Question 1

Dorothy Little purchased a mailing list of 2,000 names and addresses for her mail order business,but after scanning the list she doubts the authenticity of the list.She randomly selects five names from the list for validation.If 40% of the names on the list are non-authentic,and x is the number of non-authentic names in her sample,P(x>0)is ______________.
A)0.2172
B)0.9533
C)0.1846
D)0.9222
E)1.0000

Accepted Answer

The answer of Dorothy Little purchased a mailing list of...

Question 2

Assume that a random variable has a Poisson distribution with a mean of 5 occurrences per ten minutes.The number of occurrences per hour follows a Poisson distribution with λ equal to _________
A)5
B)60
C)30
D)10
E)20

Accepted Answer

The Poisson distribution parameter λ represents the average number of occurrences in a fixed interval of time. Given the mean is 5 occurrences per ten minutes, for an hour (which is six times ten minutes), the mean would be 6 * 5 = 30 occurrences per hour.

Question 3

Dorothy Little purchased a mailing list of 2,000 names and addresses for her mail order business,but after scanning the list she doubts the authenticity of the list.She randomly selects five names from the list for validation.If 40% of the names on the list are non-authentic,and x is the number on non-authentic names in her sample,the expected (average)value of x is ______________.
A)2.50
B)2.00
C)1.50
D)1.25
E)1.35

Accepted Answer

If 40% of the names on the entire list are non-authentic, then the probability of selecting a non-authentic name for each individual selection is also 40%. Thus, the distribution of x, the number of non-authentic names in the sample, follows a binomial distribution with n=5 and p=0.4. 
The expected value of x is given by E(x) = np = 5 * 0.4 = 2. Therefore, the best choice is B) 2.00.

Question 4

Dorothy Little purchased a mailing list of 2,000 names and addresses for her mail order business,but after scanning the list she doubts the authenticity of the list.She randomly selects five names from the list for validation.If 40% of the names on the list are non-authentic,and x is the number of non-authentic names in her sample,P(x=0)is ______________.
A)0.8154
B)0.0467
C)0.0778
D)0.4000
E)0.5000

Accepted Answer

The answer of Dorothy Little purchased a mailing list of...

Question 5

Pinky Bauer,Chief Financial Officer of Harrison Haulers,Inc.,suspects irregularities in the payroll system,and orders an inspection of a random sample of vouchers issued since January 1,2006.A sample of ten vouchers is randomly selected,without replacement,from the population of 2,000 vouchers.Each voucher in the sample is examined for errors and the number of vouchers in the sample with errors is denoted by x.If 20% of the population of vouchers contains errors,the mean value of x is __________.
A)400
B)2
C)200
D)5
E)1

Accepted Answer

The answer of Pinky Bauer,Chief Financial Officer of Harrison Haulers,Inc.,suspects...

Question 6

During a recent sporting event,a quarter is tossed to determine which team picks the starting side.Suppose the referee says the coin will be tossed 3 times and the best two out of three wins If team A calls 'heads',what is the probability that exactly two heads are observed in three tosses?
A).313
B).375
C).625
D).875
E).500

Accepted Answer

There are a total of 2^3 = 8 possible outcomes from flipping a coin 3 times. These outcomes are: HHH, HHT, HTH, THH, HTT, THT, TTH, TTT. Out of these outcomes, there are 3 that result in exactly 2 heads: HHT, HTH, and THH. The probability of getting exactly 2 heads is therefore 3/8, or 0.375, which is answer choice B.

Question 7

The number of cars arriving at a toll booth in five-minute intervals is Poisson distributed with a mean of 3 cars arriving in five-minute time intervals.The probability of 3 cars arriving over a five-minute interval is _______.
A)0.2700
B)0.0498
C)0.2240
D)0.0001
E)0.0020

Accepted Answer

Given, the mean of cars arriving in five-minute intervals is 3. 
The Poisson probability formula is P(X = k) = (e^-λ * λ^k) / k!, where λ is the mean and k is the number of occurrences we want to find the probability for. 
Here, λ = 3 and k = 3. 
Substituting these values in the formula, we get, P(X = 3) = (e^-3 * 3^3) / 3! 
= (0.0498 * 27) / 6 
= 0.2240 
Therefore, the probability of 3 cars arriving over a five-minute interval is 0.2240 or approximately 0.22. Option C is the correct choice.

