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SOLUTIONS TO MIDTERM 2
VERSION 1
1) From the normal table, Prob{ 2 < Z < 2.5 }=.4938–.4772=.0166. Answer is B.
2) Converting to z-scores, we have Prob{–1.5 < X< –0.2}=Prob{(–1.5+1)/0.5 <
Z<(–0.2+1)/0.5} = Prob{–1 < Z < 1.6} = .3413 + .4452 = .7865. Answer is D.
3) From the course handouts, we know that the 95’th percentile of a standard
normal is 1.645. We set 1.645 = (x–20)/10, so x=(10)(1.645)+20 = 36.45.
Answer is A.
4) The expected profit is (500)(.4) –(300)(.6) = 20. Answer is C.
5) We need an arrangement of 5 things taken 3 at a time. By the arrangement
rule, the answer is (5)(4)(3)=60. Answer is B.
6) Use normal approximation to binomial. Let X = #Pepsi drinkers. X is
binomial with n=100, p=.45. X is approximately normal with mean np = 45
and standard deviation (100)(.45)(.55)  4.975. By the normal
approximation to the binomial, Prob{X40} = Prob{Normal  39.5}=Prob{ Z
(39.5 – 45) / 4.975 } = Prob {Z  –1.11 }=.5 + .3665 =.8665. Answer is B.
7) For a single spin, the expected outcome is =(1/3)(1)+(1/3)(2)+(1/3)(3)=2,
and the standard deviation is
  (1 / 3)(1  2) 2  (1 / 3)( 2  2) 2  (1 / 3)(3  2) 2  2 / 3  .8165 . The
average for n=100 spins has a mean of  x    2 and a standard error of

= .8165/10 = .08165. Thus, applying the CLT,
x 
n
Prob{ x  1.8}=Prob{Z  –2.45} = .5+.4929 = .9929. Answer is B.
15 
8) There are   =(15)(14)(13)/6 = 455 possible samples of the 15 partners.
3
The number of samples consisting of two women (and one man) is
 5  10 
    =100. The number of samples consisting of three women is
 2  1 
 5
  =(5)(4)(3)/6=10. So the probability of getting more women than men in
 3
the sample is the probability that the sample contains two women plus the
probability that the sample contains three women. This probability is therefore
(100+10)/455 = .2418. Note that you cannot use the binomial distribution here
since the population changes as you do the sampling, so the trials are not
independent. Answer is B.
9) Call the offices A, B, C and D. We need to assign employees to the offices.
10 
Use the multiplication rule. There are    120 ways to assign 3 of the 10
3
employees to Office A. (The order does not matter here, so we use the formula
 7
for combinations). There are    21 ways to assign two of the remaining
 2
 5
1
employees to Office B. Then    5 ways for Office C and    1 way for
 4
1
Office D. The total number of ways to make the assignment is the product,
(120)(21)(5)(1) = 12600. Answer is A.
10) To get three aces, we need three of the 4 aces and 2 of the 48 non-aces. The
 4   48   52 
probability of getting exactly three aces therefore is     /  
 3  2   5 
=(4)(1128)/2598960= .001736. The expected winnings is therefore
(100)(.001736) –(1)(1–.001736)= –.82. Answer is B.
11) Using the CLT, we can say that the sample mean has a normal distribution
with mean 2 and standard deviation  x   n  1 / 100  0.1. Thus,
Prob{ x 2.2}=Prob{Z  (2.2–2)/0.1} = Prob{Z  2} = .5 – .4772 = .0228.
Answer is B.
12) Let X be the number of series of the next three won by the Yankees. Then X
is binomial with n=3 and p=.62. We need is p(2)+p(3)=
 3
 3
  (.62) 2 (1  .62)1   ( .62) 3  .4382  .2383  .6765 . Answer is B.
 2
 3
13) We know that np=15 and npq=3.75. Thus, q=3.75/15 = .25 and p=1 – q =
.75. Now, we find that n=15/.75 = 20. So X is binomial with n=20 and p=.75.
The probability that X is equal to 15 is
 20 
 20 
  (.75)15 (.25) 5    (.75)15 (.25) 5  .2023. Answer is B.
 15 
5
14) We need  x 

n
= 5/5 = 1. Answer is A.
15) Since x 2 has a smaller standard error than x1 , and since both are (essentially)
normally distributed with a mean equal to the population mean, x 2 is more
likely than x1 to be within one unit of the population mean. But we cannot say
with certainty which will be closer to the population mean. Answer is D.