Download Chapters 4 and 5 HW Solutions

 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).