Question 1

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

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.
 $$\begin{array} { c c } 
\text { Number of } & \
\text { New Clients } & \text { Prabability } \
0 & 0.05 \
1 & 0.10 \
2 & 0.15 \
3 & 0.35 \
4 & 0.20 \
5 & 0.10 \
6 & 0.05
\end{array}$$
-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

The answer of Exhibit 5-3
Roth is a computer-consulting firm. The...

Question 3

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 binomial distribution, the probability of success remains constant from trial to trial, while in the hypergeometric distribution, the probability of success changes from trial to trial as the number of successes or failures in the population changes. This is because the hypergeometric distribution models sampling without replacement, while the binomial distribution models sampling with replacement. The other options (A, C, and D) do not correctly describe the key difference between the two distributions.

Question 4

X is a random variable with the probability function: fX) = 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 5

Exhibit 5-1
The following represents the probability distribution for the daily demand of computers at a local store.
 $$\begin{array} { c c } 
\text { Demand } & \text { Probability } \
0 & 0.1 \
1 & 0.2 \
2 & 0.3 \
3 & 0.2 \
4 & 0 ?
\end{array}$$
-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 6

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 7

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.
 $\begin{array}{ll}\text { Cups of Coffee }&\text { Frequency }\
0 & 700 \
1 & 900 \
2 & 600 \
3 & 300\&2,500
\end{array}$
-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 answer of Exhibit 5-6
A sample of 2,500 people was...

Question 8

Exhibit 5-1
The following represents the probability distribution for the daily demand of computers at a local store.
 $$\begin{array} { c c } 
\text { Demand } & \text { Probability } \
0 & 0.1 \
1 & 0.2 \
2 & 0.3 \
3 & 0.2 \
4 & 0 ?
\end{array}$$
-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 answer of Exhibit 5-1
The following represents the probability distribution...

Question 9

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.
 $$\begin{array} { c c } 
\text { Number of } & \
\text { New Clients } & \text { Prabability } \
0 & 0.05 \
1 & 0.10 \
2 & 0.15 \
3 & 0.35 \
4 & 0.20 \
5 & 0.10 \
6 & 0.05
\end{array}$$
-Refer to Exhibit 5-3. The variance is
A) 1.431
B) 2.047
C) 3.05
D) 21

Accepted Answer

The answer of Exhibit 5-3
Roth is a computer-consulting firm. The...

Question 10

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.
 $\begin{array}{ll}\text { Cups of Coffee }&\text { Frequency }\
0 & 700 \
1 & 900 \
2 & 600 \
3 & 300\&2,500
\end{array}$
-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 11

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 can take on any value within a range of values (i.e. an interval) whereas a discrete random variable can only take on specific, isolated values. The other answer choices are not applicable as a continuous probability function may refer to the probability density function for a continuous random variable and a finite probability function refers to a discrete probability distribution with a finite number of possible outcomes.

Question 12

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 13

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.
 $$\begin{array} { c c } 
\text { Number of } & \
\text { New Clients } & \text { Prabability } \
0 & 0.05 \
1 & 0.10 \
2 & 0.15 \
3 & 0.35 \
4 & 0.20 \
5 & 0.10 \
6 & 0.05
\end{array}$$
-Refer to Exhibit 5-3. The standard deviation is
A) 1.431
B) 2.047
C) 3.05
D) 21

Accepted Answer

The answer of Exhibit 5-3
Roth is a computer-consulting firm. The...

Question 14

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

Using the binomial probability formula, we have:

P(exactly 2 females) = (8 choose 2) * 0.6^2 * 0.4^6
= 28 * 0.36 * 0.02704
= 0.0413

Therefore, the probability that among the students in the sample exactly two are female is 0.0413 or approximately 4.13%. The best choice is C.

Question 15

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 answer of Exhibit 5-4
Forty percent of all registered voters...

Question 16

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

This problem involves a binomial distribution with n = 5 and p = 0.4. Using the binomial probability formula, we can calculate the probability of getting exactly 2 females in the sample:

P(X = 2) = (5 choose 2) * 0.4^2 * 0.6^3 = 0.3456

Therefore, the correct answer is D.

Question 17

Exhibit 5-5
Probability Distribution
 $\begin{array}{ll}x&fx)\
10 & .2 \
20 & .3 \
30 & .4 \
40 & .1
\end{array}$
-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
 \(\begin{array}{ll}x&fx)\
10 & .2 \
20...

Question 18

Exhibit 5-5
Probability Distribution
 $\begin{array}{ll}x&fx)\
10 & .2 \
20 & .3 \
30 & .4 \
40 & .1
\end{array}$
-Refer to Exhibit 5-5. The expected value of x equals
A) 24
B) 25
C) 30
D) 100

Accepted Answer

The answer of Exhibit 5-5
Probability Distribution
 \(\begin{array}{ll}x&fx)\
10 & .2 \
20...

Question 19

Assume that you have a binomial experiment with p = 0.5 and a sample size of 100. The value of the standard deviation is
A) 50
B) 2
C) 25
D) 5

Accepted Answer

The standard deviation of a binomial experiment with sample size n and probability of success p is given by sqrt(np(1-p)). Therefore, in this case, the standard deviation is sqrt(100*0.5*(1-0.5)) = 5.

Question 20

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.

Quiz 5: Discrete Probability Distributions

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?

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

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

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 5-1 The following represents the probability distribution for the daily demand of computers at a local store. $\begin{array} { c c } \text { Demand } & \text { Probability } \\0 & 0.1 \\1 & 0.2 \\2 & 0.3 \\3 & 0.2 \\4 & 0 ?\end{array}$ -Refer to Exhibit 5-1. The expected daily demand is

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

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

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?

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

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 5-5 Probability Distribution $\begin{array}{ll}x&fx) \\10 & .2 \\20 & .3 \\30 & .4 \\40 & .1\end{array}$ -Refer to Exhibit 5-5. The variance of x equals

Exhibit 5-5 Probability Distribution $\begin{array}{ll}x&fx) \\10 & .2 \\20 & .3 \\30 & .4 \\40 & .1\end{array}$ -Refer to Exhibit 5-5. The expected value of x equals

Assume that you have a binomial experiment with p = 0.5 and a sample size of 100. The value of the standard deviation is

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

Quiz 5: Discrete Probability Distributions

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?

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

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

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

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

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?

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

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 5-5 Probability Distribution xfx) 10.220.330.440.1\begin{array}{ll}x&fx) \\10 & .2 \\20 & .3 \\30 & .4 \\40 & .1\end{array}x10203040​fx) .2.3.4.1​ -Refer to Exhibit 5-5. The variance of x equals

Exhibit 5-5 Probability Distribution xfx) 10.220.330.440.1\begin{array}{ll}x&fx) \\10 & .2 \\20 & .3 \\30 & .4 \\40 & .1\end{array}x10203040​fx) .2.3.4.1​ -Refer to Exhibit 5-5. The expected value of x equals

Assume that you have a binomial experiment with p = 0.5 and a sample size of 100. The value of the standard deviation is

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

Exhibit 5-5 Probability Distribution $\begin{array}{ll}x&fx) \\10 & .2 \\20 & .3 \\30 & .4 \\40 & .1\end{array}$ -Refer to Exhibit 5-5. The variance of x equals

Exhibit 5-5 Probability Distribution $\begin{array}{ll}x&fx) \\10 & .2 \\20 & .3 \\30 & .4 \\40 & .1\end{array}$ -Refer to Exhibit 5-5. The expected value of x equals