QMB

Alison has all her money invested in two mutual funds, A and B. She knows that there is a 40% chance that fund A will rise in price, and a 60% chance that fund B will rise in price given that fund A rises in price. What is the probability that both fund A and fund B will rise in price?

A|B= AnB/A .6= AnB/.4 .4*.60=.24

An instructor grades on a curve (normal distribution) and your grade for each test is determined by the following where S = your score. A-grade: S ≥ μ + 2σ B-grade: μ + σ ≤ S < μ + 2σ C-grade: μ - σ ≤ S < μ + σ D-grade: μ - 2σ ≤ S < μ - σ F-grade: S < μ − 2σ If on a particular test, the average on the test was μ = 66, the standard deviation was σ = 15. What is the range of scores for a B-grade?

66+15 = 81 < S< 66+30 81<s<96

5. If A and B are independent events with P(A) = 0.05 and P(B) = 0.65, then P(A l B) = A) 0.05 B) 0.0325 C) 0.65 D) 0.8 E) None of the above

.05/.65 = .05

You own a company that employs three salespeople, Quentin, Michael and Bubba. Quentin makes 30% of the sales, Michael 45%, and Bubba 25%. While they are all competent, they do not enjoy completing the paperwork on their sales as required by the SEC. Of the sales Quentin makes, he completes paperwork 90% of the time. When Michael makes a sale, he completes the paperwork 80% of the time. WhenBubba makes a sale, he completes paperwork 85% of the time. You were recently fined by the SEC for this reason, and have instituted tough new policies on this manner. What is the probability that Bubba makes a sale and the paperwork on the sale is completed?

.25*.85=.2125

Forty percent of all high school graduates work during the summer to earn money for college tuition for the upcoming fall term. Assuming a binomial distribution, if 6 graduates are selected at random, what is the probability that 3 of these graduates have a summer job?

.27648

An investment service is currently recommending the purchase of shares of Dollar Department Store selling at $18 per share. The price is approximately normally distributed with a mean of 20 and a standard deviation of 2. What is the probability that in a year the shares will be selling for: More than $21

.3085

An investment service is currently recommending the purchase of shares of Dollar Department Store selling at $18 per share. The price is approximately normally distributed with a mean of 20 and a standard deviation of 2. What is the probability that in a year the shares will be selling for: More than $19

.6915

1. When a particular machine is functioning properly, 80% of the items produced are non-defective. If three items are examined, what is the probability that none is non- defective? Use the binomial probability function to answer this question.

1) p = 0.80 , n = 3 P[X=0] C(3,0) * (.80)^0 * (1-.80)^3 =0.008 or by using Excel command =BINOMDIST(0,3,0.8,FALSE)

9. Mr. Jones recently purchased a used car and was advised that it breaks down at an average rate of 2.5 per week. If the time between breakdowns follows an exponential distribution, what is the probability that Jones will go more than two weeks without the car breaking down?

1- expondist(2,2.5,true) .006738

A student is applying to Harvard and Dartmouth. If the student is accepted at Dartmouth, the probability of being accepted at Harvard is 40%. If the student is not accepted at Dartmouth there is an 80% of non-acceptance at Harvard. There is a 50% chance of being accepted at Dartmouth. 1. What's the probability a student is accepted at Harvard? 2. What's the probability that a student is accepted at Harvard or accepted at Darthmouth? 3. What's the probability a student is accepted at either Harvard or Dartmouth?

1-.5=.5 1-.4= .6 1-.8= .2 #1 .4(.5)+.2(.5) =.3 #2 .3+.5-(.4)(.5) = .3+.5-.2=.6 #3 .6-(.4)(.5) .6-.2=.4

In a sample of 1000 cases, the mean of a certain test is 14 and the standard deviation is 2.5. Assume the distribution to be normal. How many students score between 12 and 15?

15-14/2.5 - 12-14/2.5 normdist(.40)- normdist(-.80) =.4435 (look it up on Z table) .6554-.2119= .4435

The starting salaries of individuals with an MBA degree are normally distributed with a mean of $40,000 and a standard deviation of $5,000. What is the probability that a randomly selected individual with an MBA degree will get a starting salary of at least $30,000?

