1) Following is the probability distribution of in how mechanical failure during a

particular production process

Number of Days 0 1 2 3 4

Probability 0.1 0.26 0.42 0.16 0.06

a) Find the mean and standard deviation of the number of mechanical failures during a

production process.

b) If it cost the company $1500 in terms of the lost output, what is the mean and standard

deviation of the cost due to the mechanical failures?

a. = 1.82 breakdowns 2 = 1.0276 = 1.0137 breakdowns

Breakdowns P(x) Mean Variance Cost P(x) Mean Variance

0 0.1 0 0.33124 0 0.1 0 745290

1 0.26 0.26 0.174824 1500 0.26 390 393354

2 0.42 0.84 0.013608 3000 0.42 1260 30618

3 0.16 0.48 0.222784 4500 0.16 720 501264

4 0.06 0.24 0.285144 6000 0.06 360 641574

1.00 1.82 1.0276 1.00 2730 2312100

S.D. 1.013706 S.D. 1520.559

b. Cost: C = 1500X

E(C) = 1500(1.82) = = $2,730

= |1500|(1.0137) = $1,520.559

2) You stop in at a local Internet café. The manager tells you that there are four computers in the

cafe, and that the probability for one of them being available at any one time is 0.4. Assume

that this probability is the same for each computer and the probability of one computer being

occupied is statistically independent of any of the other computers.

What is the probability that all four computers are occupied?

Answer: P(All computers are occupied) = (0.6)4 = 0.1296

a) What is the probability that at least one computer is available?

Answer: P(All computers are occupied) = (0.6)4 = 0.1296

P(at least one computer is available ) = 1 - P(all computers are occupied)

= 1 - 0.1296 = 0.8704

3) In a survey of top executives, it was found that 17% had traveled internationally

on business. The probability of one of these executives fluently speaking a foreign language was

found to be 10%. The probability that one of these executives neither spoke a foreign language nor had

traveled internationally was 0.81. What is the probability that an executive who speaks a foreign

language has traveled internationally?

Answer: T =Traveled internationally

S = Speaks a foreign language

S

Total

T 0.08 0.09 0.17

0.02 0.81 0.83

Total 0.10 0.90 1.00

P(T/S) = = = 0.80

4)The probability that a person has an Internet connection at home is 34%. The probability that

they have access to the Internet at work is 40%. The probability that a person who has access

to the Internet at work also has access at home is 55%. What is the probability that a person

with an Internet connection at home also has one at work?

Answer: H = Internet at home

W = Internet at work

P(H) = 0.34, P(W) = 0.40, P(H/W) = 0.55.

Hence, P(H ∩ W)=P(H/W) ∙ P(W) = (0.55)(0.40) = 0.22 and P(W/H) = P(W ∩ H)/P(H) = 0.22/0.34 = 0.647.

5) In a recent survey about capital punishment, 64% of the respondents said that they support

capital punishment. Females comprised 48% of the sample, and of the females, 46% supported

capital punishment.

What is the probability that a randomly selected person is a female and capital punishment supporter?

Answer: S = Capital punishment supporter

F = Female

P(S ∩ F) = P(S | F) ∙ P(F) = (0.46)(0.48) = 0.2208

6) Suppose we select a capital punishment supporter so that we can ask some follow up questions.

What is the probability that the person we select is female?

Answer: S = Capital punishment supporter

F = Female

P(S ∩ F) = P(S | F) ∙ P(F) = (0.46)(0.48) = 0.2208

P(F | S) = P(F ∩ S) / P(S) = 0.2208 / 0.64 = 0.345

What is the probability that a randomly selected person is a male and does support capital

punishment?

Answer: S = Capital punishment supporter

F = Female

M = Male

P(S ∩ F) = P(S | F) ∙ P(F) = (0.46)(0.48) = 0.2208

Since P(S) = P(S ∩ F) + P(S ∩ M) ⇒ 0.64 = 0.2208 + P(S ∩ M), then, P(S ∩ M) = 0.4192

7) Consider a sample space defined by events A1, A2, B1, B2. Let P(A1) = 0.40 , P(B1 ∣ A1) = 0.60

and P(B1 ∣ A2) = 0.70

What is P(A1 ∣ B1)?

Answer: P(B1 ∣ A1) = (0.60) = P(B1 ∩ A1)/P(A1) ⇒ P(B1 ∩ A1) = P(A1 ∩ B1) = (0.60)(0.40) = 0.24

P(A1) = P(A1 ∩ B1) + P(A1 ∩ B2) ⇒ P(A1 ∩ B2) = 0.40 - 0.24 = 0.16

P(B1 ∣ A2) = 0.70 = P(B1 ∩ A2)/P(A2) ⇒ P(B1 ∩ A2) = P(A2 ∩ B1) = (0.70)(0.60) = 0.42

P(B1) = P(A1 ∩ B1) + P(A2 ∩ B1) = 0.24 + 0.42 = 0.66

P(A1 ∣ B1) = P(A1 ∩ B1)/ P(B1) = 0.24 / 0.66 = 0.3636

8) A table of joint probabilities is shown below, regarding the instructors at the University of

Michigan, where the events Ai(i = 1, 2, 3) and Bj (j = 1, 2) are defined as follows: A1 = Full

professor, A2 = Associate professor, A3 = Assistant professor, B1 = Male instructor, and B2 =

Female instructor.

