# Quantitative Analysis For Management, 13th Edition Solution Manual

Preview Extract

CHAPTER 2
Probability Concepts and Applications
TEACHING SUGGESTIONS
Teaching Suggestion 2.1: Concept of Probabilities Ranging From 0 to 1.
People often misuse probabilities by such statements as, โIโm 110% sure weโre going to win the
big game.โ The two basic rules of probability should be stressed.
Teaching Suggestion 2.2: Where Do Probabilities Come From?
Students need to understand where probabilities come from. Sometimes they are subjective and
based on personal experiences. Other times they are objectively based on logical observations
such as the roll of a die. Often, probabilities are derived from historical dataโif we can assume
the future will be about the same as the past.
Teaching Suggestion 2.3: Confusion Over Mutually Exclusive and Collectively Exhaustive Events.
This concept is often foggy to even the best of studentsโeven if they just completed a course in
statistics. Use practical examples and drills to force the point home.
Teaching Suggestion 2.4: Addition of Events That Are Not Mutually Exclusive.
The formula for adding events that are not mutually exclusive is P(A or B) = P(A) + P(B) โ
P(A and B). Students must understand why we subtract P(A and B). Explain that the intersection
has been counted twice.
Teaching Suggestion 2.5: Expected Value of a Probability Distribution.
A probability distribution is often described by its mean and variance. These important terms
should be discussed with such practical examples as heights or weights of students. But students
need to be reminded that even if most of the men in class (or the United States) have heights
between 5 feet 6 inches and 6 feet 2 inches, there is still some small probability of outliers.
Teaching Suggestion 2.6: Bell-Shaped Curve.
Stress how important the normal distribution is to a large number of processes in our lives (for
example, filling boxes of cereal with 32 ounces of cornflakes). Each normal distribution depends
on the mean and standard deviation. Discuss Figures 2.7 and 2.8 to show how these relate to the
shape and position of a normal distribution.
2-1
Copyright ยฉ 2018 Pearson Education, Inc.
Teaching Suggestion 2.7: Three Symmetrical Areas Under the Normal Curve.
Figure 2.13 is very important, and students should be encouraged to truly comprehend the
meanings of ยฑ1, 2, and 3 standard deviation symmetrical areas. They should especially know that
managers often speak of 95% and 99% confidence intervals, which roughly refer to ยฑ2 and 3
standard deviation graphs. Clarify that 95% confidence is actually ยฑ1.96 standard deviations,
while ยฑ3 standard deviations is actually a 99.7% spread.
Teaching Suggestion 2.8: Using the Normal Table to Answer Probability Questions.
The IQ example in Figure 2.9 is a particularly good way to treat the subject since everyone can