(30,000-40,000/5000)=.9772

Two independent events with A= .4 and B= .2 therefore

(ANB) = A*B .4 *.2 = .08 (AUB)= A+B-ANB .4+.2-.08=.52

In a sample of 1000 cases, the mean of a certain test is 14 and the standard deviation is 2.5. Assume the distribution to be normal. How many score above 18? Group of answer choices

18-14/2.5 = 4/2.5 = 1.6Solution:Given Mean =14Standard devaition = 2.5Sample size = 1000We need to calculate P(Xbar>18)=1-P(Xbar<=18)Z = (18-14)/2.5 = 4/2.5 = 1.6 From Z table we foundP(Xbar>18) = 1-0.9452 = 0.0548So there is 5.48% score above than 18.So in 1000 cases there is (1000*5.48%) = 54.80 So its answer is D. i.e. 54.8

8. A financial analyst recommends the purchase of shares of Family Dollar department store selling at $18 per share. The price follows a normal distribution with a mean of 20 and a standard deviation of 2. What is the probability that, in the near future, the shares will be selling for less than $18?

18-20/2 z= -1 z= .1587

An investment service is currently recommending the purchase of shares of Dollar Department Store selling at $18 per share. The price is approximately normally distributed with a mean of 20 and a standard deviation of 2. What is the probability that in a year the shares will be selling for: Less than $18

18-20/2 = -1 z=-1 = .1587

2.When a particular machine is functioning properly, 80% of the items produced are non-defective. If three items are examined, what is the probability that at most one is non- defective? Use the binomial probability function to answer this question

2) p = 0.80 , n = 3 =C(3,0) *(.80)^0 *(1-.8)^3 + C(3,1) *(.80)^1 * (1-.8)^2 =0.008 + 0.096 =0.104 Also, by using Excel command =BINOMDIST(1,3,0.8,TRUE)

Suppose the mean number of hurricanes a year is 2.35 and the number can be modeled by a Poisson distribution. What is the probability that during the next 2 years, there's exactly 1 hurricane?

2*2.35= 4.70 e^-4.70 * 4.70^1/ 1! .0427

You are the only bank teller on duty and you want to take a break for 10 minutes but you don't want to miss any customers. Suppose the arrival of customers can be models by a Poisson distribution with mean of 2 customers per hour. What's the probability that no one will arrive in the next 10 minutes?

2/60 per minute 2/6 per 10 mins 1/3 per 10 mins e^-.333 .72

Over the last 100 business days, Harry had 20 customers on 30 of those days, 25 customers on 20 days, 35 customers on 30 days, 40 customers on 10 days, and 45 customers on 10 days. What is the variance of the number of Harry

30 20= .3 * 20 = 6 20* 25 = 500 *25 = 12500 30* 35= 1050 *35 = 36750 10 * 40= 400 * 40 = 16000 10* 45= 450 * 45 = 20250 100 3000. 97500 97500- 3000^2/100 / 100-1 7500/99= 75.75

The time it takes to complete an examination follows an exponential distribution with a mean of 40 minutes. What is the probability of completing the examination in 30 minutes or less?

30|0

In a culinary school of 750 students, soups and salads are the most popular classes. In the Salad classes thereare 300 students and 30 of those students also attend Soup classes. Additionally there are 360 students who areenrolled in Soup classes but not in Salad classes.a. Knowing that a student is enrolled in soup classes, what is the probability that he/she is not enrolled in saladclasses?b. Are Soup classes and Salad classes independent?

360/390= 12/13 300/750 * 390/750 both does not equal soup and salad

The time required to assemble a part of a machine follows an exponential probability distribution. The average time to assemble the part is 10 minutes. What is the probability that the part can be assembled in 7 minutes or less?

4.(a) The probability distribution is given by, f(x) = , x > 0 = 0, o.w. (Ans). (b) The probability that part can be assembled in 7 minutes or less = P(X <= 7) = = 1 - = 0.5034. (Ans). 1/10 e^-7/10 1-e^-7/10 .2442 (c) The probability of completing the assembly in 3 to 7 minutes = = - = 0.2442. (Ans).

# of hours 0-9 freq= 40 10-19 freq=50 20-29 freq=70 30-39 freq= 40 Cumulative relative frequency 10-19

40+50/200 =.45

You own a company that employs three salespeople, Quentin, Michael and Bubba. Quentin makes 30% of the sales, Michael 45%, and Bubba 25%. While they are all competent, they do not enjoy completing the paperwork on their sales as required by the SEC. Of the sales Quentin makes, he completes paperwork 90% of the time. When Michael makes a sale, he completes the paperwork 80% of the time. When Bubba makes a sale, he completes paperwork 85% of the time. You were recently fined by the SEC for this reason, and have instituted tough new policies on this manner. What is the probability of completing the paper work?

