document.write( "Question 980292: 5. Historical production data indicate that the diameter of a ball bearing is normally distributed with a mean of 0.525 cm and a standard deviation of 0.008 cm. Suppose that a sample of 16 ball bearings are randomly selected from a very large lot. Determine the probability that the average diameter of a sampled ball bearing is greater than 0.530 cm.
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document.write( "a. 0.2324
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document.write( "b. 0.4938
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document.write( "c. 0.5062
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document.write( "d. 0.0062
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document.write( "
Algebra.Com's Answer #601463 by jim_thompson5910(35256) ![]() You can put this solution on YOUR website! z = (xbar - mu)/(sigma/sqrt(n))\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "z = (0.530 - 0.525)/(0.008/sqrt(16))\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "z = 2.5\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "-------------------------------------------------------\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "P(xbar > 0.530) = P(z > 2.5)\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "P(xbar > 0.530) = 1 - P(z < 2.5)\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "P(xbar > 0.530) = 1 - 0.99379 ... use a table\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "P(xbar > 0.530) = 0.00621\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "P(xbar > 0.530) = 0.0062\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "The probability is 0.0062 \n" ); document.write( " \n" ); document.write( " |