relate to it. Students are typically curious about the chances of reaching certain scores. Go
through at least a half-dozen examples until itโs clear that everyone can use Table 2.10. Students
get especially confused answering questions such as P(X 1/2 As | regular class) = 0.25
P(>1/2 As | advanced class) = 0.50
P(>1/2 As and regular class)
= P(>1/2 As | regular ) ร P(regular)
= (0.25)(0.50) = 0.125
P(>1/2 As and advanced class)
= P(>1/2 As | advanced) ร P(advanced)
= (0.50)(0.5) = 0.25
So P(>1/2 As) = 0.125 + 0.25 = 0.375
P ( advanced > 1/2 As ) =
=
P ( advanced and > 1/2 As )
P ( > 1/ 2 As )
0.25
= 2/3
0.375
So there is a 66% chance the class tested was the advanced one.
2-5
Copyright ยฉ 2018 Pearson Education, Inc.
Alternative Example 2.8: Students in a statistics class were asked how many โawayโ football
games they expected to attend in the upcoming season. The number of students responding to
each possibility is shown below:
Number of games
Number of students
5
40
4
30
3
20
2
10
1
0
100
A probability distribution of the results would be:
Number of games
Probability P(X)
5
0.4 = 40/100
4
0.3 = 30/100
3
0.2 = 20/100
2
0.1 = 10/100
1
0.0 = 0/100
1.0 = 100/100
This discrete probability distribution is computed using the relative frequency approach.
Probabilities are shown in graph form below.
2-6
Copyright ยฉ 2018 Pearson Education, Inc.
Alternative Example 2.9: Here is how the expected outcome can be computed for the question
in Alternative Example 2.8.
5
E ( x ) = โ xi P ( X i ) = xi P ( X 1 ) + x2 P ( X 2 )
i =1
+ x3P(X3) +x4P(X4) + x5P(X5)
= 5(0.4) + 4(0.3) + 3(0.2) + 2(0.1) + 1(0)
= 4.0
Alternative Example 2.10: Here is how variance is computed for the question in Alternative
Example 2.8:
5
variance = โ ( xi โ E ( x ) ) P ( xi )
2
i =1
2
= (5 โ 4) (0.4) + (4 โ 4)2(0.3) + (3 โ 4)2(0.2) + (2 โ 4)2(0.1) + (1 โ 4)2(0)
= (1)2(0.4) + (0)2(0.3) + (โ1)2(0.2) + (โ2)2(0.1) + (-3)2(0)
= 0.4 + 0.0 + 0.2 + 0.4 + 0.0
= 1.0
The standard deviation is
ฯ = variance
= 1
=1
Alternative Example 2.11: The length of the rods coming out of our new cutting machine can
be said to approximate a normal distribution with a mean of 10 inches and a standard deviation
of 0.2 inch. Find the probability that a rod selected randomly will have a length
a. of less than 10.0 inches
b. between 10.0 and 10.4 inches
c. between 10.0 and 10.1 inches
d. between 10.1 and 10.4 inches
e. between 9.9 and 9.6 inches
f. between 9.9 and 10.4 inches
g. between 9.886 and 10.406 inches
2-7
Copyright ยฉ 2018 Pearson Education, Inc.
First compute the standard normal distribution, the Z-value:
z=
xโฮผ
ฯ
Next, find the area under the curve for the given Z-value by using a standard normal distribution
table.
a. P(x < 10.0) = 0.50000
b. P(10.0 < x < 10.4) = 0.97725 โ 0.50000 = 0.47725
c. P(10.0 < x < 10.1) = 0.69146 โ 0.50000 = 0.19146
d. P(10.1 < x < 10.4) = 0.97725 โ 0.69146 = 0.28579
