BA 1040 Seminar 2 2014 - 2015

BA 1040 Seminar 2

Question 1 (Chapter 4: Joint Probability, Complement, and Marginal Probability)

The probability that house sales will increase in the next 6 months is estimated to be 0.25. The

probability that the interest rates on housing loans will go up in the same period is estimated

to be 0.74. The probability that house sales or interest rates will go up during the next 6

months is estimated to be 0.89.

1.1 The probability that both house sales and interest rates will increase during the next 6

months is

Answer: 0.10

P(A) = the house sales will increase in the next 6 months = 0.25

P(B) = the interest rates on housing loans will go up in the same period = 0.74

P(A or B) = the house sales or interest rates will go up during the next 6 months = 0.89

P( A and B) = the both house sales and interest rates will increase during the next 6 months

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

P( A and B) = (0.25+ 0.74) - 0.89 = 0.10

1.2 The probability that neither house sales nor interest rates will increase during the next 6

months is

Answer: 0.11

(P(A or B)’ = neither house sales nor interest rates will increase during the next 6 months

(P(A or B)’ = 1- (P(A or B) = 1-0.89 = 0.11

1.3 The probability that house sales will increase but interest rates will not during the next 6

months is

Answer: 0.15

P(A and Not B) = the house sales will increase but interest rates will not during the next 6

months

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

P(A and B) + P(A and Not B) = P(A)

P(A and Not B) = P(A) - P(A and B)

P(A and Not B) = 0.25 – 0.1 = 0.15

Question 2 (Chapter 4: Empirical Classical Probability, Conditional Probability, Marginal

Probability Joint Probability, Multiplication Rule)

A survey is taken among customers of a fast-food restaurant to determine preference for

hamburger or chicken. Of 200 respondents selected, 75 were children and 125 were adults.

120 preferred hamburger and 80 preferred chicken. 55 of the children preferred hamburger.

Hamburger Chicken Total

Adults 65 60 125

Children 55 20 75

Total 120 80 200

2.1 Referring to table above, the probability that a randomly selected individual is an adult is

________.

Answer: 125/200 or 62.5%

Column percent table

Hamburger % Chicken % Total %

Adults 65 54.2% 60 75.0% 125 62.5%

Children 55 45.8% 20 25.0% 75 37.5%

Total 120 100.0% 80 100.0% 200 100.0%

2.2 Referring to table above, the probability that a randomly selected individual is a child

and prefers chicken is ________.

Answer: 20/200 or 10%

2.3 Referring to Table 4-3, assume we know the person is a child. The probability that this

individual prefers hamburger is ________.

Answer: 55/75 or 73.3%

Row percent table


Adults 65 52.0% 60 48.0% 125 100.0%

Children 55 73.3% 20 26.7% 75 100.0%

Total 120 60.0% 80 40.0% 200 100.0%

2.4 Referring to Table 4-3, assume we know that a person prefers chicken. The probability that

this individual is an adult is ________.

Answer: 60/80 or 75%

Column percent table


Adults 65 54.2% 60 75.0% 125 62.5%

Children 55 45.8% 20 25.0% 75 37.5%

Total 120 100.0% 80 100.0% 200 100.0%

Question 3 (Chapter 5: Mean, Standard Deviation, Variance, Covariance, portfolio)

Two different designs on a new line of winter jackets for the coming winter are available for

your manufacturing plants. Your profit (in thousands of dollars) will depend on the taste of the

consumers when winter arrives. The probability of the three possible different tastes of the

consumers and the corresponding profits are presented in the following table.

Probability Taste Design A Design B

0.2 more conservative 180 520

0.5 no change 230 310

0.3 more liberal 350 270

3.1 Referring to table above, what is your expected profit when Design A is chosen?

Answer: $256 thousands or $256,000

3.2 Referring to table above, what is the variance of your profit when Design B is chosen?

Answer: 8,400 × 1,0002 or 8,400,000,000

3.3 Referring to table above, what is the standard deviation of your profit when Design A is

chosen?

