Statistics assignment 5

Business Statistics (BUS 505) Assignment 5
7) A company knows that a rival is about to bring out a competing product. It believes that this
rival has three possible packaging plans (superior, normal, cheap) in mind and that all are
equally likely. Also there are three equally likely possible marketing strategies (intense media
advertising, price discount, and use coupon to reduce the price of future purchases). What is the
probability that the rival will employ superior packaging in conjunction with an intense media
advertising campaign? Assume that packaging plans and marketing strategies are determined
independently.
Answer: Let, A denote that one of the packaging plan in use
B denote that one of the marketing strategies in use
So, Probability distribution for packaging plan;
Ai A1 A2 A3
P(Ai)
1
3
1
3
1
3
And, Probability distribution for marketing strategies;
Bi B1 B2 B3
P(Bi)
1
3
1
3
1
3
P(A1 ÇB1)= P(A1)P(B1)
P((A1 ÇB1)= 1 X 3
3
1 = 9
1 =.111
8) A financial analyst was asked to evaluate earnings prospect for seven corporations over the
next year, and to rank them in order of predicted earnings growth rates.
a. How many different rankings are possible?
b. If, in fact, a specific ordering is simply guessed, what is the probability that is guess will turn
out to be correct?
Answer:
Given that, evaluate earnings prospect for seven corporations over the next year.
a) The possible number of different rankings is: 7! = 7´6´5´4´3´2´1 =5040
b) A specific ordering is simply guessed. So, the probability that this guess will turn out to be
correct is: 1/7! = 1/5040
9) A company has fifty sales representatives. It decides that the most successful representatives
during the previous year will be awarded a January vacation in Hawaii, while the second most
successful will win a vacation in Las Vegas. The other representatives will be required to attend a
conference on modern sales methods in Buffalo. How many outcomes are possible?
Answer:
Given that, a company has fifty sales representatives.
So, the possible numbers of outcomes are: 50p2 X 48c48 = 2450X1 =2450
1

10) A securities analyst claims that given a specific list of six common stocks, it is possible to
predict, in the correct order the three that will perform best during the coming year. What is the
possibility of making the correct selection by the chance?
Answer:
Given, n = 6 and, x = 3
The possible numbers of making the correct selection of 3 common stock out of the 6
6!
common stock are: 6p3 = (6 - 3)!
= 120
1
The probability of making the correct selection by chance is: P (6p3) = 120
11) A student committee has six members-four undergraduates and two graduate students. A
subcommittee of three members is to be chosen randomly, so that each possible combination of
three of the six students is equally likely to be selected. What is the probability that there will be
no graduate students on the subcommittee?
Answer:
Given that, n = 6 (four undergraduate and two graduate students)
x = 3
The total numbers of possible combination of three numbers chosen or selected from six numbers
are:
6!
- = 20
6c3 = 3!(6 3)!
Now, no graduate students are on the subcommittee. So, the 3 numbers must come from 4
4!
undergraduate members. The number of such combination is: 4c3 = 3!(4 - 3)!
= 4
The probability that there will be no graduate students on the subcommittee is:
c = 20
4 3
c
6 3
4 = 5
1 or 0.2
12) Baseball’s American league East has five teams. You are required to predict, in order the top
three teams at the end of the session. Ignoring the possibility of ties calculate the number of
different predictions you could make .what is the probability of making the current prediction by
chance?
Answer: Given that, n = 5 and x = 3
The total numbers of possible predict the top three teams chosen from the five teams are:
5!
5p3 = (5 - 3)!
= 60
1 =.017
Now, the probability of making the correct prediction by chance: P (5p3) = 60
13) A manager has four assistants-johns, George, Mary and jean-to assign to four tasks. Each
assistant will be assi8gn to one of the tasks, one assistant to each task:
2

