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August 15th, 2017, 03:57 AM  #1 
Newbie Joined: Feb 2016 From: Australia Posts: 11 Thanks: 1  Normal distribution Question Vitamin D status is checked by measuring serum 25hydroxyvitamin D [25OHD] concentration. It is believed that the serum 25OHD levels in healthy adults roughly follow a Normal distribution. Part A) Suppose that a criterion for vitamin D deficiency is a serum 25OHD 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 25OHD 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. 
August 15th, 2017, 06:28 AM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,822 Thanks: 750 
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.

August 15th, 2017, 06:51 AM  #3 
Newbie Joined: Feb 2016 From: Australia Posts: 11 Thanks: 1 
(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 
August 15th, 2017, 10:25 PM  #4 
Newbie Joined: Feb 2016 From: Australia Posts: 11 Thanks: 1 
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 
August 16th, 2017, 12:55 AM  #5 
Member Joined: May 2015 From: Australia Posts: 59 Thanks: 6 
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 31.2=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 
August 16th, 2017, 05:16 AM  #6 
Newbie Joined: Feb 2016 From: Australia Posts: 11 Thanks: 1 
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 
August 16th, 2017, 06:29 AM  #7 
Math Team Joined: Jul 2013 From: काठमाडौं, नेपाल Posts: 872 Thanks: 60 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 25OHD level that is 1.2 standard deviations below the mean of healthy adults. $\displaystyle P(Z <1.23) = P (Z <4. 2) = 0.00133457\%$ 

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