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August 15th, 2017, 02:57 AM   #1
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Normal distribution

Question
Vitamin D status is checked by measuring serum 25-hydroxyvitamin D [25-OHD] concentration. It is believed that the serum 25-OHD levels in healthy adults roughly follow a Normal distribution.

Part A) Suppose that a criterion for vitamin D deficiency is a serum 25-OHD level that is 1.2 standard deviations below the mean of healthy adults. What percent of healthy adults have vitamin D deficiency according to this criterion? For full marks, include an appropriate probability statement

Part B) Lifeguards are not ‘typical’ healthy adults since they experience much higher sun exposure. As a result, the mean serum 25-OHD for lifeguards is about +3 on the standard scale for healthy adults. Suppose that the standard deviation is the same as for healthy adults. What percent of lifeguards have vitamin D deficiency? For full marks, include an appropriate probability statement

-----------------------------------------
My attempted solution for part A:
I think we need to use the standard normal curve, where X~N(0,1)
Probability statement: We are interested in P(Z<-1.2)
I looked at the negative z score table and the value i obtained was 0.1151
Thus, 11.51% of healthy adults have vitamin deficiency.

My attempted solution for part B:
I dont really understand how to work out part b, but I tried to draw out a diagram of what i understand from the question. I attached the diagram, hopefully it is clear and readable.
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August 15th, 2017, 05:28 AM   #2
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Taking the "mean" for "normal" adults to be 0, part (b) says that the mean for life guards is +3. Translating that to the standard, z= 3- 1.2= 1.8 standard deviations. Do the same as for (a) but use 1.8 standard deviations instead of -1.2.
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August 15th, 2017, 05:51 AM   #3
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(b) Vitamin deficiency is 1.2 standard deviations below mean for healthy adults. Therefore, z = 3 - 1.2 = 1.8 standard deviations
P(Z<1.8 )
Using the positive z score table, P(Z<1.8 ) = 0.9641
Therefore, 96.41% of lifeguards have vitamin D deficiency.

I think i did something wrong because 96.41% seems too high because i expected less lifeguards with vitamin d deficiency
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August 15th, 2017, 09:25 PM   #4
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Hello again,
I've been thinking about this question and i think for part b, the distribution is X~N(3,1)

I attached a picture of how i think the solution is supposed to be. I'm not sure it is right though. The 'blue' drawings are for part a and the pink drawings are for part b.

Hopefully the pictures are clear and readable.

If I did it wrong, please tell me so we can find arrive at the right solution
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August 15th, 2017, 11:55 PM   #5
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The distribution for part (b) would be X~N(3,1)

The question says 1.2 standard deviations below the mean of 'healthy adults'.
P(X<-1.2)
using Minitab (statistical software), P(X<-1.2) = 0.0000133 or 0.00133%
This means likelihood of lifeguards having vitamin D deficiency is 0.00133%.

If you use 3-1.2=1.8,
P(X<1.8 ) = 0.115070 or 11.5%
This means likelihood of lifeguards having vitamin D deficiency is 11.5% which is probably incorrect because this is the same answer we got in part a.

If this is incorrect, then sorry
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August 16th, 2017, 04:16 AM   #6
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Thanks pianist.
I'm still not sure which one is the correct solution.
Could someone help me out or just confirm which one is the right answer? Thanks
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August 16th, 2017, 05:29 AM   #7
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Math Focus: सामान्य गणित
Part I:

$\displaystyle P (Z <1.2) = 11.50696702\%$


Part II:

The standard curve for lifeguards shifts 3 units right with respect to the 'typical' healthy, while the criterion for vitamin D deficiency is still a serum 25-OHD level that is 1.2 standard deviations below the mean of healthy adults.

$\displaystyle P(Z <-1.2-3) = P (Z <-4. 2) = 0.00133457\%$
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