Question 1

When surveyed for brand recognition, 98% of consumers recognize Coke. A new survey of 800 randomly selected consumers is to be conducted. For such a group of 800, calculate the mean and standard deviation for the number who recognize the Coke brand name. Considering as unusual a result that differs from the mean by more than two standard deviations, it (___ (is/is not) ___) unusual to get 775 consumers who recognize the Coke brand name.
A) 96, 9, is
B) 784, 3.96, is
C) 784, 9, is not
D) 96, 39.6, is not
E) 396, 78.4, is not

Accepted Answer

The answer of When surveyed for brand recognition, 98% of...

Question 2

In a large metropolitan area, past records revealed that 30 percent of all the high school graduates go to college. From 20 graduates selected at random, approximately how many will go to college?
A) 3
B) 4
C) 5
D) 6
E) 7

Accepted Answer

To estimate the number of graduates going to college, multiply the total number of graduates by the percentage that typically go to college: 20 graduates * 30% = 6 graduates.

Question 3

(i. A random variable with a Poisson distribution has one of three mutually exclusive values. (ii) For the hypergeometric distribution there are only 2 possible outcomes.
(iii) In the hypergeometric distribution the probability of a success is not the same on each trail.
A) (i), (ii), and (iii) are all correct statements
B) (i) is a correct statement but not (ii) or (iii).
C) (i) and (iii) are correct statements but not (ii).
D) (ii) and (iii) are correct statements but not (i).
E) (i), (ii), and (iii) are all false statements.

Accepted Answer

A hypergeometric distribution involves drawing from a finite population without replacement, leading to varying probabilities of success, making (iii) correct. It typically involves two outcomes (success or failure), making (ii) correct. A Poisson distribution can take an infinite number of values, not just three, making (i) incorrect.

Question 4

A marketing solutions poll of mutual funds and fund owners asked fund owners what action they took after the September 11^th2001 market drop. Sixteen percent of respondents said they bought more funds. If 600 fund owners were polled, calculate the mean and standard deviation of the number of respondents who bought more funds. Considering as unusual a result that differs from the mean by more than two standard deviations, it (___ (is/is not) ___) unusual that in one of these polls of 600 fund owners, 100 respondents bought more mutual funds. A) 96, 9, is not B) 96, 8, is C) 96, 9, is D) 95, 8, is not E) 95, 8, is

Accepted Answer

The mean number of respondents who bought more funds is calculated as 16% of 600, which is 0.16 * 600 = 96. The standard deviation for a binomial distribution is calculated using the formula $\sqrt{n \cdot p \cdot (1-p)}$, where $n$ is the number of trials (600) and $p$ is the probability of success (0.16). Thus, the standard deviation is $\sqrt{600 \cdot 0.16 \cdot (1-0.16)} = \sqrt{600 \cdot 0.16 \cdot 0.84} = \sqrt{96 \cdot 0.84} = \sqrt{80.64} \approx 9$. Considering a result unusual if it differs from the mean by more than two standard deviations, the range for usual results is mean ± 2*standard deviation, which is 96 ± 2*9 = 96 ± 18. This range is 78 to 114. Since 100 respondents fall within this range, it is not unusual.

Question 5

Elly's hot dog emporium is famous for its chilidogs. Some customers order the hot dogs with hot peppers, while many do not care for that added bit of zest. Elly's latest sales indicate that 30% of the customers ordering her chilidogs order it with hot peppers. Suppose 18 customers are selected at random. What is the probability that sixteen or more customers will want hot peppers?
A) 0.0000
B) 0.708
C) 0.015
D) 0.5
E) 0.80

Accepted Answer

Given that only 30% of customers order hot dogs with hot peppers, the probability of 16 or more out of 18 customers choosing hot peppers is extremely low, essentially 0.

Question 6

Elly's hot dog emporium is famous for its chilidogs. Some customers order the hot dogs with hot peppers, while many do not care for that added bit of zest. Elly's latest sales indicate that 30% of the customers ordering her chili dogs order it with hot peppers. Suppose 18 customers are selected at random. What is the probability that exactly nine customers will ask for hot peppers?
A) 0.0000
B) 0.708
C) 0.015
D) 0.5
E) 0.039

Accepted Answer

The probability can be calculated using the binomial probability formula, which is P(X = k) = (n choose k) * p^k * (1-p)^(n-k), where n is the number of trials (18 customers), k is the number of successes (9 customers asking for hot peppers), and p is the probability of success on a single trial (0.30). Plugging these values into the formula gives a probability of approximately 0.039.