a. How many different arrangements of assignments are possible?
b. if assignments are made at random , what is probability that Mary will be assigned to a
specific task?
4!
- =24
a) the total number of possible arrangement of assignment is: 4p4 = (4 4)!
c p
1 1*3 3
b) the probability that mary will be assigned to a specific task: 4 p
4
6 = 4
= 24
1 =.25
14) The senior management of corporation has decided that in the future, it wishes to divide its
advertising budget between two agencies. Eight agencies are currently being considered for this
work .How many different choices of two agencies are possible?
8!
- = 28
Two agencies could be select 8c2 = 2!(8 2)!
15) You are one of seven female candidates auditioning for two parts- the heroine and her best
friend – in a play. Before the auditions, you know nothing of the other candidates, and assume all
candidates have equal chances for either part.
(a) How many distinct choice are possible for casting the two parts?
(b) In how many of the possibilities in (a) would you be chosen to play the heroine?
(c) In how many of the possibilities in (a) would you be chosen to play her best friend?
(d) Use the results in (a) and (b) to find the probability you will be chosen to play the
heroine. Indicate a more direct way of finding this probability?
(e) Use the results in (a), (b), and (c) to find the probability you will be chosen to play the
one of the two parts. Indicate a more direct way of finding this probability?
Answer:
Given that, n=7 and x= 2
7!
- =42
(a) The possible distinct choices for casting two parts; 7p2 = (7 2)!
(b) The possibilities to play the heroine are; (1c1)´(6c1)= 6
(c) The possibilities to play her best friend; (1c1)´(6c1)= 6
1 p 1 ´ 6 p
1
(d) P(H) = 7 p
2
6 = 7
= 42
1
(e) P(AÈ B) = P(A)+P(B)-P(A Ç B)
1 + 7
= 7
1 - 7
0
3

2
= 7
16) A work crew for a building project is to be made up of two craftsmen and four laborers
selected from a total of five craftsmen and six laborers available.
a. How many different combinations are possible?
b. the brother of one of the craftsmen is a laborer. if the crew is selected at random, what is
the probability that both brothers will be selected.
c. what is the probability that neither brother will be selected?
Answer:
a. The possible combination for work crew is; (5c2) ´ (6c4) =150
b. The probability that both brother will be selected are; (1c1) (4c1) ´(1c1) (5c3) =40
(1 c 1)(4 c 1) ´ (1 c 1)(5 c
3)
The probability is = 40 (5 c 2)(6 c
4)
=.27
= 150
c. The probability that neither brother will be selected are; (1c0)(4c2) ´(1c0) (5c4) =6´5=30
(1 c 0)(4 c 2) ´ (1 c 0)(5 c
4)
The Probability is= 30 (5 c 2)(6 c
4)
=.2
=150
17) A mutual fund company has six funds that invest in the U. S. market, four that invest in
foreign markets. A customer wants to invest in two U.S. funds & two foreign funds.
a. How many different sets of funds from this company could the investor choose?
b. Unknown to this investor, one of the U.S. funds & one of the foreign funds will seriously
under-perform next year. If the investors select funds for purchase at random, what is the
probability that at least one of the chosen funds will seriously under-perform next year?
Answer:
Given that; Company has 6 funds invest in U.S market
Company has 4 funds invest in foreign market
(a) The possible set of funds from this company are; 6C2 * 4C2 = 90
(1 c 0)(5 c 2) ´ (1 c 0)(3 c
2)
(b) P(N)= (6 c 2)(4 c
2)
30 = 3
= 90
1
1 = 3
P(A)=1- P(N)= 1- 3
2
18) It was estimated that 30% of all seniors on a campus were seriously concerned about
employment prospects, 25% were seriously concerned about grades, and 20% were seriously
concerned about both .what is the probability that a randomly chosen senior from this campus is
seriously concerned about at least one of this two things?
Answer:
A=30% or .3; B=25% or .25; A Ç B=20% or .2
4

