Stat 427
NAME_______________________________________
Exam 2
Professor Berliner; WI 2011
Exam Score: _______/100 points
Part I Score: _______/40 points
Fill-in 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 > (5.3-2.5)/6)= P(Z > .467) = .32 ___.
6. P(-2.0<X<1.3) =P((-2-2.5)/6< Z < (1.3-2.5)/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
2
8. Suppose X ~ N(0,  ). Then P(0 < X < 1.5 ) is ___ .4332 _____.
P(0 < X< 1.5 ) =P(0 < Z < 1.5) = .9331-.5 = .4332
Part II Score: _______/60 points
9. (10 points) Tay-Sachs is a genetic disease. Assume that if both parents are carriers of
Tay-Sachs, 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?
P(Exactly 1) = 4(.25)(.75)3 = .422
b. Describe assumptions you made in part a.
Binomial assumptions: especially key in this case is independence
10. (15 points) Suppose X is a continuous random variable with pdf
f ( x) 
x 1
, 2 x4
8
a. Find P(3 < X < 4.2)
x 1
dx
3
8
 (1 / 8) [.5 x 2  x |34  (1 / 16)(42  2(4)  32  2(3))  9 / 16  .5625
P(3  X  4.2)  P(3  X  4)  

4

b. Find E(X)
x 1
dx
2
8
 (1 / 8) [ x3 / 3  x 2 / 2 |42  (1 / 48)(2(43 )  3(42 )  2(23 )  3(22 ))
4
E( X )   x


 (4 / 48)(32  12  4  3)  3.083
11. (20) Most graduate schools of business require applicants take the Graduate
Management Admission Council’s GMAT exam. Assume that GMAT scores are
normally distributed with mean 527 and standard deviation 112.
a. What percent of test takers score at least 650?
P(X > 650) = P(Z > (650-527)/112) =P(Z > 1.10) = .1357
b. Suppose that the OSU Graduate School of Business receives 500 applications in
a given year. What is the probability that at least 70 of the applications are from
people with GMAT scores of at least 650?
Let Y = # App’s with score > 650. Assuming scores among applications
are indep (reasonable), Y ~ Bin(n=500, p = .1357). Find P(Y > 70).
np = 67.85 so normal approx. is fine:
P(Y>70) ~ P(Z > (70 – 67.85)/(67.85*.8643)0.50) = P(Z > .281) = .3897 ~ .4
With cont. correction
P(Y>70) ~ P(Z > (70 – 67.85 - .5)/(67.85*.8643)0.50) = P(Z > .215) ~ .415
(exact bin prob = .40884)
(15) Suppose that a tree farm contains 25,000 trees, 500 of which have a disease. A random
sample of 200 different trees is collected. Find the probability that two or more of the
sampled trees have the disease. (Explain your steps).
Let X be the number of diseased trees in the sample. The appropriate
distribution is the hypergeometric with N = 25,000, M = 500, n = 200, but the
numbers are not managed easily, so
• Binomial approx. to the hypergeometric. This is reasonable since N is very
large and n/N is small. Let Y ~ Bin(n = 200, p = 500/25000 = .02):
P(X > 2) ~ P(Y > 2) = 1 − P(Y = 0 or 1) =
1 – [(.02)0(.98)200 + 200 (.02)1(.98)199] = .91
OR
• Since n is large and p is very small (i.e., normal approx. isn’t OK), use
Poisson approx to the binomial. Let W ~ Poi(  = 200(.02) = 4). Then
P(X > 2) ~ P(Y > 2) ~ P(W > 2) = 1 − P(W = 0 or 1)
= 1 − (e−4 + 4e−4) = .91.
(the exact hypergeometric answer is also .91 to two decimals)