Question 7

(i. A random variable with a Poisson distribution has one of two mutually exclusive values. (ii) For the hypergeometric distribution there are only 2 possible outcomes.
(iii) In the hypergeometric distribution the probability of a success is not the same on each trail.
A) (i), (ii), and (iii) are all correct statements.
B) (i) is a correct statement but not (ii) or (iii).
C) (i) and (iii) are correct statements but not (ii).
D) (ii) and (iii) are correct statements but not (i).
E) (i), (ii), and (iii) are all false statements.

Accepted Answer

The answer of (i. A random variable with a Poisson...

Question 8

Elly's hot dog emporium is famous for its chilidogs. Some customers order the hot dogs with hot peppers, while many do not care for that added bit of zest. Elly's latest sales indicate that 30% of the customers ordering her chili dogs order it with hot peppers. Suppose 18 customers are selected at random. What is the probability that between two and five people inclusive want hot peppers?
A) 0.0000
B) 0.708
C) 0.015
D) 0.521
E) 0.80

Accepted Answer

The probability can be calculated using the binomial probability formula. The probability of 2 to 5 customers inclusive wanting hot peppers is the sum of the probabilities for 2, 3, 4, and 5 customers, which results in approximately 0.521.

Question 9

Let X represent the number of children in a Canadian household. The probability distribution of X is as follows:  What is the probability that a randomly selected Canadian household will have more than 3 children? What is the expected number of children in a Canadian household?
A) 0.7311, 2
B) 0.3711, 3.2
C) 0.16, 2.25
D) 0.16, 3.2

Accepted Answer

The answer of Let X represent the number of children...

Question 10

In a Poisson distribution each trail is independent. (i) The binomial distribution and the Poisson distribution have two possible experimental outcomes.
(ii. The Poisson distribution or, the law of improbable events, has negatively skewed shape.
(iii) In a Poisson distribution each trail is dependent on the others.
A) (i), (ii), and (iii) are all correct statements.
B) (i) is a correct statement but not (ii) or (iii).
C) (i) and (iii) are correct statements but not (ii).
D) (ii) and (iii) are correct statements but not (i).
E) (i), (ii), and (iii) are all false statements.

Accepted Answer

(i) is true for both binomial and Poisson distributions. (ii) is false as the Poisson distribution has a positively skewed shape. (iii) is false as each trail is independent in a Poisson distribution.

Question 11

(i. A random variable with a Poisson distribution has one of two mutually exclusive values. (ii) For the hypergeometric distribution there are only 2 possible outcomes.
(iii) In the hypergeometric distribution the probability of a success is the same on each trail.
A) (i), (ii), and (iii) are all correct statements.
B) (i) is a correct statement but not (ii) or (iii).
C) (i) and (ii) are correct statements but not (iii).
D) (ii) and (iii) are correct statements but not (i).
E) (i), (ii), and (iii) are all false statements.

Accepted Answer

(i) is true for Poisson distribution but (ii) is not true for hypergeometric distribution as there can be multiple outcomes. 
(iii) is also not true for hypergeometric distribution as the probability of success changes on each trial depending on the remaining population.

Question 12

In a large metropolitan area, past records revealed that 30 percent of all the high school graduates go to college. From 10 graduates selected at random, calculate the probability that none will go to college.
A) 0.028%
B) 82%
C) 28%
D) 8.2%
E) 2.8%

Accepted Answer

The probability of a high school graduate going to college is 0.3. The probability of a high school graduate not going to college is 1 - 0.3 = 0.7. The probability that none of the 10 graduates will go to college is (0.7)^10 = 0.028%.

Question 13

(i. To construct a binomial distribution we need to know the total number of trials and the probability of a success. (ii) If n = 900 and $\pi$ = 1/3, the variance of this binomial distribution is 200.
(iii) If $\pi$ = 1/5 and n = 100, the standard deviation of this binomial distribution is four.
A) (i), (ii), and (iii) are all correct statements.
B) (i) is a correct statement but not (ii) or (iii).
C) (i) and (ii) are correct statements but not (iii).
D) (ii) and (iii) are correct statements but not (i).
E) (i), (ii), and (iii) are all false statements.

Accepted Answer

(i) is a general statement about binomial distributions. (ii) and (iii) are specific examples of binomial distributions with the given parameters.

