Question 1

Which one of these variables is a binomial random variable?

Accepted Answer

A binomial random variable is defined by two outcomes (success or failure) in a fixed number of trials. In choice C, the number of women taller than 68 inches in a random sample of 5 women fits this definition, where each trial (each woman's height) has two outcomes (taller than 68 inches or not) and there is a fixed number of trials (5 women).

Question 2

Consider an experiment that involves repeatedly rolling a six-sided die. Which of the following is a binomial random variable?

Accepted Answer

A binomial random variable is defined by having a fixed number of trials (in this case, six rolls), two possible outcomes (rolling a "4" or not rolling a "4"), and a constant probability of success on each trial (1/6 chance of rolling a "4"). Option B meets these criteria, making it a binomial random variable.

Question 3

Which of the following is an example of a binomial random variable?

Accepted Answer

A binomial random variable is defined by two outcomes (success or failure) for each trial, a fixed number of trials, and a constant probability of success on each trial. Choice B fits this definition because each question on a multiple-choice test has a fixed probability of being answered correctly if guessed randomly, and the number of questions is fixed, making it a binomial scenario. Choices A, C, and D do not necessarily have only two outcomes or a fixed probability of success, which is required for a binomial distribution.

Question 4

A medication produces side effects in each user with probability 0.10 and this is independent from one person to the next. If 50 people use the medication, the number who will experience side effects is

Accepted Answer

The number of people experiencing side effects follows a binomial distribution because each person independently experiences side effects with a probability of 0.10.

Question 5

The probability is p = 0.80 that a patient with a certain disease will be successfully treated with a new medical treatment. Suppose that the treatment is used on 40 patients. What is the "expected value"of the number of patients who are successfully treated?

Accepted Answer

The expected value is calculated by multiplying the probability of success (p = 0.80) by the number of trials (n = 40). Thus, the expected value is 0.80 * 40 = 32.

Question 6

Sara is a frequent business traveler. For security purposes, 10% of all people boarding airplanes are randomly selected for additional screening just prior to boarding. Define the random variable X = number of flights Sara completes before being chosen for additional screening. For instance, if she is searched boarding her next flight, then X = 0. What is the value of P(X = 2) = probability Sara completes two flights without screening and then is chosen for additional screening on the next one?

Accepted Answer

The probability Sara completes two flights without being selected for additional screening (0.9 for each flight) and then is selected on her next flight (0.1) is calculated as (0.9)^2 * (0.1).

Question 7

A landscaping company claims that 90% of the trees they plant survive (defined as being still alive one year from planting). If a tree does not survive, the company will replace the tree with a new one. A homeowner will have 5 trees planted in his yard by this landscaping company. Consider these 5 trees to be a random sample of all trees planted by this company. If the company's claim is correct, what is the probability that all 5 of the trees will survive?

Accepted Answer

The probability that each tree survives is 0.9. Since the trees are planted independently, the probability that all 5 trees survive is $0.9^5$.

Question 8

Use the following information for questions:
 Suppose that a student needs to buy 6 books for her history course. The number of books that she will be able to find used is a binomial random variable X with n = 6 and p = 0.30. In other words, the probability that she will find any given book used is 0.30, and is independent from one book to the next.
-What is the probability that she will find exactly 2 used books?

Accepted Answer

The answer of Use the following information for questions:
 Suppose...

Question 9

Use the following information for questions:
 Suppose that a student needs to buy 6 books for her history course. The number of books that she will be able to find used is a binomial random variable X with n = 6 and p = 0.30. In other words, the probability that she will find any given book used is 0.30, and is independent from one book to the next.
-What is the probability that she will find exactly 3 used books?

Accepted Answer

The probability of finding exactly 3 used books out of 6, when the probability of finding any given book used is 0.30, can be calculated using the binomial probability formula: P(X = k) = (n choose k) * p^k * (1-p)^(n-k). Here, n = 6, k = 3, and p = 0.30. Plugging these values into the formula gives P(X = 3) = (6 choose 3) * (0.30)^3 * (0.70)^3 = 0.185.