90+80+85/3 =85

The following data has been recorded regarding the rental history of a high pressure cleaner. Find the expected demand for the high pressure cleaner. Demand Frequency 0. 20 1 23 2 27 3 14 4 16

= (0*20 + 1*23 + 2*27 + 3*14 + 4*16) / (20 + 23 + 27 + 14 + 16) = 1.83

7. The time it takes to complete a Macro paper follows a normal distribution with a mean of 10 and standard deviation of 3. a. The probability that the paper can be completed between 7 and 13 days is? b. The probability that it may take more than 12 days to complete the paper is? c. The top 25% will complete the paper in approximately?

A 7-10/3 < s< 13-10/3 -1<s<1 .1587 .84134 .8413-.1587 .6826 B z> 12-10/3 1-P(z < .67) 1- .7486 .2514 C .675 = x-10/3 2.025 = x-10 12.023

A student is applying to Harvard and Dartmouth. He estimates that he has a probability of 0.5 of being accepted at Dartmouth and 0.3 of being accepted at Harvard. He further estimates the probability that he is accepted by both is 0.2. (a) What is the probability that he is accepted by at least one of the schools? (b) What is the probability that he is accepted by Dartmouth if he is accepted by Harvard? (c) What is the probability that he is accepted by Harvard if he is accepted by Dartmouth? (d) Is the event

A+B- AnB A. .5+.3-.2=.6 B. .2/.3=.333 C. .2/.5=.4 No because ANB doesnt equal AB

3. Harry recently opened a flower shop and he is reviewing the reports on the number of customers he had during a 90-day period. He had 24 customers on 18 days, 33 customers on 15 days, 30 customers on 27 days, 21 customers on 12 days, and 54 customers on 18 days. a. What is the average number of customers that Harry had during that period? b. What is the standard deviation?

A. 24*18= 432 33*15= 495 30*27= 810 21*12= 252 54*18= 972 90 2961 2961/80= 32.9

An instructor grades on a curve (normal distribution) and your grade for each test is determined by the following where S = your score. A-grade: S ≥ μ + 2σ B-grade: μ + σ ≤ S < μ + 2σ C-grade: μ - σ ≤ S < μ + σ D-grade: μ - 2σ ≤ S < μ - σ F-grade: S < μ − 2σ Using the grading scheme in the above problem, what is the range of scores for a C-grade on a test if the average was μ = 75 and the standard deviation was σ = 6?

A. 75+2*6= 87 B 75+6 = 81 75+2*6=87 81<=score<87 C 75-6= 69 75+6= 81 69<= score <81 D 75-2*6= 63 75-6=69 63<=score<69 F 75-2*6 score<63

A recent survey shows that the probability of a college student drinking alcohol is 0.6. Further, given that the student is over 21 years old, the probability of drinking alcohol is 0.8. It is also known that 30% of the college students are over 21 years old. The probability of drinking or being over 21 years old is __________.

A= .6 B= .3 A|B= .8 A|B = (ANB)/ B .8*.3= .24 A U B = A + B - ANB .6+ .3 -.24 =.66

A company recently announced that it would be going public. The usual suspects, Morgan Stanley, JPMorgan Chase, and Goldman Sachs will be the lead underwriters. The value of the company has been estimated to range from a low of $5billion to a high of $100billion, with $45billion being the most likely value. If there is a 20% chance that the price will be at the low end, a 20% chance that the price will be at the high end, and a 60% chance that the price will be in the middle, what value should the owner expect the company to price at?

Answer: - According to the given data Expected value of the given scenario is calculated by the formula = sum ( outcome value * probability) = $5*0.20 + $100*0.20 + $45*0.60 = $48 billion

Probability distribution of new clients the firm have obtained each month is given Number of new clients Probability 0 %.05 1 %.10 2 %.15 3 %.35 4 %.20 5 %.10 6 %.05

Expected number of new clients per month = 0*.05 + 1*10 + 2*.15 + 3*.35 + 4*.20 + 5*.10 + 6*.05 = 3.05 clients

You own a company that employs three salespeople, Quentin, Michael and Bubba. Quentin makes 30% of the sales, Michael 45%, and Bubba 25%. While they are all competent, they do not enjoy completing the paperwork on their sales as required by the SEC. Of the sales Quentin makes, he completes paperwork 90% of the time. When Michael makes a sale, he completes the paperwork 80% of the time. When Bubba makes a sale, he completes paperwork 85% of the time. You were recently fined by the SEC for this reason, and have instituted tough new policies on this manner. However, you recently selected a file at random and found that the paper work was not in order. Who is the most likely culprit?

