document.write( "Question 1086483: A study of nickels showed that the standard deviation of the weight of nickels is 150 milligrams. A coin counter manufacturer wishes to find the 80% confidence interval for the average weight of a nickel. How many nickels does he need to weigh to obtain an average accurate to within 10 milligrams?
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\n" ); document.write( "a. 225
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\n" ); document.write( "b. 369
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\n" ); document.write( "c. 613
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\n" ); document.write( "d. 159
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Algebra.Com's Answer #700686 by jim_thompson5910(35256)\"\" \"About 
You can put this solution on YOUR website!
We will use this formula
\n" ); document.write( "n = ( (z*sigma)/E )^2\r
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\n" ); document.write( "\n" ); document.write( "The variables are
\n" ); document.write( "n = sample size
\n" ); document.write( "z = critical value
\n" ); document.write( "sigma = population standard deviation
\n" ); document.write( "E = margin of error
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\n" ); document.write( "Use a table or calculator to find that
\n" ); document.write( "Critical value = z = 1.28 (at 80% confidence)\r
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\n" ); document.write( "\n" ); document.write( "In this case, we're given
\n" ); document.write( "sigma = 150
\n" ); document.write( "E = 10\r
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\n" ); document.write( "\n" ); document.write( "Plug those values into the formula to get:
\n" ); document.write( "n = ( (z*sigma)/E )^2
\n" ); document.write( "n = ( (1.28*150)/10 )^2
\n" ); document.write( "n = ( 192/10 )^2
\n" ); document.write( "n = ( 19.2 )^2
\n" ); document.write( "n = 368.64
\n" ); document.write( "n = 369 ... round up to the nearest whole number\r
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\n" ); document.write( "\n" ); document.write( "Final answer is choice B) 369
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