e. P(9.6 < x < 9.9) = 0.97725 โ 0.69146 = 0.28579
f. P(9.9 < x < 10.4) = 0.19146 + 0.47725 = 0.66871
g. P(9.886 < x 6) = P(r โฅ 7) = P(r = 7) + P(r = 8) + P(r = 9) + P(r = 10)
= 0.1172 + 0.0439 + 0.0098 + 0.0010
= 0.1719
2-20
Copyright ยฉ 2018 Pearson Education, Inc.
4.04753
2-33. This is a binomial distribution with n=4, p=0.7, and q=0.3.
P ( r = 3) =
4!
3
4โ3
( 0.7 ) ( 0.3) = 0.4116
3!( 4 โ 3) !
P ( r = 4) =
4!
4
4โ4
( 0.7 ) ( 0.3) = 0.2401
4!( 4 โ 4 ) !
2-34. This is a binomial distribution with n =5, p=0.1, and q=0.9.
P ( r = 1) =
5!
1
5โ1
( 0.1) ( 0.9 ) = 0.328
1!( 5 โ 1) !
P ( r = 0) =
5!
0
5โ 0
( 0.1) ( 0.9 ) = 0.590
0!( 5 โ 0 ) !
2-35. This is a binomial distribution with n=6, p=0.05, and q=0.95.
P ( r = 0) =
6!
0
6โ0
( 0.05) ( 0.95) = 0.735
0!( 6 โ 0 ) !
P ( r = 1) =
6!
1
6 โ1
( 0.05) ( 0.95) = 0.232
1!( 6 โ 1) !
2-36. This is a binomial distribution with n=6, p=0.15, and q=0.85.
P ( r = 0) =
6!
0
6โ0
( 0.15) ( 0.85) = 0.377
0!( 6 โ 0 ) !
P ( r = 1) =
6!
1
6 โ1
( 0.15) ( 0.85) = 0.399
1!( 6 โ 1) !
Probability of 0 or 1 defective = P(0) + P(1) = 0.377 + 0.399 = 0.776.
2-21
Copyright ยฉ 2018 Pearson Education, Inc.
2-37. ฮผ = 450 degrees
ฯ = 25 degrees
X = 475 degrees
Z=
X โฮผ
ฯ
=
475 โ 450
=1
25
The area to the left of 475 is 0.8413 from Table 2.9, where ฯ = 1. The area to the right of 475
is 1 โ 0.84134 = 0.15866. Thus, the probability of the oven getting hotter than 475 is 0.1587.
To determine the probability of the oven temperature being between 460 and 470, we
need to compute two areas.
X1 = 460
X2 = 470
Z1 =
460 โ 450 10
=
= 0.4
25
25
area X1 = 0.65542
Z2 =
470 โ 450 20
=
= 0.8
25
25
area X2 = 0.78814
The area between X1 and X2 is 0.78814 โ 0.65542 = 0.13272. Thus, the probability of being
between 460 and 470 degrees is = 0.1327.
2-22
Copyright ยฉ 2018 Pearson Education, Inc.
2-38. ฮผ = 4,700; ฯ = 500
a. The sale of 5,500 oranges (X = 5,500) is the equivalent of some Z value which may be
obtained from
Z=
=
X โฮผ
ฯ
5,500 โ 4, 700
500
= 1.6
The area under the curve lying to the left of 1.6ฯ = 0.94520. Therefore, the area to the right
of 1.6ฯ = 1 โ 0.94520, or 0.0548. Therefore, the probability of sales being greater than 5,500
oranges is 0.0548.
b.
Z=
area
4,500 โ 4, 700
200
=โ
= โ0.4
500
500
= 0.65542
probability = 0.65542
2-23
Copyright ยฉ 2018 Pearson Education, Inc.
c.
Z=
4,900 โ 4, 700 200
=
= 0.4
500
500
area = 0.65542 = probability
This answer is the same as the answer to part (b) because the normal curve is symmetrical.
d.
Z=
4,300 โ 4, 700
400
=โ
= โ0.8
500
500
Area to the right of 4,300 is 0.78814, from Table 2.9. The area to the left of 4,300 is 1 โ
0.78814 = 0.21186 = the probability that sales will be fewer than 4,300 oranges.
2-39. ฮผ = 87,000
ฯ = 4,000
X = 81,000
Z=
81, 000 โ 87, 000
6
= โ = โ1.5
4, 000
4
Area to the right of 81,000 = 0.93319, from Table 2.9, where Z = 1.5. Thus, the area to the left of