P (AÈ B) =P (A) +P (B) P - P (AÇ B)
=30%+25%-20%
=35%
19) A music store owner finds that 30% of customers entering the store ask an assistant for help
and that 20% make a purchase before leaving. It also found that 15% of all customers both ask
for assistant and make a purchase. What is the probability that a customer does at least one of
these two things?
Answer: Let,
A denotes “Customer entering the store ask an assistant for help”
B denotes “Customer entering the store make a purchase before leaving”
Given that,
P (A) =30% or .30, P (B) =25% or .25, P (A B) =15% or .15 and P (A B) =?
We know that,
P (A B) = P (A) + P (B) - P (A B)
= .30 +. 20 - .15
= .35
20) Refer to the information in Exercise 19, and consider the two events “customer ask for
assistance” and “customer makes purchase.” In answering the following questions, provide
reasons expressed in terms of probabilities of relevant events.
(a) Are the two events mutually exclusive?
(b) Are the two events collectively exhaustive?
(c) Are the two events statistically independent?
Answer:
a) The two events A and B are not mutually exclusive. Because, P (A B) not equal to 0.
b) The two events A and B are not collectively exhaustive. Because,
P(A B) not equal to P (S) = 1.
c) The two events A and B are not statistically independent. Because, P (A) P (B) not equal to P
(A B).
22) A mail - order firm considers three possible foul – ups in filling an order:
A: The wrong item is sent.
B: The item is list in transit.
C: The item is damaged in transit.
Assume that event A is independent of both B and C and that events B and C are mutually
exclusive. The individual event probabilities are P(A) = .02, P(B) = .01, and P(C) = .04. Find the
probability that at least one of these foul – ups occurs for a randomly chosen order.
Answer: Let,
A = “The wrong item is sent”
B = “The item is lost in transit”
5

C = “The item is damaged in transit”
Given that, P (A) = .02 P (B) = .01 P(C) = .04
Since, Event A is independent of both B and C
P (A B) = P (A) ´ P (B)
= .02 ´ .01
= .0002
And P (A C) = P (A) ´ P(C)
= .02 ´ .04
= .0008
Given that, B and C are mutually exclusive. P (B C) = 0
P (A B C) = P (A) + P (B) + P(C) - P (A B) – P (B C) – P(C A) + P (A B C)
= .02 + .01 + .04 - 0.0002 – 0 - 0.0008 + 0 = 0.069
23) A coach recruits for a college team a star player who is currently a high school senior. In
order to play next year, the senior must both complete high school with adequate grades and pass
a standardized test. The coach estimates that the probability the athlete will fail to obtain
adequate high school grader is .02, the probability the athlete will not pass the standardized test
is .15, and that these are independent events. According to these estimates, what is the probability
this recruit will be eligible to play in college next year?
Answer:
Let,
A = “The athlete received adequate grade”
B = “The athlete passes the standardize test”
So, Given that, P ( A ) = 0.02 and, P ( B ) = 0.15
P (A) = 1 – P ( A ) P (B) = 1 - P ( B )
= 1- 0.02 = 1 – 0.15
= 0.98 = 0.85
Since, A and B are independent events. So, A and B are also independent.
P (A B) = P (A) ´ P (B)
= 0.98 ´ 0.85
= 0.833
So, the probability for this recruit will be play in the next year is 0.833.
24) Market research in a particular city indicated that during a week 18% of all adults watch a
television program oriented to business and financial issues, 12% read a publication oriented to
these issues, and 10% do both.
a) What is the probability that an adult in this city, who watches a television program
oriented to business and financial issues, reads a publication oriented to these issues?
b) What is the probability that an adult in this city, who reads a publication oriented to
business and financial issues, watches a television program oriented to these issues?
Answer: Let,
A = “Adults watch a TV program oriented to business & financial issues”
B = “Adults read a publication oriented to business and financial issues”
6