Given the paperwork was not in order, the probability it was Michael is 0.5714. Therefore, Michael is the most likely culprit. (.30*.90) .27 + (.45*.80) .36 + (.25* .85) .2125 =

A campus survey of OWLs indicates that 7% are charming, 4% are modest, and 3% are both charming and modest. Find the probability that an OWL is modest or that he/she is charming.

P(Charming) = 0.07 P(Modest) = 0.04 P(Modest and charming) = 0.03 P(modest or charming) = P(Charming) + P(Modest) - P(Modest and charming) = 0.07 + 0.04 - 0.03 = 0.08

The time it takes to complete an examination follows an exponential distribution with a mean of 40 minutes. What is the probability of completing the examination in 30 to 35 minutes?

P(Completing the exam in 30 to 35 minutes) = F(35) - F(30) = 0.0555 Option A is correct.

7. What is the probability that the respondent likes Diet or Not Coke?

P(Diet or Not Coke)= .05+.88-.03=.90 or 90%

In a survey about soft drinks it was found that 5% of the respondents like diet soft drinks. 12% of all respondents liked the brand Coke. Of all the Diet soda drinkers 40% liked Coke. 6. What is the probability that a respondent likes Coke and Diet drinks?

P(Diet or Not Coke)= .05+.88-.03=.90 or 90%

On a Minnesota December day, the probability of snow is 0.30. The probability of a cold day is 0.50. The probability of snow and cold weather is 0.15. Are snow and cold weather independent events? Please show your work on how you arrived to the decision

P(snow)= .30 P(cold) = .50 P(snow and cold) = .15 snow * cold = .15 snow and cold are independent event

An accounts receivable auditor is examining accounts for a client. The accounts receivable balance can be considered as a continuous random variable that exhibits normal distribution characteristics. The mean amount due per account is 5000. The standard deviation is $1000. The auditor selects an account at random. Determine the probability that the account selected by the Auditor has a balance which is outside of the range between $6500 and $7000:

Step 1: Find area to left of 6500: =NORM.DIST(6500,5000,1000,TRUE) =0.9332 Step 2: Find area to right of 7000: =1-NORM.DIST(7000,5000,1000,TRUE) =0.0228 Step 3: Add both areas together to find area outside: 0.9332 + 0.0228 =0.956 or 0.96

If A and B are independent events with P(A) = 0.65 and P(A ∩ B) = 0.26, then P(B) = _____________ SHOW ALL WORK!

The probability of occurring of event AA is P\left( A \right) = 0.65P(A)=0.65 . The probability of intersection of the two events A\& BA&B is P\left( {A \cap B} \right) = 0.26P(A∩B)=0.26 .26/.65=.4

During lunch time, customers arrive at Joe's Lunch counter according to a Poisson distribution with an average of 2 per 30 second period. What is the probability of having more than two arrivals in a two-minute period?

X= More than 2 in 2 minutes Mean= (2 per 30 seconds)*4=8 per 2 minutes In Excel, use the Poisson Distribution function. =1-POISSON.DIST(2,8,TRUE) =.9862 or about 99%

Suppose the length of time (in days) between sales for an automobile salesperson is modeled as an exponential distribution with a mean of 2 days. What is the probability the salesperson goes more than 5 days without a sale?

X= More than 5 Lambda= 1/mean=1/2 In Excel, use the Exponential Distribution function. =1-EXPON.DIST(5,1/2,TRUE) =.0821 or about 8%

What is the probability that exactly 1 out of 10 cars experience a breakdown if the probability of a breakdown is 30%? Assume a Binomial distribution.

X=1 N=10 P=.30 In Excel, use the Binomial Distribution function. =BINOM.DIST(1,10,.30, FALSE)

A company markets two products (Product A and Product B) through mail order. The company will market them in sequence with the first mail order offer for product A. There is a 30% chance that any customer will purchase product A. Product B is offered some months later. It is felt, for product B, that there is a 30% chance of selling product B to a customer if the customer purchased product A and a 5% chance of selling product B to a customer who did not purchase product A. What is the probability of not selling product B to a customer?

a= .30 b|a= .30 b|not a= .05 (not B) b not b a .09 .21 = .30 not a .035 .665 = .70 .125 .875 (B&A)= P(B|A) * P(A) .09 = .30 * .30 P(B & Not A) = P(B| Not A) * P(not A) .035 = .05 * .70 1-.125= .875