A1 A2 A3

B1 0.15 0.25 0.20

B2 0.10 0.15 0.15

Calculate P(A1 ∣ B1).

Answer: P(A1 ∣ B1) = P(A1 ∩ B1) / P(B1) = 0.15 / 0.60 = 0.25

Show that P(A1 ∣ B1) + P(A2 ∣ B1) + P(A3 ∣ B1) = 1

Answer: Since P(A1 ∩ B1) + P(A2 ∩ B1) + P(A3 ∩ B1) = P(B1); then

P(A1 ∣ B1) + P(A2 ∣ B1) + P(A3 ∣ B1) = P(B1) / P(B1) = 1

Are the events A and B independent? Explain.

Answer: P(A2 ∣ B1) = P(A2 ∩ B1) / P(B1) = 0.25 / 0.60 = 0.4167

P(A2 ∣ B1) = 0.4167, and P(A2) = 0.40. Since P(A2 ∣ B1) ≠ P(A2), we conclude that the two events are

dependent.

9) Three airlines serve a small town in Indiana. Airline A has 60% of all the scheduled flights,

airline B has 30%, and airline C has the remaining 10%. Their on-time rates are 80%, 60%, and

49% respectively. Define event D as an airline arrives on time.

Calculate P(A and D).

Answer: P(A ∩ D) = P(D|A)P(A) = (0.80)(0.60) = 0.48

Calculate P(B ∩ D).

Answer: P(B ∩ D) = P(D|B) P(B) = (0.60)(0.30) = 0.18

Calculate the probability that a plane leaves on time.

Answer: P(D) = P(A ∩ D) + P(B ∩ D) + P(C ∩ D) = 0.48 + 0.18 + 0.049 = 0.709

If a plane has just left on time, what is the probability that it was airline A?

Answer: P(A | D) = P(A ∩ D) / P(D) = 0.48 / 0.709 = 0.677

If a plane has just left 40 minutes late, what is the probability that it was airline A?

Answer: P(A | ) = P(A ∩ ) / P( ) = (0.60)(0.20) / 0.291 = 0.412

10) A general contractor has submitted two bids for two projects; A and B. The probability of

getting project A is 0.60. The probability of getting project B is 0.75. The probability of getting

at least one of the projects is 0.85.

What is the probability that the contractor will get both projects?

Answer: P(A) = 0.60, P(B) = 0.75, and P(A ∪ B) = 0.85

P(A ∪ B) = P(A) + P(B) - P(A ∩ B) implies that 0.85 = 0.60 + 0.75 - P(A ∩ B)

Hence, P(A ∩ B) = 0.50

If the contractor gets project A, what is the probability that he will get project B?

Answer: P(B | A) = P(A ∩ B)/P(A) = 0.50/0.60 = 0.833

9_ A set of data is mounded (Bell Shaped with large number of observations), with a mean of 500

and a variance of 576.

Approximately what proportion of the observations is greater than 476?

Answer: The value 476 is one standard deviation below the mean; hence the empirical rule implies

that the area between 476 and the mean is approximately 0.68 / 2 = 0.34. Therefore the proportion of

the observations greater than 476 is approximately 0.84.

Approximately what proportion of the observations is less than 548?

Answer: The value 548 is two standard deviations above the mean; hence the Empirical Rule implies

that the area between the mean and 548 is approximately 0.95 / 2 = 0.475. Therefore the proportion of

the observations less than 548 is approximately 0.975.

10) A researcher interested in determining the average monthly expenditures of college students on

DVDs finds that for a sample of 25 students, the mean expenditure was $24.40, and the median

expenditure was $21.76. Specify the shape of the histogram for this data. Does this shape make sense?

Why?

Answer: The distribution is skewed to the right, implying that there are a few students (outliers) who

spend a lot of money on DVDs, raising the average above the typical or median student.

11) Given the following frequency distribution:

Average # Stock Trades Frequency

1 but < 4 5

4 but < 7 0

7 but < 10 2

10 but < 13 3

Construct a relative frequency and percent distributions of the data.

Answer:

Average # Stock Trades Relative Frequency Percent

1 but < 4 0.5 50

4 but < 7 0.0 0

7 but < 10 0.2 20

10 but < 13 0.3 30

Construct a cumulative relative frequency and cumulative percent distributions of the data.

Answer:

Average # Stock Trades

Cumulative

Relative Frequency

Cumulative

Percent

< 4 0.5 50

< 7 0.5 50

< 10 0.7 70

< 13 1.0 100

12) Data were collected on the number of people entering an electronics store each hour. The data

are presented below.

23 35 42 28 29 17 38 21 49 52

46 37 25 49 37 25 28 13 29 43

Construct a stem-and-leaf display of the data.

Answer:

Consider following set of data:

2 5 5 7 10 12 13 16 20

a) Find mean median and mode for the above data;

b) Find Inter-quartile range;