81,000 = 1 โ 0.93319 = 0.06681 = the probability that sales will be fewer than 81,000 packages.
2-24
Copyright ยฉ 2018 Pearson Education, Inc.
2-40.
ฮผ = 457,000
Ninety percent of the time, sales have been between 460,000 and 454,000 pencils. This means
that 10% of the time sales have exceeded 460,000 or fallen below 454,000. Since the curve is
symmetrical, we assume that 5% of the area lies to the right of 460,000 and 5% lies to the left of
454,000. Thus, 95% of the area under the curve lies to the left of 460,000. From Table 2.9, we
note that the number nearest 0.9500 is 0.94950. This corresponds to a Z value of 1.64. Therefore,
we may conclude that the Z value corresponding to a sale of 460,000 pencils is 1.64.
Using Equation 2-13, we get Z =
X โฮผ
ฯ
X = 460,000
ฮผ = 457,000
ฯ is unknown
Z = 1.64
1.64 =
460,000 โ 457,000
ฯ
1.64 ฯ = 3000
โดฯ =
3,000
1.64
= 1829.27
2-41. The time to complete the project (X) is normally distributed with ฮผ = 60 and ฯ = 4.
a) P(X โค 62) = P(Z โค (62 โ 60)/4) = P(Z โค 0.5) = 0.69146
b) P(X โค 66) = P(Z โค (66 โ 60)/4) = P(Z โค 1.5) = 0.93319
c) P(X > 65) = 1 โ P(X โค 65)
= 1 โ P(Z โค (65 โ 60)/4)
= 1 โ P(Z โค 1.25)
= 1 โ 0.89435 = 0.10565
2-25
Copyright ยฉ 2018 Pearson Education, Inc.
2-42. The time to complete the project (X) is normally distributed with ฮผ = 40 and ฯ = 5.
A penalty must be paid if the project takes longer than the due date (or if X > due date).
a) P(X > 40) = 1 โ (X โค 40) = 1 โ P(Z โค (40 โ 40)/5) = 1 โ P(Z โค 0) = 1 โ 0.5 = 0.5
b) P(X > 43) = 1 โ P(X โค 43) = 1 โ P(Z โค (43 โ 40)/5) = 1 โ P(Z โค 0.6) = 1 โ 0.72575 =
0.27425
c) If there is a 5% chance that the project will be late, there is a 95% chance the project will
be finished by the due date. So,
P(X โค due date) = 0.95 or
P(X โค _____) = 0.95
The Z-value for a probability of 0.95 is approximately 1.64, so the due date (X) should
have a Z-value of 1.64. Thus,
1.64 =
X โ 40
5
5(1.64) = X โ 40
X = 48.2.
The due date should be 48.2 weeks from the start of the project
2-43. ฮป = 5/day; eโฮป = 0.0067 (from Appendix C)
a. P ( 0 ) =
ฮป x eโ ฮป
X!
=
(1)( 0.0067 ) = 0.0067
1
P (1) =
( 5)( 0.0067 ) = 0.0335
P ( 2) =
25 ( 0.0067 )
= 0.0837
2
P ( 3) =
125 ( 0.0067 )
= 0.1396
6
P ( 4) =
625 ( 0.0067 )
= 0.1745
24
P ( 5) =
3125 ( 0.0067 )
= 0.1745
120
1
b. These sum to 0.6125, not 1, because there are more possible arrivals. For example, 6 or
7 patients might arrive in one day.
2-26
Copyright ยฉ 2018 Pearson Education, Inc.
2-44. P(X > 3) = 1 โ P(X โค 3) = 1 โ [P(0) + P(1) + P(2) + P(3)]
= 1 โ [0.0067 + 0.0335 + 0.0837 + 0.1396]
= 1 โ 0.2635 = 0.7365 = 73.65%
2-45. ฮผ = 3/hour
a. Expected time
1
=
ฮผ
=
1
hour
3
= 20 minutes
b. Variance =
1
ฮผ
2
=
1
9
2-46. Let S = steroids present
N = steroids not present
TP = test is positive for steroids
TN = test is negative for steroids
P(S) = 0.02
P(N) = 0.98
P(TP | S) = 0.95
P(TN | S) = 0.05
P(TP | N) = 0.10
P(TN | N) = 0.90