Answer: $64.37391 thousands or $64,373.91

SD (A) =SQRT (4,144) = 64.37

SD (B) =SQRT (8,400) = 91.65

3.4 Referring to table above, what is the covariance of the profits from the two different

designs?

Answer: -4,320 × 1,0002 or -4,320,000,000

3.5 Referring to table above, what is the expected profit if you increase the shift of your

production lines and choose to produce both designs?


3.6 Referring to table above, if you decide to choose Design A for half of the production lines

and Design B for the other half, what is your expected profit?


E(P) = (0.5)(256)+(0.5)(340) = 298


and Design B for the other half, what is the risk of your investment?

Answer: $31.241 thousands or $31,241


and Design B for the other half, what is the coefficient of variation of your investment?

Answer: 10.48%

CV =

X 100

CV= (31.24/298) X 100 = 10.48%

Question 4 (Chapter 5: Binomial Distribution)

A cosmetics consumer research report shows that 65% of female customers prefer Brand A

Body lotion. A sample of 4 female customer is to be selected

4.1 The probability that at most 2 prefer brand A is ________

4.2 The probability that more than 3 prefer brand A is ________

4.3 The average number that you would expect to prefer brand A is ________

4.4 The variance of the number that prefer brand A is ________

Answer:

4.1 P(X=0,1,2|n=4, =0.65) = 0.0150 + 0.1115 + 0.3105 = 0.4370

4.2 P(X=4 |n=4, =0.65) = 0.1785

4.3 E(X) = n = 4(0.65) = 2.6

4.4 = = 0.91

Question 5 (Chapter 5: Poisson distribution)

A major hotel chain keeps a record of the number of mishandled bags per 1,000 customers.

In a recent year, the hotel chain had 4.06 mishandled bags per 1,000 customers. Assume that

the number of mishandled bags has a Poisson distribution.

X

(bags/1000Customers) λ

0 4.06

1 4.06

2 4.06

3 4.06

4 4.06

5.1 What is the probability that in the next 1,000 customers, the hotel chain will have no

mishandled bags?

Answer: 0.0172

5.2 What is the probability that in the next 1,000 customers, the hotel chain will have at least

two mishandled bags?

Answer: 0.5298

P(X>= 2) = 0.1421 + 0.1924 + 0.1953 = 0.5298

5.3 What is the probability that in the next 1,000 customers, the hotel chain will have no more

than three mishandled bags?

Answer: 0.4218

P(X<=3) = 0.0172 + 0.07002 + 0.1422+ 0.1924 = 0.4218

Question 6 (Chapter 6 : Normal Distribution, Probability

You were told that the mean score on a statistics exam is 75 with the scores normally

distributed. In addition, you know the probability of a score between 55 and 60 is 4.41% and

that the probability of a score greater than 90 is 6.68%.

6.1 What is the probability of a score between 90 and 95?

Answer: 4.46% or 0.0446

6.2 What is the probability of a score between 75 and 90?

Answer: 43.32% or 0.4332

6.3 The middle 86.64% of the students will score between which two scores?

Answer: 59.078 and 90.093

Question 7 (Chapter6: standardized normal distribution, probability)

7.1 Suppose Z has a standard normal distribution with a mean of 0 and standard deviation of

1. The probability that Z is between -0.88 and 2.29 is ________.

Answer: 0.7996

Mean= 0

SD= 1

P(X=-0.88)= 0.1894

P(X=2.29)= 0.9890

P(-0.88<X<2.29)= 0.7996


1. The probability that Z values are larger than ________ is 0.3483

Answer:

1-P(X)=0.3483

P(X)=0.6517

Z=0.39

X=0+(0.39*1)


1. So, 85% of the possible Z values are smaller than ________.

Answer: 1.036

P(X)=0.85

Z=1.036

X=0+(1.036X1)=1.036


1. So, 50% of the possible Z values are between ________ and ________ (symmetrically

distributed about the mean).

