18.2
a. I would expect that this histogram would be unimodal and symmetric centered
around .1 or 10%.
b. no, this could not be approximated by a normal model because it fails the
success/failure condition there are not 10 expected negative outcomes
c. the center should be at .1
d. sqrt(.1*.9/50
18.4
a. it is appropraiate to use a normal model to describe the distribution of green
MandM’s because drawing a green MandM is independent of drawing a different color,,
the number of bags of MandMs taken are going to be less than 10% of the population of
bags because there is no way that they could sample 10% of the total bags of MandM’s in
the world and it passes the success/failure condition, the expected outcome of failures and
success is more than 10% of the population p*200 = 20, q*200 = 180.
b. sd = .0212, 68, 2.12.% lie within 12.127.88% green, 95% lie within 14.24%
5.76% green, and 99.7 lie within 14.36%3.64% green.
c. the larger model would have a smaller sd, and thus the range of percentages
within each percentile in the normal model would be smaller. The distribution of greens
would be closer to .1.
18.6
a. sd (P) = sqrt((.1*.9)/500)
= .0134 = 13.4%. z= .12.1/.0134 = 1.49 P(z>1.49)
=
.93141 = 0686. this is a very unusual amount of greens as the probability of getting
that many is .0686
19.22.
a.
sqrt(.2367*.7633)/207 = .029 *1.65sds that contain 90% = .04785+.2609 = .
3088
.2609.04785 = .21305.
b.
I am 90% confident that between 30.88% and 21.31% of 40 year old women
who were previously unable to conceive had live births at this san diego
clinic.
c.
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 Spring '07
 RichardD.DeVeaux
 Normal Distribution

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