P (TP S ) P ( S )
P ( S TP ) =
=
P (TP S ) P ( S ) + P (TP N ) P ( N )
0.95 ( 0.02 )
= 0.16
0.95 ( 0.02 ) + 0.10 ( 0.98 )
2-47. Let G = market is good
P = market is poor
PG = test predicts good market
PP = test predicts poor market
P(G) = 0.70
P(P) = 0.30
P(PG | G) = 0.85
P(PP | G) = 0.15
P(PG | P) = 0.20
P(PP | P) = 0.80
P ( G PG ) =
=
P ( PG G ) P ( G )
P ( PG G ) P ( G ) + P ( PG P ) P ( P )
0.85 ( 0.70 )
= 0.91
0.85 ( 0.70 ) + 0.20 ( 0.30 )
2-27
Copyright ยฉ 2018 Pearson Education, Inc.
2-48. Let W = candidate wins the election
L = candidate loses the election
PW = poll predicts win
PL = poll predicts loss
P(W) = 0.50
P(L) = 0.50
P(PW | W) = 0.80
P(PL | W) = 0.20
P(PW | L) = 0.10
P(PL | L) = 0.90
P (W PW ) =
=
P ( L PL ) =
=
P ( PW W ) P (W )
P ( PW W ) P (W ) + P ( PW L ) P ( L )
0.80 ( 0.50 )
0.80 ( 0.50 ) + 0.10 ( 0.50 )
= 0.89
P ( PL L ) P ( L )
P ( PL L ) P ( L ) + P ( PL W ) P (W )
0.90 ( 0.50 )
= 0.82
0.90 ( 0.50 ) + 0.20 ( 0.50 )
2-49. Let S = successful restaurant
U = unsuccessful restaurant
PS = model predicts successful restaurant
PU = model predicts unsuccessful restaurant
P(S) = 0.70
P(U) = 0.30
P(PS | S) = 0.90
P(PU | S) = 0.10
P(PS | U) = 0.20
P(PU | U) = 0.80
P ( S PS ) =
=
P ( PS S ) P ( S )
P ( PS S ) P ( S ) + P ( PS U ) P (U )
0.90 ( 0.70 )
= 0.91
0.90 ( 0.70 ) + 0.20 ( 0.30 )
2-28
Copyright ยฉ 2018 Pearson Education, Inc.
2-50. Let D = Default on loan; D’ = No default; R = Loan rejected; Rโฒ = Loan approved Given:
P(D) = 0.2
P(D’) = 0.8
P(R | D) = 0.9
P(R’ | D’) = 0.7
(a) P(R | D’) = 1 โ 0.7 = 0.3
(b) P ( Dโฒ R ) =
=
P ( R D โฒ ) P ( Dโฒ )
P ( R Dโฒ ) P ( D โฒ ) + P ( R D ) P ( D )
0.3 ( 0.8 )
= 0.57
0.3 ( 0.8) + 0.9 ( 0.2 )
2-51. (a) F0.05, 5, 10 = 3.33
(b) F0.05, 8.7 = 3.73
(c) F0.05, 3, 5 = 5.41
(d) F0.05, 10. 4 = 5.96
2-52. (a) F0.01, 15, 6 = 7.56
(b) F0.01, 12, 8 = 5.67
(c) F0.01, 3, 5 = 12.06
(d) F0.01, 9, 7 = 6.72
2-53. (a) From the appendix, P(F3,4 > 6.59) = 0.05, so P(F > 6.8) must be less than 0.05.
(b) From the appendix, P(F7,3 > 8.89) = 0.05, so P(F > 3.6) must be greater than 0.05.
(c) From the appendix, P(F20,20 > 2.12) = 0.05, so P(F > 2.6) must be less than 0.05.
(d) From the appendix, P(F7,5 > 4.88) = 0.05, so P(F > 5.1) must be less than 0.05.
(e) From the appendix, P(F7,5 > 4.88) = 0.05, so P(F 15.52) = 0.01, so P(F > 14) must be greater than 0.01.
(b) From the appendix, P(F6,3 > 27.91) = 0.01, so P(F > 30) must be less than 0.01.
(c) From the appendix, P(F10,12 > 4.30) = 0.01, so P(F > 4.2) must be greater than 0.01.
(d) From the appendix, P(F2,3 > 30.82) = 0.01, so P(F > 35) must be less than 0.01.
(e) From the appendix, P(F2,3 > 30.82) = 0.01, so P(F < 35) must be greater than 0.01.
2-55. Average time = 4 minutes = 1/ยต. So ยต = ยผ = 0.25
(a) P(X < 3) = 1 โ e-0.25(3) = 1 โ 0.4724 = 0.5276
(b) P(X < 4) = 1 โ e-0.25(4) = 1 โ 0.3679 = 0.6321
(c) P(X 5) = e-0.25(5) = 0.2865