Question 8

A large industrial firm allows a discount on any invoice that is paid within 30 days.Of all invoices,10% receive the discount.In a company audit,15 invoices are sampled at random.The mean (average)value of the number of the 15 sampled invoices that receive discount is _______
A)1
B)3
C)1.5
D)2
E)10

Accepted Answer

The probability of an invoice receiving the discount is 10%, or 0.1. The number of sampled invoices receiving the discount follows a binomial distribution with parameters n=15 and p=0.1. The mean value of a binomial distribution is np, so the mean value of the number of sampled invoices receiving the discount is 15 x 0.1 = 1.5. Therefore, the answer is C.

Question 9

On Monday mornings,customers arrive at the coffee shop drive thru at the rate of 6 cars per fifteen-minute interval.Using the Poisson distribution,the probability that five cars will arrive during the next fifteen-minute interval is _____________.
A)0.1008
B)0.0361
C)0.1339
D)0.1606
E)0.5000

Accepted Answer

The answer of On Monday mornings,customers arrive at the coffee...

Question 10

A large industrial firm allows a discount on any invoice that is paid within 30 days.Of all invoices,10% receive the discount.In a company audit,10 invoices are sampled at random.The probability that fewer than 3 of the 10 sampled invoices receive the discount is approximately_______________.
A)0.1937
B)0.057
C)0.001
D)0.3486
E)0.9298

Accepted Answer

The answer of A large industrial firm allows a discount...

Question 11

Pinky Bauer,Chief Financial Officer of Harrison Haulers,Inc.,suspects irregularities in the payroll system,and orders an inspection of a random sample of vouchers issued since January 1,2006.A sample of ten vouchers is randomly selected,without replacement,from the population of 2,000 vouchers.Each voucher in the sample is examined for errors and the number of vouchers in the sample with errors is denoted by x.If 20% of the population of vouchers contain errors,P(x>0)is _______________.
A)0.8171
B)0.1074
C)0.8926
D)0.3020
E)1.0000

Accepted Answer

The answer of Pinky Bauer,Chief Financial Officer of Harrison Haulers,Inc.,suspects...

Question 12

Pinky Bauer,Chief Financial Officer of Harrison Haulers,Inc.,suspects irregularities in the payroll system,and orders an inspection of a random sample of vouchers issued since January 1,2006.A sample of ten vouchers is randomly selected,without replacement,from the population of 2,000 vouchers.Each voucher in the sample is examined for errors and the number of vouchers in the sample with errors is denoted by x.If 20% of the population of vouchers contains errors,the standard deviation of x is ______.
A)1.26
B)1.60
C)14.14
D)3.16
E)0.00

Accepted Answer

The answer of Pinky Bauer,Chief Financial Officer of Harrison Haulers,Inc.,suspects...

Question 13

Pinky Bauer,Chief Financial Officer of Harrison Haulers,Inc.,suspects irregularities in the payroll system,and orders an inspection of a random sample of vouchers issued since January 1,2006.A sample of ten vouchers is randomly selected,without replacement,from the population of 2,000 vouchers.Each voucher in the sample is examined for errors and the number of vouchers in the sample with errors is denoted by x.If 20% of the population of vouchers contain errors,P(x = 0)is _______________.
A)0.8171
B)0.1074
C)0.8926
D)0.3020
E)0.2000

Accepted Answer

Since 20% of the population of vouchers has errors, the probability that any one voucher in the population contains an error is 0.20. 
The sample size is 10, so the probability that any one of the 10 sampled vouchers does not have an error is 0.80. 
Therefore, the probability that all 10 vouchers in the sample do not have errors is:
0.80^10 = 0.1074
Thus, P(x=0) is 0.1074.

Question 14

A student randomly guesses the answers to a five question true/false test.If there is a 50% chance of guessing correctly on each question,what is the probability that the student misses exactly 1 question?
A)0.200
B)0.031
C)0.156
D)0.073
E)0.001

Accepted Answer

The probability of guessing correctly is 0.5, so the probability of guessing incorrectly is also 0.5. The probability of missing exactly 1 question can be calculated using the binomial distribution formula: P(X=1) = (5 choose 1) * (0.5)^1 * (0.5)^4 = 5 * 0.5 * 0.0625 = 0.15625 ≈ 0.156. Therefore, the answer is C.

Question 15

A student randomly guesses the answers to a five question true/false test.If there is a 50% chance of guessing correctly on each question,what is the probability that the student misses no questions?
A)0.000
B)0.200
C)0.500
D)0.031
E)1.000

Accepted Answer

The probability of getting a question right is 0.5, therefore the probability of getting a question wrong is also 0.5. To find the probability of a student missing no questions, we need to find the probability that the student gets all 5 questions correct, which is (0.5)^5 = 0.031. Therefore, the answer is D, 0.031.

