Mathematical Statistics for Econometrics and Business - Springer

138 Pages·2013·1.1 MB·English

by  Ron Mittelhammer

Student Answer Key
Odd Questions Only
All rights reserved ©
For information contact Ron C. Mittelhammer, WSU, [email protected]
For corrections, please contact Danielle Engelhardt [email protected]
Chapter 1 – Student Answer Key – Odd Numbers Only
Elements of Probability Theory

1. a) {0, 1, 2, ..., 100}

b) {0, 1, 2, ..., N}, N = number of ads circulated

c) x,y:x y, x and y459.67, 212 (while perhaps conservative, the temperature
range has been designated to be between absolute zero and the boiling point).

d) x:0 x c, where c is delivery capacity of the jobber in a given week.

e) {w: w = $x.yz, x is a nonnegative integer, y and z ϵ{0, 1, 2,..., 9}}.

f) {x: x is a nonnegative integer}

3. a) A = {Karen, Wendy, Brenda}
P(A) = 3/8.

b) A = {Tom, Richard}
P(A) = 1/4.

c) A = {Karen, Frank, Scott}
P(A) = 3/8.

d) A = {Frank, Scott}
P(A) = 1/4.

e) A = {∅}
P(A) = 0.

f) A = {Tom, Karen, Frank, Eric, Wendy, Brenda, Scott, Richard}
P(A) = 1.

5. Let S x ,x ,x :x ,x ,x are three jokes chosen randomly from an inventory of 12
1 2 3 1 2 3
jokes}.

S is a finite sample space, has equally likely outcomes, and

12!
N(S)   220,
3!9!
Chapter 1 – Student Answer Key – Odd Numbers Only 2
represents the different combination of 12 jokes taken 3 at a time.

a) We need to count the number of ways in which 2 or all three jokes can be different from the
previous month’s three jokes, say x0,x0,x0.
1 2 3

9
Let A = 2 jokes are different. Then N(A)  3   108.
2

9
Let B = all three are different. Then N(B)   84.
 
3
N(AB) 192
So, PAB  .8727.
220 220

N(B) 84
b) PB   .3818.
220 220

7. NOTE: Each staff member has violated the axioms and theorems of probability, and there is
more than one way to characterize the violation. In each case we identify one such
characterization.

PA A   PA  or PA 
Tom: 1 2 1 2 .
.9  .5 or .3

PA A A   1
Dick: 1 2 3 .
1.5  1

PA  0
Harry: 3 .
.4 0

P(A  A) .3
Sally: PA |A 3 1  1.51.
3 1 P(A) .2
1

9. .05(1000) (cid:3404) 50 chips will be inspected and there are 5 detectives in total.

Let A  {event of picking a nondefective chip on ith draw}
i
Elements of Probability Theory Mathematical Statistics for Economics and Business
Chapter 1 – Student Answer Key – Odd Numbers Only 3
50  i1 
P(accepting shipment)  P(A) P(A | A)  PA | A ,
1 2 1 i j
 j1 
i3
995 994 946 995! 1000! 950! 1000!
 • • ...•            ,
1000 999 951 945!  950!  945!  995! 
 .7734.

11. To show Axioms 1-3 are satisfied: PA|B0 A since PAB0, PB0, and
B
thus PAB/ PB0.

PB|B PBB/ PB PB/ PB1.

Let A,iI be disjoint subsets of B. Then
i
    
P A  B P A B
PA |B iI i  iI i
iI i P(B) P(B)

 PA B
 iI i  PA |B.
P(B) i
iI

since A B,iI, are disjoint events.
i

13. Let D (cid:3404) {secretary dissatisfied}
A (cid:3404) {dislikes supervisor}
1
A (cid:3404) {not paid enough}
2
A (cid:3404) {dislikes type of work}
3
A (cid:3404) {conflict with other employees}
4

Note: A  D, A A  for i  j, and 4 A  D, so the A 's form a partition of D.
i i j i1 i i

Let Q D be the event that a dissatisfied secretary quits her job.

We are given the following:

P(D) (cid:3404) .20 P(Q|A ) (cid:3404) .20
1
P(A |D) (cid:3404) .55 P(Q|A ) (cid:3404) .30
1 2
P(A |D) (cid:3404) .30 P(Q|A ) (cid:3404) .90
2 3
P(A |D) (cid:3404) .10 P(Q|A ) (cid:3404) .05
3 4
P(A |D) (cid:3404) .05
4

a) Need to determine the PA |Q values; therefore, using Bayes rule, we have:
i

Elements of Probability Theory Mathematical Statistics for Economics and Business
Chapter 1 – Student Answer Key – Odd Numbers Only 4
PQ| APA
PA |Q i i ,
i 4 PQ| APA
i1 i i

PA D PA
Now, PA |D i  i  PA PA |DPD, allowing us to rewrite
i PD PD i i
Bayes rule as:

PQ| APA |DPD
PA |Q i i ,
i 4 PQ| APA |DPD
i1 i i

where 4 PQ| APA |D(.20)(.55)  (.30)(.30)  (.90)(.10)  (.05)(.05)  .2925.
i1 i i

Then

PQ| A PA |D (.20)(.55)
PA |Q 1 1   .3761
1 .2925 .2925
(.30)(.30)
PA |Q  .3077  The most probable reason is that she disliked her supervisor.
2 .2925

(.90)(.10)
PA |Q  .3077
3 .2925
(.05)(.05)
PA |Q  .0085
4 .2925

b) A  D  PA  PA |DPD(.30)(.20).06.
2 2 2

c)
PQ|D PQD/PD
  4 
 PQA / PD,
 i1 i
4 PQA
  i (since  distributive and QA, i 1,...,4 are disjoint),
PD i
i1
4 PQA PA
  i i ,
PA PD
i1 i
4
  PQ| APA |D since PA PA |DPD,
i i i i
i1
 .2925.

     
15. a) Yes. Using DeMorgan’s Laws, note that P A A .18P A P A ,
1 2 1 2
Elements of Probability Theory Mathematical Statistics for Economics and Business
Chapter 1 – Student Answer Key – Odd Numbers Only 5
           
P A A .2475P A P A , and P A A .22P A P A , so that Theorem
1 3 1 3 2 3 2 3
1.12 can then be applied.

b) No. Note that PA A | A .20, but PA A  PA PA PA A 
2 3 1 2 3 2 3 2 3
.60.45.78.27, so conditional and unconditional probabilities are different.

c)
PA A A  PA A | A PA 
PA A |A  1 2 3  2 3 1 1
1 2 3 PA  PA 
3 3
(.20)(.55)
 .244
.45

d) Yes, different since PA A PAPA PA A .33. Theorem 1.5 can be
1 2 1 2 1 2
extended by letting A A, BA A ,  PA A A  PA PA PA 
1 2 3 1 2 3 1 2 3
PA A PA A PA A PA A A ,
1 2 1 3 2 3 1 2 3

or

PA A A  PA PA PA PA PA PA A 
1 2 3 1 2 3  1 2 1 2 
PA PA PA A PA PA PA A 
 1 3 1 3   2 3 2 3 
PA A A 
1 2 3
 PA A A PA A PA A PA A 
1 2 3 1 2 2 3 1 3
PA PA PA 
1 2 3
.11.82.78.7525.55.60.45
.8625.

17. P(B|I) (cid:3404) .04, P(B|II) (cid:3404) .01.

P(I) (cid:3404) .85, P(II) (cid:3404) .15.

a)
P(B|II )P(II)
P(II |B)  (Bayes Rule)
P(B|II ) P(II)  P(B|I )P(I )

(.01)(.15)
  .04225.
(.01)(.15)  (.04)(.85)

b) PA|B1PA|B1.985.015.

Elements of Probability Theory Mathematical Statistics for Economics and Business
Chapter 1 – Student Answer Key – Odd Numbers Only 6
c)
PA|BPB
PB| A (Bayes Rule)
PA|BPB  PA|BPB

(.985)(.04)
  .73234
(.985)(.04)  (.015)(.96)

The testing device is not especially accurate.

19. a)
P(AB)  .20,
P(A)P(B)  (.40)(.40)  .16,
P(AB)  P(A)P(B)  A and B are not pairwise independent.

P(AC)  .15
P(A)P(C)  (.40)(.375)  .15
P(AC)  P(A) P(C)  A and C are pairwise independent.

P(BC)  .15
P(B)P(C)  (.40)(.375)  .15
P(BC)  P(B) P(C)  B and C are pairwise independent.

b) No, pairwise independence is a necessary (not sufficient) condition for joint independence.
Since A and B are not pairwise independent, we know immediately that A, B, and C are not
jointly independent.

c) P(A∩B) = .20

P((AB)C) .10
PAB|C   .267
P(C) .375

d) Given DABC DA, then PDA0. But
PDPA.05.40.020 D and A are not independent. Given
PDABC0, then PDPABC.05.775.038750 D and
ABC are not independent.

P(C(AB)) .10
e) PC|AB   .50
P(AB) .20

Elements of Probability Theory Mathematical Statistics for Economics and Business
Chapter 1 – Student Answer Key – Odd Numbers Only 7

21. a) The numerator of this function will always be (cid:3410) 0 for any A⊂S (it will equal 0 when A(cid:3404)∅).
The denominator is always positive. This implies P(A) (cid:3410) 0, for any event A⊂S, thus Axiom
1.1 holds.

P(S) = (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8)/36 = 1, thus Axiom 1.2 holds.

Let {A: iϵI} be a collection of disjoint events contained in S. Then
i

P A   x/36  (x/36) (since A 's disjoint) P(A),
 i i i
iI
xA iI xAi iI
iI i
Axiom 1.3 holds .