Given that, P (A) = .18 P (B) = .12 P (A B) = .10
P A B
( ) = .10
=
P A
a) P (B/A) = 0.56
.18
( )
P A B
( ) = .10
=
P B
b) P (A/B) = 0.83
.12
( )
25) An inspector examines item coming from an assembly line .A review of her record reveals
that she accepts only 8% of all defect items. It was also found that 1% of all items from the
assembly line are both defective and accepted by the inspector. What is the probability that a
randomly chosen item from this assembly line is defective?
Answer:
Let
A denotes “An assembly line product is defective”.
B denotes “An assembly line product is accepted”.
So,
P(B / A) =.08 P(AÇB) =.01
We Know that,
P B A P A B
( / ) ( )
P A
P A = P A Ç
B
( ) ( )
( / )
( ) .01
.08
=
P A
( ) 0.125
( )
=
= Ç
P A
P B A
So, the probability of randomly chosen item from this assemble line product is defect is 0.125
26) An analyst is presented with least of four stocks and five bonds .He is asked to predict, in
order the two stocks that will yield that highest return over the next year and the two bonds that
will have the highest return over the next year. Suppose that these predictions are made randomly
and independently of each other. What is the probability that the analyst will be successful in at
least one of the two tasks?
Answer: Let,
S is denote Stock and B is denote Bond
S=4p2=12 P (S) = 12
1 B=5p2=20 P (B) = 1
20
7

1 ´ 20
P (SÇ B) = 12
1 = 1
240
P (SÈ B) =P (A) +P (B) -P (AÇ B)
1 + 20
=12
1 - 1
240
20 +12 +1
=.129
= 240
27) A bank classifies borrowers as high risk or low risk. Only 15% of its loans are made to those
in the high risk category .of all its loans, 5% are in default, and 40% of those in default are to
high risk borrowers. What is the probability that a high risk borrower will default?
Answer:
Let,
H denote ‘High risk category borrowers’
D denote ‘In default are to high-risk borrowers’
So, P(H)=.15 P(D)=.05 P(H I D)=.40
Now,
P ( HID ) P ( D
)
P(D I H)= .40´.05 P ( H
)
=.133
= .15
28) A conference began at noon with two parallel sessions. The session on portfolio management
was attended by 40% of the delegates, while the session on Chartism was attended by 50%. The
evening session consisted of a talk titled “is the random walk dead?” This was attended by 80%
of delegates.
a) If attendance at the session on portfolio management and Chartism was mutually exclusive,
what is the probability that a random chosen delegate attended at least one of the session?
b) If attendance of the portfolio management and evening sessions are statistically
independent, what is the probability that a randomly chosen delegate attended at least one of the
session?
c) Of those attending the Chartism session, 75% also attended the evening session. What is the
probability that a randomly chosen delegate attended at least one of these two sessions?
Answer:
Let
A denote be the probability on portfolio management
B denote be the probability on Chartism
C denote be the probability on all delegates
Given that, P(A)=.40 P(B)=.50 P(C)=.80
(a) P(A B)= P(A)+P(B)-P(A B)
= .40+ .50 -0
= .90
8

(B) the portfolio management and evening sessions are statistically independent
So, P(A C)= P(A)+P(C)-P(A C)
=.40+.80-P(A)P(C)
=1.20-(.40*.80)
=1.20-.32
=.88
(C) Here, Of those attending the Chartism session, 75% also attended the evening session.
So, P(C I B)= .75
P(B C)= P(B)+P(C)-P(B C)
=.50+.80- P(B)P(C)
=1.30- (.50 X .75)
=1.30- .375
=.925
=92.5%
29) A stock market analyst claims expertise in picking stocks that will outperform the
corresponding industry norms. This analyst is presented with a list of five high technology stocks
and a list of five airlines stock, and she is invited to nominate, in order the three stocks that will
do best on each of these two lists over the next year. The analyst claims that success in just one of
these two tasks would be a substantial accomplishment .If, in fact, the choices were made
randomly and independently, what would be the probability of success in at least one of the two
tasks merely by chance? Given this result, what do you think of the analysts claim?
Answer:
1
A=5p3=60 P (A) = 60
1
B=5p3=60 P (B) = 60
1 ´ 60
(AÇ B)=P (A) +P (B) = 60
1 = 1
3600
P (AÈ B) =P (A) +P (B) P - P (AÇ B)
1 + 60
= 60
1 - 1
3600
119 or .033055555
= 3600
I think that the analyst claim is not right.
30) A quality control manager found that 30% of worker- related problems occurred on
Mondays, and that 20% occurred on the last hour of a day’s shift. It was also found that 4% of
worker -related problem occurred in the last hour of Monday’s shift.
a) What is the probability that a worker- related problem that occurs on a Monday does not
occur in the last hour of the days shift?
b) Are the events “Problems occur on Monday” and “problem occurs on last hour of the day’s
shift” statistically independent?
9