2-56. Average number per minute = 5. So ฮป = 5
(a) P(X is exactly 5) = P(5) = (55e-5)/5! = 0.1755
(b) P(X is exactly 4) = P(4) = (54e-5)/4! = 0.1755
(c) P(X is exactly 3) = P(3) = (53e-5)/3! = 0.1404
(d) P(X is exactly 6) = P(6) = (56e-5)/6! = 0.1462
(e) P(X< 2) = P(X is 0 or 1) = P(0) + P(1) = 0.0067 + 0.0337 = 0.0404
2-57. The average time to service a customer is 1/3 hour or 20 minutes. The average number that
would be served per minute (ยต) is 1/20 = 0.05 per minute.
P(time < ยฝ hour) = P(time < 30 minutes) = P(X < 30) = 1 โ e-0.05(30) = 0.7769
P(time < 1/3 hour) = P(time < 20 minutes) = P(X < 20) = 1 โ e-0.05(20) = 0.6321
P(time < 2/3 hour) = P(time < 40 minutes) = P(X < 40) = 1 โ e-0.05(40) = 0.8647
These are exactly the same probabilities shown in the example.
2-30
Copyright ยฉ 2018 Pearson Education, Inc.
2-58. X = 280
ฮผ = 250
ฯ = 25
โดz =
X โฮผ
ฯ
280 โ 250
25
30
=
25
=
= 1.20 standard deviations
From Table 2.9, the area under the curve corresponding to a Z of 1.20 = 0.88493. Therefore, the
probability that the sales will be less than 280 boats is 0.8849.
2-31
Copyright ยฉ 2018 Pearson Education, Inc.
2-59. The probability of sales being over 265 boats:
X = 265
ฮผ = 250
ฯ = 25
265 โ 250
25
15
=
25
Z=
= 0.60
From Table 2.9, we find that the area under the curve to the left of Z = 0.60 is 0.72575. Since we
want to find the probability of selling more than 265 boats, we need the area to the right of Z =
0.60. This area is 1 โ 0.72575, or 0.27425. Therefore, the probability of selling more than 265
boats = 0.2743.
For a sale of fewer than 250 boats:
X = 250
ฮผ = 250
ฯ = 25
However, a sale of 250 boats corresponds to ฮผ = 250. At this point, Z = 0. The area under the
curve that concerns us is that half of the area lying to the left of ฮผ = 250. This area = 0.5000.
Thus, the probability of selling fewer than 250 boats = 0.5.
2-32
Copyright ยฉ 2018 Pearson Education, Inc.
2-60. ฮผ = 0.55 inch (average shaft size)
X = 0.65 inch
ฯ = 0.10 inch
Converting to a Z value yields
Z=
X โฮผ
ฯ
0.65 โ 0.55
0.10
0.10
=
0.10
=
=1
We thus need to look up the area under the curve that lies to the left of 1ฯ. From Table 2.9, this
is seen to be = 0.84134. As seen earlier, the area to the left of ฮผ is = 0.5000.
We are concerned with the area between ฮผ and ฮผ + 1ฯ. This is given by the difference
between 0.84134 and 0.5000, and it is 0.34134. Thus, the probability of a shaft size between 0.55
inch and 0.65 inch = 0.3413.
2-61. Greater than 0.65 inch:
area to the left of 1ฯ = 0.84134
area to the right of 1ฯ = 1 โ 0.84134
= 0.15866
Thus, the probability of a shaft size being greater than 0.65 inch is 0.1587.
2-33
Copyright ยฉ 2018 Pearson Education, Inc.
The shaft size between 0.53 and 0.59 inch:
X2 = 0.53 inch
X1 = 0.59 inch
ฮผ = 0.55 inch
Converting to scores:
Z1 =
X1 โ ฮผ
ฯ
0.59-0.55
=
0.10
0.04
=
0.10
=0.4
Z2 =
X2 โ ฮผ
ฯ
0.53 โ 0.55
=
0.10
โ0.02
=
0.10
= โ0.2
