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Chapters 4 and 5 SOLUTIONS 1. A certain assay for serum alanine aminotransferase (ALT) is rather imprecise. The results of repeated assays of a single specimen follow a normal distribution with mean equal to the ALT concentration for that specimen and standard deviation equal to 4 U/Li (as in Exercise 4.40). Suppose a hospital lab measures many specimens every day, and specimens with reported ALT values of 40 or more are flagged as “unusually high.” If a patient's true ALT concentration is 35 U/Li find the probability that his specimen will be flagged as “unusually high” a. if the reported value is the result of a single assay The result of a single assay is like a random observation Y from the population of assays. A value Y ≥ 40 will be flagged as “unusually high”. pnorm(40,35,4,lower.tail=FALSE) = 0.1056 –OR‐ 1 – pnorm(40,35,4) = 0.1056 Thus, Pr{specimen will be flagged as “unusually high”} = 0.1056. b. Find the 99th percentile of this distribution. qnorm(.99, 35, 4) = 44.31 U/Li c. if the reported value is the mean of three independent assays of the same specimen The reported value is the mean of three independent assays, which is like the mean Y of a sample of size n = 3 from the population of assays. A value y ≥ 40 will be flagged as “unusually high.” We are concerned with the sampling distribution of y for a sample of size n = 3 from a population with mean μY = 35 and standard deviation σY = 4. From Theorem in “Sampling Distributions” lecture notes), the mean of the sampling distribution of y is μY = 35, the standard deviation is σY = = 2.309, and the shape of the distribution is normal because the √ population distribution of Y is normal. pnorm(40,35,4/sqrt(3),lower.tail=FALSE) –OR‐ 1 – pnorm(40,35, 4/sqrt(3)) = 0.0152 2. The heights of men in a certain population follow a normal distribution with mean 69.7 inches and standard deviation 2.8 inches. a. If a man is chosen at random from the population, find the probability that he will be more than 72 inches tall. pnorm(72,69.7,2.8,lower.tail=FALSE) = 0.2057 b. Find the 95th percentile of this distribution. qnorm(.95, 69.7, 2.8) = 74.306 inches c. If two men are chosen at random from the population, find the probability that their mean height will be more than 72 inches. pnorm(72,69.7, 2.8/sqrt(2),lower.tail=FALSE) = 0.1227 d. Iff two men are chosen n at random from the population, find th he probability that bo oth of them m will be m more than 72 inches tall. Usin ng the bino omial distribution, Pr{b both are > 72} = dbinom(2,2,0.2 2057) = 0.0 0423 ‐or‐ Thin nk of the probability rule on ind dependentt events: If two even nts are indeependent (say A and d B), tthen the probability tthey both occur is th he probabi lity of one times the probabilitty of the otheer (P(A and d B) = P(A)P(B) when A and B are indepenndent). So, Pr{both are > 72} = 0 0.20572 = 0 0.0423 e is measurred by counting emisssions from m a radioacctively 3. The activityy of a certain enzyme ue specime en, the couunts in consecutive 10‐second ttime labeeled moleccule. For a given tissu periiods may b be regarded (approximately) ass repeated independent observvations from a norm mal distrib bution (as in Exercise 4.26). Sup ppose the m mean 10‐ssecond cou unt for a ceertain tissu ue specime en is 1,200 0 and the standard de eviation is 35. For that specimeen, let Y reepresent a 10‐ssecond cou unt and lett Y represe ent the mean of six 110‐second ccounts. Find P{11 175 ≤ Y ≤ 1 1225}and P P{1175 ≤ Y ≤ 1225}an nd comparee the two. Does the compariso on indicate thatt counting for one minute and dividing byy 6 would ttend to givve a more precise ressult than merrely counting for a sin ngle 10‐seccond time period? How? Y == 1,200; σY = 35 Pr{1 1175 ≤ Y ≤ 1225} = pn norm(1225 5, 1200, 35 5) ‐ pnorm((1175, 12000, 35) = 0.525 Pr{1 1175 ≤ Y ≤ 1225} = pn norm(1225 5, 1200, 35 5/sqrt(6)) ‐ pnorm(11175, 1200, 35/sqrt(6))) = 0.9198 8 COM MPARISON N 0.91 189 > 0.525 5. This sho ows that th he mean off 6 counts iis more precise, in th hat it is mo ore likely to b be near the e correct vaalue (1200 0) than is a single couunt. 4. Find the following crittical pointss. 2 a. Z..10 qnorm((.9) = 1.282 b. Z.025 qnorm(.975) = 1.96 n a study o of the fruitffly Drosoph hila melangoster, a 5. In singgle experim mental fly w was observved for thre ee minutes while e in a cham mber with tten other flies of the me sex. The e observer recorded tthe timing of each sam epissode ("bou ut") of pree ening by th he experim mental fly. The experiment was rep plicated 15 times with h male mes with ffemale fliess (differentt flies fliess and 15 tim each h time). Th he research her would like to use e a stattistical testt to evaluatte the pree ening time es for the males that rrequires th he assumpttion the daata come fem from m a normal populatio on. Use the QQplo ot to assesss normalitty. e definitelyy making a systematic departurre from thee line. They are makiing a "U" The points are pe. The po oints are to oo high at tthe high an nd low endds of the diistribution, indicatingg a skew shap right. This is a clear violaation of the e normality assumpttion. 6. 4.45(modifiied). The fo ollowing th hree normal probability plots, ((a), (b), and (c), weree histogramss I, II, and IIII. Which n normal pro obability generated from the distributions sshown by h h which hisstogram? H How do yo ou know? plott goes with Corrrect: Histtogram II iss skewed to the rightt, so it goess with plott (c). Histoggram I lookks normal,, so it goes with h plot (b). H Histogram III has long tails, so iit goes witth plot (a).