Answer:
Let,
A denote “worker related problem occur on Monday”
B denote “worker related problem in last hour of a day shift”
So, P(A)= .30 P(B)=.20 P(A B)=.04
P A B Here, P(B) = 1-P(B) =1- .20 =.80
( )
P A
(a) P (B A) = ( )
.26 We know, P (A) = P(A B)U
= .30
P(A B)
=.867 So, .30 = .04+ P(A B)
Þ P(A B) = .30-.04
Þ P(A B) =.26
(b) P(A B)=P(A) P(B)
P(A B)= .30X.20
P(A B)= .06
Þ.04 ¹ .06
So, P(A B) ¹ P(A)P(B)
So, the two events are not statistically independent. Because P(AÇB) ¹P(A)´P(B) .
31) A corporation was concerned about the basic educational skills of its workers and decided to
offer a selected group of them separate classes in reading and practical mathematics. Forty
percent of these workers signed up for the reading classes, and 50% for the practical
mathematics classes. Of those signing up for reading classes, 30% signed up for the mathematics
classes.
(a)What is the probability that a randomly selected worker signed up for both classes?
(b)What is the probability that a randomly selected worker who signed up for the
mathematics classes also signed up for the reading classes?
(c)What is the probability that a randomly chosen worker signed up for at least one of
these two classes?
(d)Are the events “Signs up for reading classes” and “Signs up for mathematics classes”
statistically independent?
Answer: Let, A denotes that “Classes in reading.”
B denotes that “Classes in practical mathematics.”
Given that, P(A)= .40 P(B)= .50 P(B I A)=.30
P A B = ]
( )
P B
(a) P(A B)= P(A) P(B I A) [P(A I B)= ( )
= .40 ´ .30
10

=.12
P A B =
( )
P B
(b) P(A I B) ( )
P BIA P A
( ) ( )
P B
= ( )
.30´.40 =.24
= .50
(c) P(AUB)= P(A)+P(B)-P(A B)
=.40+.50-P(A)P(B I A)
=.90- (.40´.30)
=.90-.12 =.78
(d) P(A B)=P(A)P(B)
P(A B)=.40´.50
P(A B)=.20
But, .12 ¹ .20
So, this two events are not statistically independent because they are not P(A B)= P(A)P(B).
32) A lawn care service makes telephone solicitation, seeking customers for the coming season. A
review of the records indicated that 15%of these solicitations produced new customers, and that,
of these new customers. 80% had used some rival service in the previous year. It was also
estimated that, of all solicitation call made. 60% were to people who had used a rival service the
previous year.
What is the probability that a call to a person who used a rival service the previous year will
produce a new customer for the lawn care service?
Answer:
Given that, P(A)= .15 P(B I A)= .80 P(B)= .60
P ( A B )
P(A I B) =
P ( B
)
P A P B A
( ) ( )
P B
= ( )
.15´.80 = .60
= .60
.12 =.20
33) An editor may use all, some, or none of three possible strategies to enhance the sales of a
book:
A: An expensive prepublication promotion
B: An expensive cover design
C: A bonus for sales representatives who meet predetermined sales levels
In the past, these three strategies have been applied simultaneously to only 2% of the company’s
books. Twenty percent of the books have had expensive cover designs, and of these, 80% have
had expensive prepublication promotion. A rival editor learns that a new books is to have both
11

expensive prepublication promotion and cover design and now wants to know how likely it is that
a bonus scheme for sales representatives will be introduced. Compute the probability of interest
to the rival editor.
Answer:
Given,
12

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Statistics assignment 5