Answer: -0.67 and 0.67 or -0.68 and 0.68

P(X)=0.5

P(X-Lower)=0.25

P(X-Upper)=0.75

Z-Lower=-0.6745

Z-Upper=0.6745

X-Lower=0+(-0.6745X1)= -0.6745

X-Upper=0+(0.6745X1)= 0.6745

Question 8 (Chapter 7: Sampling Distribution, Mean, Unbiased, Central Limit Theorem,

Standard Error)

A manufacturer of power tools claims that the mean amount of time required to assemble

their top-of-the-line table saw is 80 minutes with a standard deviation of 40 minutes. Suppose

a random sample of 64 purchasers of this table saw is taken.

8.1 The standard deviation of the sampling distribution of the sample mean is ________

minutes.

Answer: 5

8.2 The probability that the sample mean will be between 77 and 89

Answer: 0.6898

8.3 The middle 95% of the sample means based on sample sizes of 64 will be between

________ and ________.

Answer: 70.2 and 89.8 minutes

8.4 The probability that the sample mean will be greater than 88 minutes is ________.

Answer: 0.0548

Question 9 (Chapter 8: Confidence Interval, Proportion, Properties, Width, Mean, Sample Size

Determination)

9.1 The width of a confidence interval estimate for a proportion will be

A) narrower for 99% confidence than for 95% confidence.

B) wider for a sample size of 100 than for a sample size of 50.

C) narrower for 90% confidence than for 95% confidence.

D) narrower when the sample proportion is 0.50 than when the sample proportion is 0.20.

Answer: C

9.2 When determining the sample size for a proportion for a given level of confidence and

sampling error, the closer to 0.50 that π is estimated to be, the sample size required

A) is smaller.

B) is larger.

C) is not affected.

D) can be smaller, larger or unaffected.

Answer: B

9.3 Suppose a 95% confidence interval for μ turns out to be (1,000, 2,100). To make more

useful inferences from the data, it is desired to reduce the width of the confidence interval.

Which of the following will result in a reduced interval width?

A) Increase the sample size.

B) Increase the confidence level.

C) Increase the population mean.

D) Increase the sample mean.

Answer: A

9.4In the construction of confidence intervals, if all other quantities are unchanged, an

increase in the sample size will lead to a ________ interval.

A) narrower

B) wider

C) less significant

D) biased

Answer: A

9.5 A 99% confidence interval estimate can be interpreted to mean that

A) if all possible sample sizes of n are taken and confidence interval estimates are

developed, 99% of them would include the true population mean somewhere within their

interval.

B) we have 99% confidence that we have selected a sample whose interval does include

the population mean.

C) Both of the above.

D) None of the above.

Answer: C

Question 10 (Chapter 8: Mean, Sample Size Determination, Confidence Interval, Mean,

Standardized Normal Distribution)

The managers of a company are worried about the morale of their employees. In order to

determine if a problem in this area exists, they decide to evaluate the attitudes of their

employees with a standardized test. They select the Fortunato test of job satisfaction, which

has a known standard deviation of 24 points.

10.1 Referring to the given information, they should sample ________ employees if they want

to estimate the mean score of the employees within 5 points with 90% confidence.

Answer: 62.33

10.2 Referring to the given information, due to financial limitations, the managers decide to

take a sample of 45 employees. This yields a mean score of 88.0 points. A 90% confidence

interval would go from ________ to ________.

Answer: 82.12 to 93.88

10.3 True or False: Referring to the given information, this confidence interval is only valid if the

scores on the Fortunato test are normally distributed.

Answer: FALSE

Explanation: With a sample size of 45, this confidence interval will still be valid if the scores

are not normally distributed due to the central limit theorem.

Difficulty: Difficult

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BA 1040 Seminar 2 2014 - 2015