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Hi. I am taking a course in college level statistics. I have a problem with an assignment in the "Sampling Distribution" chapter. It reads:

* Blood potassium levels**. Judy's doctor is concerned that she may have hypokalemia (low potassium in her blood, measured in millimoles per liter, mmol/l). There is variation both in the actual potassium level and in the blood test that measures the level. Judy's measured potassium level varies according to the Normal distribution with mean (μ) = 3.8 mmol/l and standard deviation (σ) = 0.2 mmol/l. A patient is classified as a hypokalemic if the potassium level is less than 3.5 mmol/l.*

*a) If a single potassium measurement is made, what is the probability that Judy is diagnosed as hypokalemic?*

*b) If measurement ar made instead on 4 separate days and the mean result is compared with the citerion of 3.4 mmol/l, what is the probability that Judy is diagnosed as hypokalemic?*

My problem:

Solving a) went well. Since it says that the distribution is normal. I calculated the confidence based on the normal distribution. We are looking for a observation at 3.5 mmol/l or lower. So 3.5 - 3.8 mmol/l that is the mean is -0.3. Then we divide it by the standard deviation 0.2 and gets a z-score of 1.5, checking the probability table in my book it gives a value of .0668. the facet to the assigment back in the book says the value is 0.067. That is really close so I dont know if I calculated wrong, or they just adjusted the desimals.

b) Here I am a little lost. I tried to devide the probalitity of 0.668 or 0.667 by number of trials (n). That my book says is the formula for finding the probability of different independent triales, that in this case is 4. I get a score of 0.01675, but the facet says 0.0013. I am really confused and tried all the formulas in the chapter. Even the square root of 1-p devided by 4, but that gives a really farfetched number. I would really like to find the formula and undterstand why this is the correct formula to use in this exact problem.

Thank you for your help!

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