document.write( "Question 1206804: A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 110.3 cm and a standard deviation of 1.7 cm. For shipment, 7 steel rods are bundled together.\r
\n" ); document.write( "\n" ); document.write( "Find the probability that the average length of a randomly selected bundle of steel rods is more than 112 cm.
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Algebra.Com's Answer #844470 by MathLover1(20850)\"\" \"About 
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\n" ); document.write( " mean: \"mu=110.3cm\"\r
\n" ); document.write( "\n" ); document.write( "\"n=7\" steel rods are bundled together\r
\n" ); document.write( "\n" ); document.write( "a standard deviation: \"sigma=1.7cm\"\r
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\n" ); document.write( "\n" ); document.write( "find the standard error:\r
\n" ); document.write( "\n" ); document.write( "\"s+=+sigma%2Fsqrt%28n+%29=+1.7%2Fsqrt%287%29+%E2%89%88+0.6425\"\r
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\n" ); document.write( "\n" ); document.write( "now find the \"z\"-score using the sample standard deviation to standardize the normal distribution:\r
\n" ); document.write( "\n" ); document.write( "\"z+=+%28mu-sigma%29%2Fs+=%28112-110.3%29%2F0.6425=2.6459\"\r
\n" ); document.write( "\n" ); document.write( "Using the standard normal distribution table or calculator, we can find the probability that the average length of a randomly selected bundle of steel rods is more than \"112cm\"
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\n" ); document.write( "\"P%28M%3E112%29+=+P%28Z%3E2.6459%29=0.00407369\"\"0.0041\"\r
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