Question 10

Use the following information for questions:
 Suppose that a student needs to buy 6 books for her history course. The number of books that she will be able to find used is a binomial random variable X with n = 6 and p = 0.30. In other words, the probability that she will find any given book used is 0.30, and is independent from one book to the next.
-What is the expected number of used books she will find, E(X)?

Accepted Answer

The expected value E(X) of a binomial distribution is calculated as n*p, where n is the number of trials and p is the probability of success on each trial. Here, n = 6 and p = 0.30, so E(X) = 6 * 0.30 = 1.8.

Question 11

Use the following information for questions:
 Suppose that a student needs to buy 6 books for her history course. The number of books that she will be able to find used is a binomial random variable X with n = 6 and p = 0.30. In other words, the probability that she will find any given book used is 0.30, and is independent from one book to the next.
-What is the variance for the number of used books she will find, V(X)?

Accepted Answer

The answer of Use the following information for questions:
 Suppose...

Question 12

Use the following information for questions:
 A child is observing squirrels in the park and notices that some are black and some are gray. For the next five squirrels she sees, she counts X = the number of black squirrels. Suppose X is a binomial random variable with n = 5 and p = 0.50.
-What is the probability that she will see exactly one black squirrel out of the five?

Accepted Answer

The probability of seeing exactly one black squirrel out of five, when each squirrel has a 0.50 chance of being black, can be calculated using the binomial probability formula: P(X = k) = (n choose k) * p^k * (1-p)^(n-k). Here, n = 5, k = 1, and p = 0.50. Thus, P(X = 1) = (5 choose 1) * (0.50)^1 * (0.50)^(5-1) = 5 * 0.50 * 0.0625 = 0.156.

Question 13

Use the following information for questions:
 A child is observing squirrels in the park and notices that some are black and some are gray. For the next five squirrels she sees, she counts X = the number of black squirrels. Suppose X is a binomial random variable with n = 5 and p = 0.50.
-What is the probability that she will see exactly two black squirrels out of the five?

Accepted Answer

The probability of seeing exactly two black squirrels out of five, when each squirrel seen has a 0.50 chance of being black, can be calculated using the binomial probability formula: P(X = k) = (n choose k) * p^k * (1-p)^(n-k). For this scenario, n = 5, k = 2, and p = 0.50, so P(X = 2) = (5 choose 2) * (0.50)^2 * (0.50)^(5-2) = 10 * 0.25 * 0.125 = 0.3125, which rounds to 0.313.

Question 14

Use the following information for questions:
 A child is observing squirrels in the park and notices that some are black and some are gray. For the next five squirrels she sees, she counts X = the number of black squirrels. Suppose X is a binomial random variable with n = 5 and p = 0.50.
-What is the expected number of black squirrels she will see, E(X)?

Accepted Answer

The expected number of black squirrels, E(X), for a binomial distribution is calculated as n*p, where n is the number of trials (5 squirrels) and p is the probability of success (seeing a black squirrel, 0.50). Therefore, E(X) = 5 * 0.50 = 2.5.

Question 15

Use the following information for questions:
 A child is observing squirrels in the park and notices that some are black and some are gray. For the next five squirrels she sees, she counts X = the number of black squirrels. Suppose X is a binomial random variable with n = 5 and p = 0.50.
-What is the variance for the number of black squirrels she will see, V(X)?

Accepted Answer

The variance of a binomial distribution is given by V(X) = n*p*(1-p), where n is the number of trials and p is the probability of success on each trial. Here, n = 5 and p = 0.50, so V(X) = 5*0.50*(1-0.50) = 1.25.

Question 16

Use the following information for questions:
 In a family with 4 children, the number of children with blue eyes is a binomial random variable X with n = 4 and p = .20.
-What is the probability that all 4 children will have blue eyes?

Accepted Answer