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)

And, Probability distribution for marketing strategies;

Bi B1 B2 B3

P(Bi)

P = P(A1)P(B1)

P( = X = =.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 =5040b) 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

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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 = 3The possible numbers of making the correct selection of 3 common stock out of the 6 common

stock are: 6p3 = = 120

The probability of making the correct selection by chance is: P (6p3) =

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 = 3The total numbers of possible combination of three numbers chosen or selected from six numbers are:

6c3 = = 20

Now, no graduate students are on the subcommittee. So, the 3 numbers must come from 4

undergraduate members. The number of such combination is: 4c3 = = 4

The probability that there will be no graduate students on the subcommittee is:

= = 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:

5p3 = = 60

Now, the probability of making the correct prediction by chance: P (5p3) = =.017

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:

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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?


a) the total number of possible arrangement of assignment is: 4p4 = =24

b) the probability that mary will be assigned to a specific task: = = =.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?


Two agencies could be select 8c2 = = 28

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

(a) The possible distinct choices for casting two parts; 7p2 = =42

(b) The possibilities to play the heroine are; (1c1) (6c1)= 6

(c) The possibilities to play her best friend; (1c1) (6c1)= 6

(d) P(H) = = =

(e) P(A B) = P(A)+P(B)-P(A B)

= + -

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=

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) =150b. The probability that both brother will be selected are; (1c1) (4c1) (1c1) (5c3) =40

The probability is = = =.27

c. The probability that neither brother will be selected are; (1c0)(4c2) (1c0) (5c4) =6 5=30

The Probability is= = =.2

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

(b) P(N)= = =

P(A)=1- P(N)= 1- =

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 B=20% or .2

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P (A B) =P (A) +P (B) P - P (A 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”

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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 ( ) = 0.02 and, P ( ) = 0.15

P (A) = 1 – P ( ) P (B) = 1 - P ( ) = 1- 0.02 = 1 – 0.15 = 0.98 = 0.85Since, and 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”

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B = “Adults read a publication oriented to business and financial issues”

Given that, P (A) = .18 P (B) = .12 P (A B) = .10

a) P (B/A) =

b) P (A/B) =

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,

We Know that,

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

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S=4p2=12 P (S) = B=5p2=20 P (B) =

P (S B) = =

P (S B) =P (A) +P (B) -P (A B)

= + -

=

=.129

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)=.40Now,

P(D I H)= = =.133

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 managementB denote be the probability on ChartismC 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)

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= .40+ .50 -0 = .90

(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:

A=5p3=60 P (A) =

B=5p3=60 P (B) =

(A B)=P (A) +P (B) = =

P (A B) =P (A) +P (B) P - P (A B)

= + -

= or .033055555

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?

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b) Are the events “Problems occur on Monday” and “problem occurs on last hour of the day’s shift” statistically independent?

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

(a) P = Here, = 1-P(B) =1- .20 =.80

= We know, P (A) = U

=.867 So, .30 = .04+ = .30-.04 =.26

(b) P(A B)=P(A) P(B) P(A B)= .30X.20 P(A B)= .06

.04 .06 So, P(A B) P(A)P(B)

So, the two events are not statistically independent. Because .

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

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(a) P(A B)= P(A) P(B I A) [ P(A I B)= ]

= .40 .30 =.12

(b) P(A I B)

= = =.24

(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 .50P(A B)=.20 But, .12 .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 I B)

= = = =.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 promotionB: An expensive cover design C: A bonus for sales representatives who meet predetermined sales levels

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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 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,

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Education

Statistics assignment 5