document.write( "Question 1083275: A donut baker wants to ensure that the radius of its donuts is within +- 0.25cm. Based on past statistical studies, she knows that the donut’s radius has a standard deviation of 1.25cm. How many measurements should she take to ensure a 90 percent confidence in the radius of the donuts are within 0.25cm, rather than measure each and every donut that is baked? \n" ); document.write( "
Algebra.Com's Answer #697281 by jim_thompson5910(35256)\"\" \"About 
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Use a stats table to find that the critical value is roughly z = 1.645
\n" ); document.write( "To find this value, locate \"90%\" at the bottom (in the Confidence Level section) and the value just above this entry is 1.645\r
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\n" ); document.write( "\n" ); document.write( "The population standard deviation sigma is given to be sigma = 1.25\r
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\n" ); document.write( "\n" ); document.write( "The margin of error desired is 0.25, so E = 0.25\r
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\n" ); document.write( "\n" ); document.write( "We will plug z = 1.645, sigma = 1.25 and E = 0.25 into the formula below to find n.\r
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\n" ); document.write( "\n" ); document.write( "n = ( (z*sigma)/E )^2\r
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\n" ); document.write( "\n" ); document.write( "n = ( (1.645*1.25)/0.25 )^2\r
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\n" ); document.write( "\n" ); document.write( "n = ( 2.05625/0.25 )^2\r
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\n" ); document.write( "\n" ); document.write( "n = ( 8.225 )^2\r
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\n" ); document.write( "\n" ); document.write( "n = 67.650625\r
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\n" ); document.write( "\n" ); document.write( "n = 68 ... round up to the nearest whole number\r
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\n" ); document.write( "\n" ); document.write( "Min Sample Size: n = 68\r
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\n" ); document.write( "\n" ); document.write( "So this means that she can use a sample size of 68 or larger.\r
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\n" ); document.write( "\n" ); document.write( "She should take at least 68 measurements to ensure that the radius of the donuts is within +- 0.25cm.
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