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Suppose that blood chloride concentration (mmol/L) has a
Chapter 5, Problem 37E(choose chapter or problem)
Suppose that blood chloride concentration (mmol/L) has a normal distribution with mean 104 and standard deviation 5 (information in the article “Mathematical Model of Chloride Concentration in Human Blood,” ?J. of Med. Engr. and Tech?., 2006: 25–30, including a normal probability plot as described in Section 4.6, supports this assumption). a.? ?What is the probability that chloride concentration equals 105? Is less than 105? Is at most 105? b.? ?What is the probability that chloride concentration differs from the mean by more than 1 standard deviation? Does this probability depend on the values of and ? c.? ?How would you characterize the most extreme .1% of chloride concentration values?
Questions & Answers
QUESTION:
Suppose that blood chloride concentration (mmol/L) has a normal distribution with mean 104 and standard deviation 5 (information in the article “Mathematical Model of Chloride Concentration in Human Blood,” ?J. of Med. Engr. and Tech?., 2006: 25–30, including a normal probability plot as described in Section 4.6, supports this assumption). a.? ?What is the probability that chloride concentration equals 105? Is less than 105? Is at most 105? b.? ?What is the probability that chloride concentration differs from the mean by more than 1 standard deviation? Does this probability depend on the values of and ? c.? ?How would you characterize the most extreme .1% of chloride concentration values?
ANSWER:Answer : Step 1 : Given Suppose that blood chloride concentration (mmol/L) has a normal distribution with mean 104 and standard deviation 5. Here mean = 104 and standard deviation = 5. a). Now we have to find the probability that chloride concentration equals 105 Is less than 105 Is at most 105. x=105. The formula is given by 2 1 (x2) P(X = x) = 2 e 2 Substitute and . (105104) P(X = 105) = 1 e 2 (5) 2 (5) (1) 1 2 (25) P(X = 105) = 2 (5) e 2 (105104) P(X = 105) = 1 e (50) 2 (3.14) (5) (105104) P(X = 105) = 1 e (50) (6.28) (5) P(X = 105) = 0.078 Then we have to less than 105. So x50. The formula is given by, x P(Xx) = (z < ) We substitute the x, and values. P(X50) = (z < 1010404) P(X50) = (z < 0.2) Using area under normal curve. P(X50) = 0.5793 Therefore P(X50) = 0.5793. Now we have to find the P(X<50). We know that. P(X = 105) = 0.078and P(X50) = 0.5793. P(X<50)=P(X = 105) P(X 50) P(X<50)= 0.078-0.5793 P(X<50)= 0.5013 Therefore the P(X<50) = 0.5013