Question 16

It is known that screws produced by a certain company will be defective with probability .01 independently of each other.The company sells the screws in packages of 25 and offers a money-back guarantee that at most 1 of the 25 screws is defective.Using Poisson approximation for binomial distribution,the probability that the company must replace a package is approximately _________
A)0.01
B)0.1947
C)0.7788
D)0.0264
E)0.2211

Accepted Answer

The number of defective screws in a package follows a binomial distribution with parameters n=25 and p=.01. 
Using the Poisson approximation for binomial distribution, we have 
λ = np = 25 x .01 = .25 
Let X be the number of defective screws in a package. 
Then P(X > 1) = 1 - P(X ≤ 1) 
≈ 1 - [P(X=0) + P(X=1)] (using Poisson approximation) 
≈ 1 - [e^(-.25) (1 + .25)] 
≈ 1 - (.7788) 
≈ 0.2212 
Therefore, the probability that the company must replace a package is approximately 0.0264 (closest to option D).

Question 17

Dorothy Little purchased a mailing list of 2,000 names and addresses for her mail order business,but after scanning the list she doubts the authenticity of the list.She randomly selects five names from the list for validation.If 40% of the names on the list are non-authentic,and x is the number of non-authentic names in her sample,P(x<2)is ______________.
A)0.3370
B)0.9853
C)0.9785
D)0.2333
E)0.5000

Accepted Answer

The answer of Dorothy Little purchased a mailing list of...

Question 18

In a certain communications system,there is an average of 1 transmission error per 10 seconds.Assume that the distribution of transmission errors is Poisson.The probability of 1 error in a period of one-half minute is approximately ________
A)0.1493
B)0.3333
C)0.3678
D)0.1336
E)0.03

Accepted Answer

The answer of In a certain communications system,there is an...

Question 19

The number of cars arriving at a toll booth in five-minute intervals is Poisson distributed with a mean of 3 cars arriving in five-minute time intervals.The probability of 5 cars arriving over a five-minute interval is _______.
A)0.0940
B)0.0417
C)0.1500
D)0.1008
E)0.2890

Accepted Answer

The answer of The number of cars arriving at a...

Question 20

Twenty five individuals are randomly selected out of 100 shoppers leaving a local bedding store.Each shopper was asked if they made a purchase during their visit.Each of the shoppers has the same probability of answering 'yes' to having made a purchase.The probability that exactly four of the twenty-five shoppers made a purchase could best be found by _______.
A)using the normal distribution
B)using the binomial distribution
C)using the Poisson distribution
D)using the exponential distribution
E)using the uniform distribution

Accepted Answer

The situation described here fits the criteria for a binomial distribution, where each shopper has the same probability of success (i.e. making a purchase) and the number of trials (i.e. shoppers surveyed) is fixed at 25. The probability of exactly 4 successes can be calculated using the binomial probability formula.

Quiz 5: Discrete Distributions

Assume that a random variable has a Poisson distribution with a mean of 5 occurrences per ten minutes.The number of occurrences per hour follows a Poisson distribution with λ equal to _________

During a recent sporting event,a quarter is tossed to determine which team picks the starting side.Suppose the referee says the coin will be tossed 3 times and the best two out of three wins If team A calls 'heads',what is the probability that exactly two heads are observed in three tosses?

The number of cars arriving at a toll booth in five-minute intervals is Poisson distributed with a mean of 3 cars arriving in five-minute time intervals.The probability of 3 cars arriving over a five-minute interval is _______.

A large industrial firm allows a discount on any invoice that is paid within 30 days.Of all invoices,10% receive the discount.In a company audit,15 invoices are sampled at random.The mean (average) value of the number of the 15 sampled invoices that receive discount is _______

On Monday mornings,customers arrive at the coffee shop drive thru at the rate of 6 cars per fifteen-minute interval.Using the Poisson distribution,the probability that five cars will arrive during the next fifteen-minute interval is _____________.

A large industrial firm allows a discount on any invoice that is paid within 30 days.Of all invoices,10% receive the discount.In a company audit,10 invoices are sampled at random.The probability that fewer than 3 of the 10 sampled invoices receive the discount is approximately_______________.

A student randomly guesses the answers to a five question true/false test.If there is a 50% chance of guessing correctly on each question,what is the probability that the student misses exactly 1 question?

A student randomly guesses the answers to a five question true/false test.If there is a 50% chance of guessing correctly on each question,what is the probability that the student misses no questions?

In a certain communications system,there is an average of 1 transmission error per 10 seconds.Assume that the distribution of transmission errors is Poisson.The probability of 1 error in a period of one-half minute is approximately ________

The number of cars arriving at a toll booth in five-minute intervals is Poisson distributed with a mean of 3 cars arriving in five-minute time intervals.The probability of 5 cars arriving over a five-minute interval is _______.