All three axioms hold implying this is a valid probability set function.

b) The integrand ex 0x, so that the integral of ex over any event in the event space must
be (cid:3410)0, which implies Axiom 1.1 holds.

S  [0,), so P(S) exdxlim (en  e0)1  lim en1
0 n n
 Axiom 1.2 holds .

PA    ex dx    exdx (valid given the A’s are disjoint)=  PA 
 i i i
iI
x Ai iI xAi iI
iI 
 Axiom 1.3 holds .

All three axioms hold implying this is a valid probability set function.

c) This is not a probability set function. To see why, consider Axiom 1.2, i.e., P(S) =1.
 1 
P(S)   x2/105   x2  ,
105
i1 i1
 Axiom 1.2 does not hold .

d) Consider the integrand f(x)(cid:3404)12x(1(cid:3398)x)2. Notice that f(x) is a positive valued function over the
domain S (cid:3404) (0,1), implying that any integral of this function over events in the event space
will always be nonnegative. Thus Axiom 1.1 holds.

Axiom 1.2 holds, since P(S)112x1x2dx 6x2 8x3 3x4|1 1.
0
0
Finally,

PA   12x1x2dx    12x1x2dx   P(A)
 i i
iI
xiIAi iI xAi iI
 Axiom 1.3 holds.
Elements of Probability Theory Mathematical Statistics for Economics and Business
Chapter 1 – Student Answer Key – Odd Numbers Only 8

All three axioms hold, implying this is a valid probability set function.

23. a) S (cid:3404) {(x,y): x and y ϵ{1, 2, 3, 4}}

b) A (cid:3404) {(x,y): x = 1 or 3, yϵ{1, 2, 3, 4}} (First disk is a 1 or 3).

B (cid:3404) {(x,y): y = 1 or 2, xϵ{1, 2, 3, 4}} (Second disk is a 1 or 2).

N(S) (cid:3404) 16, N(A) (cid:3404)8, N(B) (cid:3404) 8, N(A∩B) (cid:3404)4.

4  8  8 
P(A∩B) (cid:3404) =    (cid:3404) P(A) P(B) ⇒ independence.
16 1616

c) A (cid:3404) {(x, y): x (cid:3404) 1, yϵ{1, 2, 3, 4}}

B (cid:3404) {(x, y): y (cid:3404)1, xϵ{1, 2, 3, 4}}

N(A) (cid:3404)4, N(B) (cid:3404) 4, N(AB) (cid:3404)1

1  4  4 
P(A∩B) (cid:3404) =    (cid:3404)P(A) P(B)  independence.
16 1616

d) A (cid:3404) {(1,1)}

B (cid:3404) {(x, y): x and y ϵ{1, 2}}
(neither 3 or 4 is chosen in the selection process)

N(A) = 1, N(B) = 4, N(A∩B) = 1

1 1 4
P(A∩B) =   =P(A) P(B)  not independent.
16 16 16

25. a) PA|B1PA|B.02.

PA|BPB (.98)(.05)
b) PB| A 
PA|BPBPA|BPB (.98)(.05)  (.02)(.95)

P(B|A)  .7206.

c) From Bayes Rule

r (.05)
.95   r = .9972.
r (.05)  (1r)(.95)
Elements of Probability Theory Mathematical Statistics for Economics and Business
Chapter 1 – Student Answer Key – Odd Numbers Only 9

27. a) S = {(x, y): x and y {0, 1, 2, 3, 4}}.

b) Yes. Let f(x, y) represent the probability of selling x number of printers and y number of
computers, as displayed in the table. Then if A is an event contained in S,

P(A) =   f(x, y).
(x,y)A

c) Let A = {(x, y): y = 3 or 4, x{0, 1, 2, 3, 4}} (more than two computers sold)

P(A) =  f(x, y) = .56.
(x,y)A

Let B = {(x, y): x = 3 or 4, y{0, 1, 2, 3, 4}} (more than two printers sold)

P(B) =   f(x, y) = .50.
(x,y)B

P(x  2,y  2) .40
d) P(x > 2 | y > 2) = = = .7143.
P(y  2) .56

e) P(x > 2, y > 2) = .10 + .10 + .05 + .15 = .40.

f) f(0, 0) = .03,

P(x 0, y 0) .03
P(x=0  y = 0) = = = .375.
P(y 0) .08

29. a) Total number of invoices = 529.
Number of invoices from sales team C = 129.
Probability that a randomly selected invoice is from sales team C = 129/529 =0.2439.

b) Total number of invoices = 529.
Number of invoices over 180 days old = 72.
Probability that a randomly selected invoice is from sales team C = 72/529 =0.1361.

c) Total number of invoices = 529.
Number of invoices over 180 days old and from sales team C= 19.
Probability that a randomly selected invoice is over 180 days old and from sales team
C=19/529 =0.0354.

Sales team A has the lowest probability (0.1493) of being associated with a randomly
selected invoice.

Elements of Probability Theory Mathematical Statistics for Economics and Business