Normal Glucose Mg Dl
What is the distribution and probability?
In pharmalogical research a variety of clinical chemistry measurements are routinely monitored closely for evidence of side effects of the medication under study.
Supposed typical blood-glucose levels are normally distributed with mean 90mg/dL and standard deviation 38 mg/dL. Suppose that 100 normal (i.e typical , healthy) individuals are tested.
a) Specify the distribution of the number of the 100 individuals with blood-glucose levels greater than 120 mg/dL
b) Find the probability that at least 30 of the 100 individuals ahve blood-glucose levels greater than 120 mg/dL using normal approximation.
And also, just for interest, are there any assumptions made for these calculations??
Thanks.!
Using the z-score convert this from a Normal Distribution problem to a Standard Normal Distribution problem.
z = ( x – mu ) / sigma.
For this problem mu = 90 and sigma = 38
So for (a) z = (120-90)/38 = 0.789 ~ 0.79
So you need to find the probability for a Standard Normal Distribution (SND) that P(z >= 0.79) = 0.2148 (I am assuming that you know how to find the probability for a SND.
So the distribution of the number of the 100 individuals with blood-glucose levels greater than 120 mg/dL would be 100* 0.2148 = 21.48 = 21.
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