Question 14

(i. If $\pi$ = 1/3 and n = 900, the mean of this binomial distribution is 300. (ii) If n = 900 and $\pi$ = 1/3, the variance of this binomial distribution is 200.
(iii) If $\pi$ = 1/5 and n = 100, the standard deviation of this binomial distribution is 16.
A) (i), (ii), and (iii) are all correct statements.
B) (i) is a correct statement but not (ii) or (iii).
C) (i) and (ii) are correct statements but not (iii).
D) (ii) and (iii) are correct statements but not (i).
E) (i), (ii), and (iii) are all false statements.

Accepted Answer

(i) is true because the mean of a binomial distribution with parameters n and p is n*p. 
(ii) is true because the variance of a binomial distribution with parameters n and p is n*p*(1-p). 
(iii) is false because the standard deviation of a binomial distribution with parameters n and p is sqrt(n*p*(1-p)), so if n=100 and F1F1F1S1F1F1F10 = 1/5, the standard deviation would be sqrt(100*(1/5)*(4/5)) = 4, not 16.

Question 15

Elly's hot dog emporium is famous for its chilidogs. Some customers order the hot dogs with hot peppers, while many do not care for that added bit of zest. Elly's latest sales indicate that 30% of the customers ordering her chili dogs order it with hot peppers. Suppose 18 customers are selected at random. This situation is an example of what type of distribution?
A) Binomial distribution
B) Hypergeometric distribution
C) Poisson distribution
D) Chi-squared distribution

Accepted Answer

This situation can be modeled as a binomial distribution since there are a fixed number of trials (18 customers selected) and each trial has only two outcomes (ordering with or without hot peppers). Additionally, the trials are independent and the probability of success (ordering with hot peppers) remains constant at 30%.

Question 16

(i. If $\pi$ = 1/3 and n = 900, the mean of this binomial distribution is 300. (ii) If n = 900 and $\pi$= 1/3, the variance of this binomial distribution is 200.
(iii) If $\pi$ = 1/5 and n = 100, the standard deviation of this binomial distribution is four.
A) (i), (ii), and (iii) are all correct statements.
B) (i) is a correct statement but not (ii) or (iii).
C) (i) and (ii) are correct statements but not (iii).
D) (ii) and (iii) are correct statements but not (i).
E) (i), (ii), and (iii) are all false statements.

Accepted Answer

(i) Mean = n * p = 900 * 1/3 = 300(ii) Variance = n * p * q = 900 * 1/3 * 2/3 = 200(iii) Standard deviation = sqrt(n * p * q) = sqrt(100 * 1/5 * 4/5) = 4

Question 17

The arrival of customers at a service desk follows a Poisson distribution. If they arrive at the rate of two every five minutes, what is the probability that no customers arrive in a five-minute period?
A) 0.7311
B) 0.3711
C) 0.1353
D) 0.16

Accepted Answer

The mean rate of customers arriving in a 5-minute period is:

λ = (2/5) * 5 = 2

Using the Poisson probability formula, the probability of zero customers arriving in a 5-minute period is:

P(0) = (e^-2) * (2^0) / 0! = e^-2 = 0.1353

Therefore, the answer is (C) 0.1353.

Question 18

(i. The binomial distribution and the Poisson distribution have two possible experimental outcomes. (ii) The Poisson distribution or, the law of improbable events, has negatively skewed shape.
(iii) In a Poisson distribution each trail is independent.
A) (i), (ii), and (iii) are all correct statements.
B) (i) is a correct statement but not (ii) or (iii).
C) (i) and (iii) are correct statements but not (ii).
D) (ii) and (iii) are correct statements but not (i).
E) (i), (ii), and (iii) are all false statements.

Accepted Answer

(i) is a correct statement, as both binomial and Poisson distributions have two possible outcomes: success (usually denoted as 1) and failure (usually denoted as 0). 
(ii) is a false statement, as the Poisson distribution is positively skewed, not negatively skewed. 
(iii) is a correct statement, as each trial in a Poisson distribution is assumed to be independent from all other trials.

Question 19

(i. To construct a binomial distribution we need to know the total number of trials and the probability of a success. (ii) If n = 900 and $\pi$ = 1/3, the variance of this binomial distribution is 200.
(iii) If $\pi$ = 1/5 and n = 100, the standard deviation of this binomial distribution is 20.
A) (i), (ii), and (iii) are all correct statements.
B) (i) is a correct statement but not (ii) or (iii).
C) (i) and (ii) are correct statements but not (iii).
D) (ii) and (iii) are correct statements but not (i).
E) (i), (ii), and (iii) are all false statements.

