document.write( "Question 189250: A standard test for diabetes is based on glucose levels in the blood after fasting for a prescribed period. For untreated diabetics the mean fasting glucose is found to be 11.74mmol/L, and the standard deviation is 3.50mmol/L. The level of glucose appears to be approximately normally distributed.
\n" ); document.write( "To operate a simple diagnostic test based on fasting glucose levels, we need to set a cut off point, C, so that if the patients fasting glucose level is at least C we say they have diabetes. If it is lower we say they do not have diabetes. \r
\n" ); document.write( "\n" ); document.write( "(i) For the moment, suppose we use C=6.5. What is the probability that a diabetic is incorrectly diagnosed as not having diabetes? \r
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\n" ); document.write( "\n" ); document.write( "NOTE: I have attempted this question, but got stuck, because I'm not sure if what I'm doing is correct. I found that the P(6.5 < x < 11.74) = 0.4325, but don't know where to go from here...
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Algebra.Com's Answer #141978 by stanbon(75887)\"\" \"About 
You can put this solution on YOUR website!
A standard test for diabetes is based on glucose levels in the blood after fasting for a prescribed period. For untreated diabetics the mean fasting glucose is found to be 11.74mmol/L, and the standard deviation is 3.50mmol/L. The level of glucose appears to be approximately normally distributed.
\n" ); document.write( "To operate a simple diagnostic test based on fasting glucose levels, we need to set a cut off point, C, so that if the patients fasting glucose level is at least C we say they have diabetes. If it is lower we say they do not have diabetes.
\n" ); document.write( "(i) For the moment, suppose we use C=6.5. What is the probability that a diabetic is incorrectly diagnosed as not having diabetes?
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\n" ); document.write( "Those with glucose above 6.5 will be diagnosed \"diabetic\"
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\n" ); document.write( "Those below 6.5 will be diagnosed \"non-diabetic\"
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\n" ); document.write( "Find the z-value of 6.5.
\n" ); document.write( "z(6.5) = (6.5-11.74)/3.5 = -1.4971
\n" ); document.write( "P(a person is diagnosed non-diabetic) = P(z < -1.4971) = 0.0672
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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