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Here is the whole problem. Assume body temperatures are normally distributed with mean of 98.2 and a standard dev. of 0.62 Part A) Assuming a hospital defines a fever as 100 or over, determine the percentage of a normal healthy person considered to have a fever using Z table Answer: 100-98.2/.62=2.90 =.19% rounded. Solution. Denote the body temperature by X. Then by hypothesis, X follows normal distribution with mean of 98.2 and a standard dev. of 0.62, i.e., X ~ N (98.2,0.62 2 ) . Now we need to find the probability .2 10098.2 X 98.2 Pr( X 100) Pr( X0.98 62 0.62 ) Pr( 0.62 2.9032) .2 Note 1: If X ~ N (98.2,0.62 2 ) , then X0.98 62 ~ N (0,1) Using a website at http://people.hofstra.edu/faculty/Stefan_Waner/RealWorld/stats/normaldist.html or looking up Z table, we get .2 10098.2 X 98.2 Pr( X 100) Pr( X0.98 62 0.62 ) Pr( 0.62 2.9032) 0.00185 0.19% Note 2: Using the above website, you can see a window which is a probability calculator. You only need to input 2.9032 in the lower bound of X and nothing in the upper bound, then click ”calculate probability”, you will get 0.001847 Part B) I have the procedure right but I can't get the right answer. z=x-u/o Determine the minimum fever if doctors at the hospital want only 3.0% of healthy persons to be diagnosed as having a fever. Use closest z score. Answer is 99.37. Solution. By part A), we know that if a hospital defines a fever as 100 or over, then the percentage of a normal healthy person considered to have a fever is 0.19% Now we need to re-define a fever, for instance, as x 0 or over. We need to find this x 0 . How to find this x 0 ? We know that if we use this new definition, then the percentage of a normal healthy person considered to have a fever is 3%, namely, we have x0 98.2 .2 Pr( X x0 ) Pr( X0.98 62 0.62 ) 3% By looking up Z table or using these two websites http://duke.usask.ca/~rbaker/Tables.html and http://people.hofstra.edu/faculty/Stefan_Waner/RealWorld/stats/normaldist.html, we can get Pr( Z 1.88) 3% (*) where Z~N(0,1). .2 We know that if X ~ N (98.2,0.62 2 ) , then X0.98 62 ~ N (0,1) . So, in order to find x 0 such that x0 98.2 .2 (**) Pr( X x0 ) Pr( X0.98 62 0.62 ) 3% By the above fact (*) and note that (**), we get x0 98.2 0.62 1.88 So, x0 98.2 1.88 * 0.62 99.3656 99.37 Another part B, different problem. How does 100.1-98.2/.62=.11 in part A and part B = 99.55 wanting 1.5% of healthy to be diagnosed? Solution. (I) Denote the body temperature by X. Then by hypothesis, X follows normal distribution with mean of 98.2 and a standard dev. of 0.62, i.e., X ~ N (98.2,0.62 2 ) . Now we need to find the probability .2 100.198.2 X 98.2 Pr( X 100.1) Pr( X0.98 62 0.62 ) Pr( 0.62 3.0645) .2 Note 1: If X ~ N (98.2,0.62 2 ) , then X0.98 62 ~ N (0,1) Using a website at http://people.hofstra.edu/faculty/Stefan_Waner/RealWorld/stats/normaldist.html or looking up Z table, we get .2 100.198.2 X 98.2 Pr( X 100.1) Pr( X0.98 62 0.62 ) Pr( 0.62 3.0645) 0.00109 0.11% Note 2: Using the above website, you can see a window which is a probability calculator. You only need to input 3.0645 in the lower bound of X and nothing in the upper bound, then click ”calculate probability”, you will get 0.00109 x0 98.2 .2 (II) We want to find x 0 such that Pr( X x0 ) Pr( X0.98 . 62 0.62 ) 1.5% By looking up Z table or using these two websites http://duke.usask.ca/~rbaker/Tables.html and http://people.hofstra.edu/faculty/Stefan_Waner/RealWorld/stats/normaldist.html, we can get Pr( Z 2.17) 1.5% where Z~N(0,1). So, we get x0 98.2 0.62 1.88 So, x0 98.2 2.17 * 0.62 99.545 99.55 A different B: 99.2-98.2/.62 =5.37 for part A, and B wanting 2.5% = 99.42.?? I have been taking 1-the %s (3.0%, 2.5% et cetera as above) but I'm still not getting the right answers. Solution. (I) Denote the body temperature by X. Then by hypothesis, X follows normal distribution with mean of 98.2 and a standard dev. of 0.62, i.e., X ~ N (98.2,0.62 2 ) . Now we need to find the probability .2 99.298.2 X 98.2 Pr( X 99.2) Pr( X0.98 62 0.62 ) Pr( 0.62 1.61) .2 Note 1: If X ~ N (98.2,0.62 2 ) , then X0.98 62 ~ N (0,1) Using a website at http://people.hofstra.edu/faculty/Stefan_Waner/RealWorld/stats/normaldist.html or looking up Z table, we get .2 99.298.2 X 98.2 Pr( X 99.2) Pr( X0.98 62 0.62 ) Pr( 0.62 1.61) 0.053699 5.37% Note 2: Using the above website, you can see a window which is a probability calculator. You only need to input 1.61 in the lower bound of X and nothing in the upper bound, then click ”calculate probability”, you will get 0.053699 x0 98.2 .2 (II) We want to find x 0 such that Pr( X x0 ) Pr( X0.98 . 62 0.62 ) 2.5% By looking up Z table or using these two websites http://duke.usask.ca/~rbaker/Tables.html and http://people.hofstra.edu/faculty/Stefan_Waner/RealWorld/stats/normaldist.html, we can get Pr( Z 1.96) 2.5% where Z~N(0,1). So, we get x0 98.2 0.62 1.96 So, x0 98.2 1.96 * 0.62 99.4152 99.42 Note: This website http://duke.usask.ca/~rbaker/Tables.html can tell you the critical value of many distributions. For instance, if Z~N(0,1) standard normal distribution, then P( Z Z ) , we call Z -critical value. If we choose 2.5% 0.025 , then you can use this website by choosing the critical values for z-test. Then choose 2.5% 0.025 , we get Z 1.96 Hope this helps. If you have any questions, please let me know.