Quiz 2 Pool - professor mary schindlebeck PDF

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At a computer-consulting firm, the number of new clients that they have obtained each month has ranged from 0 to 6. The number of new clients has the probability distribution that is shown below. Determine the expected number of new clients per month.

A. B. C. D. E.

# of Clients

Probability

0 1 2 3 4 5 6

0.05 0.10 0.15 0.35 0.20 0.10 0.05

3.05 0 4.7 21 none of the answers is correct

Determine the variance of the new clients per month. 2.0475 = ((A1-3.05)^2*(B1)) Other words column 1 # clients- ev ^2 *(prob)) 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? = Low price range × chance percentage + high price range × chance percentage + most likely price range × chance percentage = $5 billion × 20% + $100 billion × 20% + $45 billion × 60% = $1 + $20 + $27 A. B. C. D. E.

42.5 48.0 66.0 49.5 38.5

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 10% chance that the price will be at the high end, and a 70% chance that the price will be in the middle, what value should the owner expect the company to price at? = $5B*20% + $100B*10% + $45*70% = $1 + $10 + $31.5 = 42.5B

When a particular machine is functioning properly, 80% of the items produced are non-defective. Use the binomial probability function to answer this question. A. 0.0080 B. 0.0555 C. 0.2800 D. 0.0960 E. 0.1040 1. If three items are examined, what is the probability that none is non- defective?  0.0080 = BINOM.DIST(0,3,0.8,TRUE) 2. If three items are examined, what is the probability that at most one is non- defective?  0.104 = BINOM.DIST(1,3,0.8,TRUE) Assuming a binomial distribution, four percent of the customers of a mortgage company default on their payments. A sample of five customers is selected. a. 0.9994 b. 0.9588 c. 0.7408 d. 0.0006 e. 0.0142 1. What is the probability that more than 2 customers in the sample will default on their payments?  =1-BINOM.DIST(2,5,0.04,TRUE) = 0.0006 2. What is the probability that at most two customers in the sample will default on their payments?  =BINOM.DIST(2,5,0.04,TRUE) = 0.9994 3. What is the probability that exactly two customers in the sample will default on their payments?  =BINOM.DIST(2,5,0.04,FALSE) = 0.0142 Assuming a Poisson distribution, on the average, 6 cars arrive at the drive-up window of a bank every hour. A. 0.2800 B. 0.1385 C. 0.9161 D. 0.096 E. 0.3406 1. Compute the probability that no more than 5 cars will arrive in the next half-hour.  =POISSON.DIST(5,3,TRUE) = 0.9161 2. Compute the probability that no more than 5 cars will arrive in the next hour.  =POISSON.DIST(5,6,TRUE) = 0.4457 3. Compute the probability that exactly 5 cars will arrive in the next hour.  =POISSON.DIST(5,6.7,FALSE) = 0.1385

A certain university awards scholarship to athletes and to superior academic candidates. Eight percent of the students received a scholarship. 20% of the students were athletes. What percent of the students did not receive a scholarship? 92% (100-8=92%)

The weight of football players is normally distributed with a mean of 200 pounds and a standard deviation of 25 pounds. 1.

What percent of players weigh between 180 and 220 pounds? A. 57.62% B. 28.81% C. 34.13% D. 2.28% E. 95.05% 2. The probability of a player weighing more than 241.25lbs is  0.0495 =1-NORM.DIST(241.25,200,25,TRUE) 3. The probability of a player weighing more than 250lbs is  0.0228 =1-NORM.DIST(250,200,25,TRUE) 4. The probability of a player weighing less than 250lbs is  0.9772 =NORM.DIST(250,200,25,TRUE) A random variable X is exponentially distributed with a mean of 25. 1. What is the probability that X is between 25 and 30? =EXPON.DIST(30,1/25,TRUE)-EXPON.DIST(25,1/25,TRUE) A. 0.0667 B. 0.4493 C. 0.3022 D. 0.7504 2. What is the probability that X is between 15 and 35? 0.3022 =EXPON.DIST(35,1/25,TRUE)-EXPON.DIST(15,1/25,TRUE) X is a normally distributed random variable with a mean of 22 and a standard deviation of 5. 1. The probability that X is less than 9.7 is  0.0069 =NORM.DIST(9.7,22,5,TRUE) 2. The probability that X is more than 9.7 is  0.9931 =1-NORM.DIST(9.7,22,5,TRUE) 3. The probability that X is between 17 and 22 is  0.3413 =NORM.DIST(22,22,5,TRUE)-NORM.DIST(17,22,5,TRUE) 4. The probability that X is between 22 and 27 is  0.3413 =NORM.DIST(27,22,5,TRUE)-NORM.DIST(22,22,5,TRUE) 5. The probability that X is between 9.7 and 22 is  0.4931 =NORM.DIST(22,22,5,TRUE)-NORM.DIST(9.7,22,5,TRUE) 6. The probability that X is between 17 and 27 is  0.6826 =NORM.DIST(27,22,5,TRUE)-NORM.DIST(17,22,5,TRUE) In a binomial experiment the probability of success is 0.06. 1. What is the probability of at most two successes in seven trials? 0.9937 =BINOM.DIST(2,7,0.06,TRUE) 2. What is the probability of two successes in seven trials? 0.555 =BINOM.DIST(2,7,0.06,FALSE)

An experiment consists of four outcomes with P(E1) = 0.2, P(E2) = 0.3, and P(E3) = 0.4. The probability of outcome E4 is 0.10 =1-0.2-0.3-0.4 1. If A and B are independent events with P(A) = 0.65 and P(A ∩ B) = 0.26, then, P(B) = .26/.65= .40 2. If A and B are independent events with P(A) = 0.05 and P(B) = 0.65, then P(A|B) = P(A)*P(B)/P(B)= .05*.65/ .65= 0.0325/.65 = 0.05 3. If A and B are independent events with P(A) = 0.05 and P(B) = 0.65, then P(B|A) = P(B)*P(A)/P(A)= .05*.65/ .05= 0.0325/.05 = 0.65 4. If A and B are independent events with P(A) = 0.4 and P(B) = 0.6, then P(A ∩ B) .4*.8 = 0.24 5. If A and B are mutually exclusive events with P(A)= 0.3 and P(B)= 0.5, then P(A ∪ B) 0.3+0.5 = 0.8 6. If P(A) = 0.68, P(A ∪ B) = 0.91, and P(A ∩ B) = 0.35, then P(B) = 0.58 Use the table below which contains data on 1200 students in a law class. GPA 0.0 - 2.0

GPA 2.01 - 3.0

GPA 3.01 - 4.0

Total

Pass

100

200

540

840

Fail

200

120

40

360

Total

300

320

580

1200

1. What percent of the students failed the class or had a GPA greater than 3.0? 

580/1200= 0.4833, 360/1200=0.3, 40/1200=0.0333 so .4833+.3-.033 =.7533 or 75.33%

2. How many students in the law class have a GPA above 2.0? 

1200 - 300 = 900

3. What percent of the students failed the class? 

360/1200 = 30%

4. What percent of the students have a GPA less than or equal to 2.0? 

300/1200 = 25%

5. If a student passed the course, what is the probability that their GPA was greater than 3.0? 

540/840=.64285 or 64.29%

6. What percent of the students failed the class and had a GPA greater than 3.0? 

40/1200 = 3.33%...