Since Table 2.9 handles only positive Z values, we need to calculate the probability of the shaft
size being greater than 0.55 + 0.02 = 0.57 inch. This is determined by finding the area to the left
of 0.57, that is, to the left of 0.2ฯ. From Table 2.9, this is 0.57926. The area to the right of 0.2ฯ
is 1 โ 0.57926 = 0.42074. The area to the left of 0.53 is also 0.42074 (the curve is symmetrical).
The area to the left of 0.4ฯ is 0.65542. The area between X1 and X2 is 0.65542 โ 0.42074 =
0.23468. The probability that the shaft will be between 0.53 inch and 0.59 inch is 0.2347.
Under 0.45 inch:
X = 0.45
ฮผ = 0.55
ฯ = 0.10
X โฮผ
Z=
ฯ
0.45 โ 0.55
=
0.10
โ0.10
=
0.10
= โ1
2-34
Copyright ยฉ 2018 Pearson Education, Inc.
Thus, we need to find the area to the left of โ1ฯ. Again, since Table 2.9 handles only positive
values of Z, we need to determine the area to the right of 1ฯ. This is obtained by 1 โ 0.84134 =
0.15866 (0.84134 is the area to the left of 1ฯ). Therefore, the area to the left of โ1ฯ = 0.15866
(the curve is symmetrical). Thus, the probability that the shaft size will be under 0.45 inch is
0.1587.
โnโ
2-62. โ โ p x q nโ x
โ xโ
x=3
n=4
q = 15/20 = .75
p = 5/20 = .25
3
โ 4โ
โ โ (.25) (.75) =
โ 3โ
4!
3
(.25) (.75) =
3!( 4 โ 3) !
(4)(.0156)(.75) =
.0468 [probability that Marie will win 3 games]
โ 4โ
4
0
โ โ (.25) (.75 ) = .003906 [Probability that Marie will win all four games against Jan]
โ 4โ
Probability that Marie will be number one is .04698 + .003906 = .050886.
2-35
Copyright ยฉ 2018 Pearson Education, Inc.
2-63. Probability one will be fined =
P(2) + P(3) + P(4) + P(5)
= 1 โ P(0) โ P(1)
โ 5โ
โ 5โ 1
0
5
4
= 1 โ โ โ (.5) (.5) โ โ โ (.5) (.5)
โ 0โ
โ1โ
= 1 โ .03125 โ .15625
= .08125
Probability of no foul outs = P(0) = 0.03125
Probability of foul out in all 5 games = P(5) = 0.03125
2-64.
X
P(X)
0
โ 5โ
0
5
โ โ (.2 ) (.8) = .327
โ 0โ
.327
1
โ 5โ 1
5โ1
โ โ (.2 ) (.8) = .410
โ1โ
.410
2
โ 5โ
2
5โ 2
โ โ (.2 ) (.8) = .205
โ 2โ
.205
3
โ 5โ
3
5โ3
โ โ (.2 ) (.8) = .051
โ 3โ
.051
4
โ5โ
4
5โ 4
โ โ (.2 ) (.8) = .0064
โ 4โ
.0064
5
.00032
โ 5โ
5
5 โ5
.2
.8
.00032
=
(
)
(
)
โ โ
โ 5โ
1.0
XP(X)
0.0
.41
.41
.153
.024
.0015
.9985
X โ E(X)
(X โ E(X))2
(X โ E(X))2P(X)
โ.9985
.0015
1.0015
2.0015
3.0015
4.0015
.997
0
1.003
4.006
9.009
16.012
.326
0
.2056
.2043
.0541
.0048
.7948
E(X) = .9985 โ
1.0
ฯ2 = ฮฃ(X โ E(X))2P(X)
= .7948 โ
0.80
2-36
Copyright ยฉ 2018 Pearson Education, Inc.
Using the formulas for the binomial:
E(X) = np
= (5)(.2) = 1.0
2
ฯ = np(1 โ p)
= 5(.2)(.8)
= 0.80
The equation produced equivalent results.
2-65. a. n = 10; p = .25; q = .75;
โ10 โ x 10โ x
โ โp q
โxโ
P(X)
X
โ10 โ
0
10 โ0
โ โ (.25) (.75) =
โ0โ
.0563
0
โ10 โ
1
10โ1
โ โ (.25) (.75) =
โ1โ
.1877
1
โ10 โ
2
10โ 2
โ โ (.25) (.75) =
โ2โ
.2816
2
โ10 โ
3
10 โ3
โ โ (.25 ) (.75 ) =
โ3โ
.2503
3
โ10 โ
4
10โ 4
โ โ (.25) (.75) =
โ4โ
.1460
4
โ10 โ
5
10 โ5
โ โ (.25 ) (.75) =
โ5โ
.0584
5
โ10 โ
6
10 โ6
โ โ (.25) (.75) =
