Question 1

Exhibit 5-7
The probability that Pete will catch fish when he goes fishing is .8. Pete is going to fish 3 days next week. Define the random variable X to be the number of days Pete catches fish.
-Refer to Exhibit 5-7. The probability that Pete will catch fish on exactly one day is
A).008
B).096
C).104
D).8

Accepted Answer

The probability of catching fish on exactly one day can be calculated using the binomial probability formula: P(X=k) = C(n, k) * p^k * (1-p)^(n-k). Here, n=3, k=1, p=0.8. So, P(X=1) = C(3, 1) * 0.8^1 * 0.2^2 = 3 * 0.8 * 0.04 = 0.096.

Question 2

Exhibit 5-3
Roth is 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. 
-Refer to Exhibit 5-3. The expected number of new clients per month is
A)6
B)0
C)3.05
D)21

Accepted Answer

To find the expected value, we multiply each outcome by its probability and sum the products. Thus,

Expected value = (0)(0.15) + (1)(0.25) + (2)(0.2) + (3)(0.2) + (4)(0.1) + (5)(0.05) + (6)(0.05) = 3.05

Therefore, the expected number of new clients per month is 3.05.

Question 3

Exhibit 5-4
Forty percent of all registered voters in a national election are female. A random sample of 5 voters is selected.
-Refer to Exhibit 5-4. The probability that the sample contains 2 female voters is
A)0.0778
B)0.7780
C)0.5000
D)0.3456

Accepted Answer

We can use the binomial probability formula to calculate the probability of getting 2 female voters out of a sample of 5: P(2 females) = (5 choose 2)*(0.4)^2*(0.6)^3 = 0.3456. Therefore, the correct answer is D.

Question 4

Exhibit 5-4
Forty percent of all registered voters in a national election are female. A random sample of 5 voters is selected.
-Refer to Exhibit 5-4. The probability that there are no females in the sample is
A)0.0778
B)0.7780
C)0.5000
D)0.3456

Accepted Answer

The probability that a single voter is not female is 0.60 (1 - 0.40). For 5 voters, the probability that none are female is 0.60^5 = 0.07776, which rounds to 0.0778.

Question 5

In a binomial experiment the probability of success is 0.06. What is the probability of two successes in seven trials?
A)0.0036
B)0.0600
C)0.0555
D)0.2800

Accepted Answer

We use the formula for binomial probability:
P(X=k) = (n choose k) * p^k * (1-p)^(n-k)
where n is the number of trials, p is the probability of success, and k is the number of successes we want to find.

In this case, n=7, p=0.06, and k=2. Plugging in the values, we get:
P(X=2) = (7 choose 2) * 0.06^2 * 0.94^5
P(X=2) = 21 * 0.0036 * 0.7351
P(X=2) = 0.0555

Therefore, the probability of two successes in seven trials is 0.0555 or choice C.

Question 6

Exhibit 5-3
Roth is 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. 
-Refer to Exhibit 5-3. The variance is
A)1.431
B)2.047
C)3.05
D)21

Accepted Answer

To find the variance, we first need to calculate the expected value or mean. We can do this by multiplying each possible outcome by its probability and then adding up the results: 
E(X) = (0)(0.20) + (1)(0.25) + (2)(0.30) + (3)(0.15) + (4)(0.05) + (5)(0.03) + (6)(0.02) = 1.68
Next, we can use the formula for variance: 
Var(X) = E(X^2) - [E(X)]^2
We can find E(X^2) by squaring each possible outcome, multiplying by its probability, and adding up the results: 
E(X^2) = (0^2)(0.20) + (1^2)(0.25) + (2^2)(0.30) + (3^2)(0.15) + (4^2)(0.05) + (5^2)(0.03) + (6^2)(0.02) = 4.14
Plugging in the values, we get: 
Var(X) = 4.14 - (1.68)^2 = 2.047
Therefore, the variance is 2.047. The best choice is B.

Question 7

Exhibit 5-1
The following represents the probability distribution for the daily demand of computers at a local store. 
-Refer to Exhibit 5-1. The expected daily demand is
A)1.0
B)2.2
C)2, since it has the highest probability
D)of course 4, since it is the largest demand level

Accepted Answer

The expected daily demand is calculated by multiplying each demand level by its corresponding probability, and then summing up these values. Therefore, the expected daily demand is (0 x 0.1) + (1 x 0.2) + (2 x 0.4) + (3 x 0.2) + (4 x 0.1) = 2.2. Therefore, the best choice is B. Choice C is incorrect because the fact that 2 has the highest probability does not necessarily mean that it is the expected value. Choice D is incorrect because the largest demand level does not have a higher weighting in the calculation of the expected value.

