Download Midterm, Version 2, Solutions

SOLUTIONS TO MIDTERM
VERSION 2
1) Let A={Employee smokes}, B={Employee listens to music}. We have
P(A)=.2, P(B)=.8 and P(A|B)=.1. We need P(A∩B)=P(B)P(A|B)=(.8)(.1)=.08.
Answer is B.
2) We have P(B|A)=P(A∩B) /P(A)=P({4})/P({1,2,4})=(1/6)/(3/6)=1/3. Answer
is D.
3) The events are not mutually exclusive, since A∩B=A, which is not empty.
The events are not independent, since P(B|A)=P(A∩B)/P(A)=P(A)/P(A)=1,
which is not the same as P(B)=1/2. The knowledge that a six was thrown
gives us information as to whether the outcome is an even number. The events
are not complementary, since the complement of B is {An odd number is
thrown}, and this is different from A. Since the events are not independent,
they are dependent. Answer is C.
4) Let X be the profit, in millions. of Dollars. We have
E(X)=(0)(.4)+(–1)(.1)+(1)(.5)=.4. Answer is C.
5) X is binomial with n=100, p=.Prob{At most 2}=Prob({1,2})=1/3. The
standard deviation of X is   npq  (100)(1 / 3)(2 / 3)  4.71. Answer is
A.
6) Converting to z-scores, we have Prob{X<5}=Prob{Z<(5–3)/2} = Prob{ Z <1}
= .5 +.3413 = .8413. Answer is C.
7) We need to find σ, the standard deviation of X. Since X has mean µ=0, it is
clear that P(X< –4) is an area in the left tail of a normal distribution.
Standardizing, we have .0228=P(Z< (–4–0)/σ}, where Z is standard normal.
We see from the normal table that Prob{Std Normal< –2}=.5–.4772=.0228.
Thus, we need –4/σ = –2, and therefore σ = 2. Answer is B.
8) Since P(B|A)=0 but P(B)>0, we have P(B|A)≠P(B). Therefore, A, B are
dependent. Answer is B. Note that A, B need not be complementary events.
For example, in tossing a die once, let A={2}, B={3}. Then we have P(A)>0,
P(B)>0 and P(B|A)=0, but A, B are not complementary events. Answer is B.
9) There are four equally likely possibilities for the outcomes: (0,0), (0,1), (1,0),
(1,1). Each has probability 1/4. If X is the larger of the two numbers,
P(X=0)=P{(0,0)}=1/4, and P(X=1)=P{(0,1),(1,0),(1,1)}=3/4. Therefore,
E[X]=0(1/4)+1(3/4)=3/4. Answer is D.
10) Let's first identify the answers that are incorrect. Since the population need not
be normal, there is no reason for the values in the sample to have a normal
distribution. Since the standard error of the mean is  x 

we see that the
n
standard error approaches zero, and not the population standard deviation σ.
Since  x   , the sample mean is always unbiased. Since its bias is always
zero, this bias doesn't get smaller as n increases. We have eliminated A), B),
and D). Based on the explanation of the Central Limit Theorem in the course
handouts, we see that Answer is C.
11) First, let's solve for x 2 . Note that p( x 2 )=1–p( x1 )=.25. We have
E[X]=0= x1 p( x1 )+ x 2 p( x 2 )=(–3)(.75)+ x 2 (.25), so x 2 = (3)(.75)/(.25)= 9.
Next, we need to get the standard deviation of X, which is
  (3  0) 2 (.75)  (9  0) 2 (.25)  27  5.196. Finally, the z-score
corresponding to x 2 is z=( x 2 –µ)/σ = 9/5.196 = 1.73. Answer is C.
12) First, calculate P(X<0)=P(Z<(0–1)/1)=P(Z< –1) = .5 – .3413 = .1587. Since
X, Y, Z are independent, the number of them that turn out to be positive has a
binomial distribution with n=3, p=.1587. Thus, q=1–.1587 =.8413. We need
 3
p(1)    (.1587)1 (.8413) 2  (3)(.1587)(.8413) 2  .3370. Answer is B.
1
13) Let Xi = Demand from i'th customer. E[Xi] =µ= .85, SD(Xi) = σ =.6, n=625.
Prob(Total Demand ≤ 500) = Prob (X1 + …+X625 ≤ 500)= Prob ( X ≤ .8). By
the Central Limit Theorem, X is normal with mean .85 and standard deviation

.6
=.6/25 =.024. Thus, Prob{ X < .8}=Prob{Z < (.8–.85)/  x }
x 

n
625
=Prob{ Z  2.08} =.5–.4812=.0188. Answer is B.
14) Since Y= –X, we have P(X>2 and Y< –2) = P(X>2 and –X < –2)=P(X>2 and
X>2) = P(X>2) = .5–.4772 = .0228. Answer is B.
15) Since n=1, the sample mean is just a single observation drawn at random from
the population. Therefore, the sampling distribution of the sample mean is the
same as the distribution of the population. This distribution need not be
normal, the standard error of the sample mean will be equal to (not less than)
the standard deviation of the population since  / n   / 1   , and the
standard error of the sample mean need not be zero. Answer is D.