1. During the basketball game between Spurs (home team) and Warriors (visitors) the sports Analysts debated whether or not a home team could win if they are ahead at half time. When the home team is ahead at half time they win the game sixty percent of the times. Based on past data the probability that a home team can win and be ahead at half time is 45%. By half time, what is the probability that the Spurs have the highest score?

a= .60 P(a and b) = .45 .45/.60= .75

A quality control department finds that it accepted only 5% of all bad items and it rejected only 1% of good items. A supplier has just delivered a shipment of a certain item. Past records show that only 90% of the parts of that supplier are good. If the department accepts an item, what is the probability that the item is bad? Round your answer to five decimal places.

accepted not accepted good .0009 .891 .9 not good .005 .095 .1 .014 .986 .005 = .05 *.10 .009= .01 *.90 .00558 = .005/.896

5. Based on a survey conducted by Tele.com, 20% of cable customers are willing to switch companies. If a binomial process is assumed with a sample of 40 cable customers, what is the probability that no more than 10 customers are willing to switch companies?

binom dist(10,40,.2,true) .84

10. Suppose the serving time (in minutes) at Le Bistro during lunch hours is modeled as an exponential distribution with a mean of 15 minutes. What is the probability that an efficient server at Le Bistro would take between 8 and 12 minutes to serve customers?

e ^ -8/15 - e ^-12/15 .137317 using excel expondist(x,15,true) for 8 and 12

Suppose the mean number of hurricanes a year is 2.35 and the number can be modeled by a Poisson distribution with this mean. What is the probability of no hurricanes next year?

e^-2.35*2.35^0/1

The salespeople at Owl Realty sell up to 9 houses per month. The probability distribution of a salesperson selling (n) houses in a month is as follows: Sales Probability 0 .05 1 .10 2 .15 3 .20 4 .15 5 6 .10 7 .05 8 .05 9 .05 What are the expected value and the standard deviation for the number houses sold by the salespeople per month, respectively?

expected value: sales * probabilty add them all together to get expected value

Suppose that for a certain football game the probability that the home team will be ahead at half time is 0.60. The probability that the home team will be ahead at half time and also win the game is 0.45. What is the probability that the home team will win this game given that it is ahead at half time?

half time=.60 halftime and final = .45 final | halftime= ? .45/.60 =.75

This problem is about students who may or may not take advantage of the opportunities provided in QMB such as homework. Some of the students pass the course, and some of them do not pass. Research indicates that 40% of the students do the assigned homework. Of the students who do homework, there is an 80% chance they will pass the course. The probability of not passing if the student does not do the homework is 90%. What is the probability of a student not doing homework or passing?

hwk= .40 Pass hwk = .80 not pass not hwk= .90 paass hwk* p(hwk)= pass &hwk Pass= .32+.06 not pass= .08+.54 hwk= .32+.08 nothwk= .06+.54 .80*.40= .32 not pass not hwk * not hwk = not pass and not hwk .90*.60= .54 not hwk or pass = not hwk + pass - not hwk & pass .92= .60+.38 -.06

Given the joint probability table above what is P[Not A|B]?

not a / b get the info from not a divided by B answer .05/ .20 5/20 = 1/4

A campus survey of OWLs indicates that 7% are charming, 4% are modest, and 3% are both charming and modest. Given that a student isn't modest, what is the probability that they are not charming? .5264 .4286 .9583 .5628 .08

not being modest 1-.04=.96 not being charming and not being modest 1-(.07+.04-.03) = .92 Probability of not being charming if student is not modest = 92%/96% = 95.83%

The probability that house sales will increase over the next six months is estimated at 0.25. It is also estimated that the probability is 0.74 that 30 year fixed-loan mortgage rates will increase over this period. Economists estimate that the probability is 0.89 that either housing sales or interest rates will increase. The probability that both house sales and interest rates will increase is estimated at:

sales= .25 rates= .74 sales inc= .89 .89 = .25 + .74 -x .100

Assume the speed of vehicles along a stretch of I-10 has an approximately normal distribution with a mean of 71 mph and a standard deviation of 8 mph. What proportion of the vehicles would be going less than 50 mph?

x-71/8 < 50-71/8 (z< -2.63) .0043

In a sample of 1000 cases, the mean of a certain test is 14 and the standard deviation is 2.5. Assume the distribution to be normal. The top 20% of the students will score how many points above the mean?

x=.20 x-14/2.5= .84 2.5* .84+14 2.1 +14 16.1

In a sample of 1000 cases, the mean of a certain test is 14 and the standard deviation is 2.5. Assume the distribution to be normal. How many score below 18?

z< 18-14/2.5 Z< 1.6 .9452*1000 945.2