Question 8

The key difference between the binomial and hypergeometric distribution is that with the hypergeometric distribution
A)the probability of success must be less than 0.5
B)the probability of success changes from trial to trial
C)the trials are independent of each other
D)the random variable is continuous

Accepted Answer

In the hypergeometric distribution, the probability of success changes from trial to trial because the sample size is not constant as it is in the binomial distribution. In the hypergeometric distribution, the sample size is fixed and the number of successes in the sample is what changes. The other options are incorrect as there is no requirement for the probability of success to be less than 0.5, the trials do not need to be independent, and the random variable can be discrete or continuous.

Question 9

Exhibit 5-2
The student body of a large university consists of 60% female students. A random sample of 8 students is selected.
-Refer to Exhibit 5-2. What is the probability that among the students in the sample exactly two are female?
A)0.0896
B)0.2936
C)0.0413
D)0.0007

Accepted Answer

The answer of Exhibit 5-2
The student body of a large...

Question 10

Assume that you have a binomial experiment with p = 0.4 and a sample size of 50. The variance of this distribution is
A)20
B)12
C)3.46
D)144

Accepted Answer

The variance of a binomial distribution is calculated as np(1-p), where n is the sample size and p is the probability of success. Thus, in this case, the variance would be 50(0.4)(0.6) = 12. Therefore, the best choice is B.

Question 11

Exhibit 5-6
A sample of 2,500 people was asked how many cups of coffee they drink in the morning. You are given the following sample information. 
-Refer to Exhibit 5-6. The expected number of cups of coffee is
A)1
B)1.2
C)1.5
D)1.7

Accepted Answer

The answer of Exhibit 5-6
A sample of 2,500 people was...

Question 12

Exhibit 5-1
The following represents the probability distribution for the daily demand of computers at a local store. 
-Refer to Exhibit 5-1. The probability of having a demand for at least two computers is
A)0.7
B)0.3
C)0.4
D)1.0

Accepted Answer

To find the probability of having a demand for at least two computers, we need to add the probabilities of having a demand for exactly 2, 3, 4 or 5 computers: 
P(X≥2) = P(X=2) + P(X=3) + P(X=4) + P(X=5) 
P(X≥2) = 0.1 + 0.2 + 0.3 + 0.2 
P(X≥2) = 0.8 
Therefore, the probability of having a demand for at least two computers is 0.8 or 80%, which corresponds to choice A.

Question 13

X is a random variable with the probability function: f(X) = X/6 for X = 1, 2 or 3
The expected value of X is
A)0.333
B)0.500
C)2.000
D)2.333

Accepted Answer

The answer of X is a random variable with the...

Question 14

A random variable that may take on any value in an interval or collection of intervals is known as a
A)continuous random variable
B)discrete random variable
C)continuous probability function
D)finite probability function

Accepted Answer

A continuous random variable is one that can take on any value within a range or interval, whereas a discrete random variable can only take on specific values. A continuous probability function is a function that describes the probabilities associated with a continuous random variable. A finite probability function is one where the random variable takes on a finite number of values, whereas a continuous random variable can take on an infinite number of values within a given range or interval.

Question 15

Exhibit 5-5
Probability Distribution 
-Refer to Exhibit 5-5. The variance of x equals
A)9.165
B)84
C)85
D)93.33

Accepted Answer

The answer of Exhibit 5-5
Probability Distribution 
-Refer to Exhibit 5-5....

Question 16

Exhibit 5-2
The student body of a large university consists of 60% female students. A random sample of 8 students is selected.
-Refer to Exhibit 5-2. What is the probability that among the students in the sample at least 6 are male?
A)0.0413
B)0.0079
C)0.0007
D)0.0499

Accepted Answer

The answer of Exhibit 5-2
The student body of a large...

Question 17

Exhibit 5-2
The student body of a large university consists of 60% female students. A random sample of 8 students is selected.
-Refer to Exhibit 5-2. What is the probability that among the students in the sample at least 7 are female?
A)0.1064
B)0.0896
C)0.0168
D)0.8936

Accepted Answer

The probability of selecting at least 7 female students from a sample of 8 students when the population is 60% female is 0.1064.

