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August 16th, 2017, 06:20 AM   #1
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Obtaining lower and upper limit

Question: What are safe cholesterol levels? Measurements of cholesterol for males 20 to 29 years old are approximately Normally distributed with mean 4.6 mmol/L and standard deviation 0.93 mmol/L. In medical diagnosis, physicians often use reference limits for judging results as ‘normal’ or else requiring further investigation or treatment. These are typically the limits
within which we would expect to find 95% of healthy people. Obtain the lower and upper limit for that range in the case of cholesterol for males 20 to 29 years old. According to your range, at what cholesterol levels would we suspect a health problem?

My answer:
95% of values under curve lies within two standard deviations of mean.
Lower limit:
μ-(2 x σ)
=4.6 – (2 x 0.93)
=2.74 mmol/L
Upper limit:
μ+(2× σ)
4.6 + (2 x 0.93)
= 6.46 mmol/L
The lower limit is 2.74 mmol/L and the upper limit is 6.46 mmol/L for the range we expect to find 95% of healthy people.
According to this range, we'd suspect a health problem for cholesterol levels below 2.74 mmol/L or above 6.46 mmol/L.

Is that method of solving the problem correct? Thanks in advance
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August 16th, 2017, 10:25 AM   #2
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I get slightly different values:

Lower limit:
μ-(2 x σ)
=4.6 – (1.96 x 0.93)
=2.7772 mmol/L

Upper limit:
μ+(2× σ)
4.6 + (1.96 x 0.93)
= 6.4228 mmol/L
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August 16th, 2017, 11:11 AM   #3
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Ah, lim soup.
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August 16th, 2017, 02:52 PM   #4
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Hi, im confused as to why you changed the 2 to 1.96.

Lower limit:
μ-(2 x σ)
=4.6 – (1.96 x 0.93)
=2.7772 mmol/L
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August 16th, 2017, 08:27 PM   #5
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My solution for this question is based off of the image i just attached which basically says that 95% of values under curve lies within two standard deviations of mean.

This is why I wrote 2 x 0.93 throughout the solution. I'm not sure why we should use 1.96 x 0.93?

Lower limit:
μ-(2 x σ)
=4.6 – (2 x 0.93)
=2.74 mmol/L
Upper limit:
μ+(2× σ)
4.6 + (2 x 0.93)
= 6.46 mmol/L
Attached Images
File Type: jpg afdads.jpg (19.4 KB, 2 views)
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August 16th, 2017, 08:57 PM   #6
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The curve I referred suggested the value of 95% in between -1.96 to 1.96
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