โ6โ
.0162
6
โ10 โ
7
10 โ7
โ โ (.25) (.75) =
โ7โ
.0031
7
โ10 โ
8
10 โ8
โ โ (.25) (.75) =
โ8โ
.0004
8
โ10 โ
9
10โ9
โ โ (.25) (.75) =
โ9โ
.00003
9
โ10 โ
10
10โ10
=
โ โ (.25) (.75)
โ10 โ
.0000
10
2-37
Copyright ยฉ 2018 Pearson Education, Inc.
b. E(X) = np = (10).25 = 2.5
ฯ2 = npq = (10)(.25)(.75)
= 1.875
c. Expected weekly profit:
$125.
ร 2.5
$312.50
SOLUTION TO WTVX CASE
1. The chances of getting 15 days of rain during the next 30 days can be computed by using the
binomial theorem. The problem is well suited for solution by the theorem because there are two
and only two possible outcomes (rain or shine) with given probabilities (70% and 30%,
respectively). The formula used is:
Probability of r successes =
n!
p r ( q nโr )
r !( n โ r ) !
where
n = the number of trials (in this case, the number of days = 30),
r = the number of successes (number of rainy days = 15),
p = probability of success (probability of rain = 70%), and
q = probability of failure (probability of shine = 30%).
n!
30!
15
15
p r ( q nโr ) =
(.70 ) (.30 ) = .0106
r !( n โ r ) !
15!(15!)
The probability of getting exactly 15 days of rain in the next 30 days is 0.0106 or a 1.06%
chance.
2. Joeโs assumptions concerning the weather for the next 30 days state that what happens on one
day is not in any way dependent on what happened the day before; what this says, for example, is
that if a cold front passed through yesterday, it will not affect what happens today.
But there are perhaps certain conditional probabilities associated with the weather (for
example, given that it rained yesterday, the probability of rain today is 80% as opposed to 70%).
Not being familiar with the field of meteorology, we cannot say precisely what these are.
However, our contention is that these probabilities do exist and that Joeโs assumptions are
fallacious.
2-38
Copyright ยฉ 2018 Pearson Education, Inc.

## Document Preview (38 of 187 Pages)

You are viewing preview pages of the document. Purchase to get full access instantly.

-37%

### Quantitative Analysis For Management, 13th Edition Solution Manual

$18.99 ~~$29.99~~Save:$11.00(37%)

24/7 Live Chat

Instant Download

100% Confidential

Store

##### Benjamin Harris

0 (0 Reviews)

## Best Selling

The World Of Customer Service, 3rd Edition Test Bank

$18.99 ~~$29.99~~Save:$11.00(37%)

Chemistry: Principles And Reactions, 7th Edition Test Bank

$18.99 ~~$29.99~~Save:$11.00(37%)

Test Bank for Hospitality Facilities Management and Design, 4th Edition

$18.99 ~~$29.99~~Save:$11.00(37%)

Solution Manual for Designing the User Interface: Strategies for Effective Human-Computer Interaction, 6th Edition

$18.99 ~~$29.99~~Save:$11.00(37%)

Data Structures and Other Objects Using C++ 4th Edition Solution Manual

$18.99 ~~$29.99~~Save:$11.00(37%)

2023-2024 ATI Pediatrics Proctored Exam with Answers (139 Solved Questions)

$18.99 ~~$29.99~~Save:$11.00(37%)