Stat 427
NAME_______________________________________
Exam 2
Professor Berliner; WI 2011
Exam Score:
_______/100 points
Part I Score:
_______
/
40 points
Fillin the blanks:
Each question is worth 5 points (no partial credit)
For problems 1 through 4, let
Z
be a standard normal RV and find
1.
P(1.00 < Z < 2.25)
= _.
9878.8413=
.1465
_.
2.
P(1.50 < Z < 2.50) =
_.9938.0668=
.927
___.
3.
P(Z >  1.54) = ___
P(Z<1.54) = .9382
__.
4.
A value z such that P(Z < z)
= .65 is z =
___.385
____.
For problems 5 through 7, assume that
X
is a normal RV with mean 2.5 and variance 36. Find
5.
P(X > 5.3) =
P(Z > (5.32.5)/6)= P(Z > .467)
= .32
_
__.
6.
P(2.0<X<1.3
) =P((22.5)/6< Z < (1.32.5)/6)
= P(.75 < Z < .2)=
.1941__.
7.
A value x such that P(X 2.5 > x) = .72 is ___
2.16
_______.
P(X 2.5 < x) = .28
P(Z < x/6) = .28
P(Z < x/6) =.64
x/6 = .36, so x = 6(.36) =2.16
8.
Suppose X ~ N(0,
2
).
Then P(0 < X < 1.5
) is ___
.4332
_____.
P(0 < X< 1.5
) =P(0 < Z < 1.5) = .9331.5 = .4332
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Part II Score:
_______
/
60 points
9.
(10 points) TaySachs is a genetic disease. Assume that if both parents are carriers of
TaySachs, a child of theirs has probability 0.25 of being born with the disease.
a.
If such a couple has four children, what is the probability that exactly one of
them has the disease?
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 Spring '08
 Staff
 Normal Distribution, Probability, Probability theory, continuous random variable, GMAT scores, normal approx., exact hypergeometric answer

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