Using stattrek.com to check  Results\r\n" );
document.write( "until You are familiar with Using Your Calculator.\r\n" );
document.write( "Trusting Your Use of Your calculator is key to success working with \r\n" );
document.write( "Distributions.\r\n" );
document.write( "Wish You the Best in your Studies.\r\n" );
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