Accepted Answer

(i) is a correct statement as binomial distribution requires the total number of trials and probability of success. 
(ii) is also correct because the variance of a binomial distribution is np(1-p), where n is the total number of trials and p is the probability of success. Here, n=900 and p=1/3, so variance = 900*(1/3)*(2/3) = 200. 
(iii) is incorrect because the standard deviation of a binomial distribution is the square root of variance, which in this case would be 15. However, the statement implies that the standard deviation is 20.

Question 20

The arrival of customers at a service desk follows a Poisson distribution. If they arrive at the rate of four every five minutes, what is the probability that more than four customers arrive in a five minute period?
A) 0.7311
B) 0.3711
C) 0.5
D) 0.16

Accepted Answer

If customers arrive at the rate of four every five minutes, the average arrival rate can be calculated as λ = 4/5 = 0.8 customers per minute. The probability of more than four customers arriving in a five minute period can be calculated using the Poisson distribution as follows:

P(X > 4) = 1 - P(X ≤ 4)
= 1 - (e^-λ ((λ^0)/0!) + e^-λ ((λ^1)/1!) + e^-λ ((λ^2)/2!) + e^-λ ((λ^3)/3!) + e^-λ ((λ^4)/4!))
= 1 - (e^-0.8 (1 + 0.8 + 0.64 + 0.341 + 0.136))
= 1 - (e^-0.8 (2.921))
= 0.3711

Therefore, the best choice is B, with the explanation given above.

Quiz 5: Discrete Probability Distributions

In a large metropolitan area, past records revealed that 30 percent of all the high school graduates go to college. From 20 graduates selected at random, approximately how many will go to college?

(i. A random variable with a Poisson distribution has one of three mutually exclusive values. (ii) For the hypergeometric distribution there are only 2 possible outcomes. (iii) In the hypergeometric distribution the probability of a success is not the same on each trail.

(i. A random variable with a Poisson distribution has one of two mutually exclusive values. (ii) For the hypergeometric distribution there are only 2 possible outcomes. (iii) In the hypergeometric distribution the probability of a success is not the same on each trail.

Let X represent the number of children in a Canadian household. The probability distribution of X is as follows: What is the probability that a randomly selected Canadian household will have more than 3 children? What is the expected number of children in a Canadian household?

(i. A random variable with a Poisson distribution has one of two mutually exclusive values. (ii) For the hypergeometric distribution there are only 2 possible outcomes. (iii) In the hypergeometric distribution the probability of a success is the same on each trail.

In a large metropolitan area, past records revealed that 30 percent of all the high school graduates go to college. From 10 graduates selected at random, calculate the probability that none will go to college.

(i. To construct a binomial distribution we need to know the total number of trials and the probability of a success. (ii) If n = 900 and $\pi$ = 1/3, the variance of this binomial distribution is 200. (iii) If $\pi$ = 1/5 and n = 100, the standard deviation of this binomial distribution is four.

(i. If $\pi$ = 1/3 and n = 900, the mean of this binomial distribution is 300. (ii) If n = 900 and $\pi$ = 1/3, the variance of this binomial distribution is 200. (iii) If $\pi$ = 1/5 and n = 100, the standard deviation of this binomial distribution is 16.

(i. If $\pi$ = 1/3 and n = 900, the mean of this binomial distribution is 300. (ii) If n = 900 and $\pi$ = 1/3, the variance of this binomial distribution is 200. (iii) If $\pi$ = 1/5 and n = 100, the standard deviation of this binomial distribution is four.

The arrival of customers at a service desk follows a Poisson distribution. If they arrive at the rate of two every five minutes, what is the probability that no customers arrive in a five-minute period?

(i. The binomial distribution and the Poisson distribution have two possible experimental outcomes. (ii) The Poisson distribution or, the law of improbable events, has negatively skewed shape. (iii) In a Poisson distribution each trail is independent.

(i. To construct a binomial distribution we need to know the total number of trials and the probability of a success. (ii) If n = 900 and $\pi$ = 1/3, the variance of this binomial distribution is 200. (iii) If $\pi$ = 1/5 and n = 100, the standard deviation of this binomial distribution is 20.

The arrival of customers at a service desk follows a Poisson distribution. If they arrive at the rate of four every five minutes, what is the probability that more than four customers arrive in a five minute period?