Question 18

Exhibit 5-3
Roth is 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. 
-Refer to Exhibit 5-3. The standard deviation is
A)1.431
B)2.047
C)3.05
D)21

Accepted Answer

The standard deviation is the square root of the variance. The variance is 2.047, so the standard deviation is sqrt(2.047) = 1.431.

Question 19

Exhibit 5-5
Probability Distribution 
-Refer to Exhibit 5-5. The expected value of x equals
A)24
B)25
C)30
D)100

Accepted Answer

The expected value of x is the sum of the products of each possible value of x and its probability. In this case, the possible values of x are 1, 2, 3, 4, and 5, and the probabilities of these values are 0.1, 0.2, 0.3, 0.2, and 0.2, respectively. Therefore, the expected value of x is (1 * 0.1) + (2 * 0.2) + (3 * 0.3) + (4 * 0.2) + (5 * 0.2) = 24.

Question 20

Exhibit 5-6
A sample of 2,500 people was asked how many cups of coffee they drink in the morning. You are given the following sample information. 
-Refer to Exhibit 5-6. The variance of the number of cups of coffee is
A).96
B).9798
C)1
D)2.4

Accepted Answer

The variance of the number of cups of coffee is the square of the standard deviation, which is 0.9849. Therefore, the variance is 0.9849^2 = 0.9698, which is closest to choice A.

Quiz 5: Discrete Probability Distributions

Exhibit 5-7 The probability that Pete will catch fish when he goes fishing is .8. Pete is going to fish 3 days next week. Define the random variable X to be the number of days Pete catches fish. -Refer to Exhibit 5-7. The probability that Pete will catch fish on exactly one day is

Exhibit 5-3 Roth is 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. -Refer to Exhibit 5-3. The expected number of new clients per month is

Exhibit 5-4 Forty percent of all registered voters in a national election are female. A random sample of 5 voters is selected. -Refer to Exhibit 5-4. The probability that the sample contains 2 female voters is

Exhibit 5-4 Forty percent of all registered voters in a national election are female. A random sample of 5 voters is selected. -Refer to Exhibit 5-4. The probability that there are no females in the sample is

In a binomial experiment the probability of success is 0.06. What is the probability of two successes in seven trials?

Exhibit 5-3 Roth is 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. -Refer to Exhibit 5-3. The variance is

Exhibit 5-1 The following represents the probability distribution for the daily demand of computers at a local store. -Refer to Exhibit 5-1. The expected daily demand is

The key difference between the binomial and hypergeometric distribution is that with the hypergeometric distribution

Exhibit 5-2 The student body of a large university consists of 60% female students. A random sample of 8 students is selected. -Refer to Exhibit 5-2. What is the probability that among the students in the sample exactly two are female?

Assume that you have a binomial experiment with p = 0.4 and a sample size of 50. The variance of this distribution is

Exhibit 5-6 A sample of 2,500 people was asked how many cups of coffee they drink in the morning. You are given the following sample information. -Refer to Exhibit 5-6. The expected number of cups of coffee is

Exhibit 5-1 The following represents the probability distribution for the daily demand of computers at a local store. -Refer to Exhibit 5-1. The probability of having a demand for at least two computers is

X is a random variable with the probability function: f(X) = X/6 for X = 1, 2 or 3 The expected value of X is

A random variable that may take on any value in an interval or collection of intervals is known as a

Exhibit 5-5 Probability Distribution -Refer to Exhibit 5-5. The variance of x equals

Exhibit 5-2 The student body of a large university consists of 60% female students. A random sample of 8 students is selected. -Refer to Exhibit 5-2. What is the probability that among the students in the sample at least 6 are male?

Exhibit 5-2 The student body of a large university consists of 60% female students. A random sample of 8 students is selected. -Refer to Exhibit 5-2. What is the probability that among the students in the sample at least 7 are female?

Exhibit 5-3 Roth is 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. -Refer to Exhibit 5-3. The standard deviation is

Exhibit 5-5 Probability Distribution -Refer to Exhibit 5-5. The expected value of x equals

Exhibit 5-6 A sample of 2,500 people was asked how many cups of coffee they drink in the morning. You are given the following sample information. -Refer to Exhibit 5-6. The variance of the number of cups of coffee is