Question 1

Exhibit 5-10
The probability that Pete will catch fish on a particular day when he goes fishing is 0.8. Pete is going fishing 3 days next week.
-Refer to Exhibit 5-10. The probability that Pete will catch fish on one day or less is
A).008
B).096
C).104
D).8

Accepted Answer

We can use the binomial distribution formula to calculate the probability of at most one success in three independent trials with probability of success 0.8. P(X ≤ 1) = P(X = 0) + P(X = 1) = (0.2)^3 + 3(0.8)(0.2)^2 = 0.008 + 0.096 = 0.104. Therefore, the correct answer is (C).

Question 2

Exhibit 5-11
The random variable x is the number of occurrences of an event over an interval of ten minutes. It can be assumed that the probability of an occurrence is the same in any two time periods of an equal length. It is known that the mean number of occurrences in ten minutes is 5.3.
-Refer to Exhibit 5-11. The probability that there are 8 occurrences in ten minutes is
A).0241
B).0771
C).1126
D).9107

Accepted Answer

The answer of Exhibit 5-11
The random variable x is the...

Question 3

Exhibit 5-11
The random variable x is the number of occurrences of an event over an interval of ten minutes. It can be assumed that the probability of an occurrence is the same in any two time periods of an equal length. It is known that the mean number of occurrences in ten minutes is 5.3.
-Refer to Exhibit 5-11. The random variable x satisfies which of the following probability distributions?
A)normal
B)Poisson
C)binomial
D)Not enough information is given to answer this question.

Accepted Answer

The given scenario fits the Poisson probability distribution as it involves counting the number of occurrences of an event over a fixed interval of time (ten minutes), with a known average rate (mean) of occurrence (5.3), and assuming the probability of occurrence is constant across any equal interval of time. The Poisson distribution models the probability of a certain number of events occurring in a fixed time interval, given the average rate of occurrence.

Question 4

Exhibit 5-9
Forty percent of all registered voters in a national election are female. A random sample of 5 voters is selected.
-Refer to Exhibit 5-9. What is the random variable in this experiment?
A)the 40% of female registered voters
B)the random sample of 5 voters
C)the number of female voters out of 5
D)the number of registered voters in the nation

Accepted Answer

The random variable in this experiment is the number of female voters out of 5, as this is what is being observed and counted in the sample. The percentage of female registered voters and the total number of registered voters are not random variables in this particular experiment. The random sample of 5 voters is important in determining the number of female voters, but it is not the random variable being measured.

Question 5

Exhibit 5-8
The student body of a large university consists of 60% female students. A random sample of 8 students is selected.
-Refer to Exhibit 5-8. 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 5-8
The student body of a large...

Question 6

Exhibit 5-10
The probability that Pete will catch fish on a particular day when he goes fishing is 0.8. Pete is going fishing 3 days next week.
-Refer to Exhibit 5-10. The probability that Pete will catch fish on exactly one day is
A).008
B).096
C).104
D).8

Accepted Answer

The answer of Exhibit 5-10
The probability that Pete will catch...

Question 7

Exhibit 5-11
The random variable x is the number of occurrences of an event over an interval of ten minutes. It can be assumed that the probability of an occurrence is the same in any two time periods of an equal length. It is known that the mean number of occurrences in ten minutes is 5.3.
-Refer to Exhibit 5-11. The appropriate probability distribution for the random variable is
A)discrete
B)continuous
C)either a or b depending on how the interval is defined
D)not enough information is given

Accepted Answer

The fact that x is the number of occurrences of an event indicates a discrete random variable. The information given about the mean number of occurrences in ten minutes further supports a discrete probability distribution.

Question 8

Exhibit 5-8
The student body of a large university consists of 60% female students. A random sample of 8 students is selected.
-Refer to Exhibit 5-8. 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 5-8
The student body of a large...

Question 9

Exhibit 5-11
The random variable x is the number of occurrences of an event over an interval of ten minutes. It can be assumed that the probability of an occurrence is the same in any two time periods of an equal length. It is known that the mean number of occurrences in ten minutes is 5.3.
-Refer to Exhibit 5-11. The expected value of the random variable x is
A)2
B)5.3
C)10
D)2.30

Accepted Answer

The expected value of a random variable is equal to its mean, in this case, the mean number of occurrences in ten minutes is given to be 5.3, so the expected value of x is also 5.3.

Question 10

Exhibit 5-8
The student body of a large university consists of 60% female students. A random sample of 8 students is selected.
-Refer to Exhibit 5-8. What is the random variable in this experiment?
A)the 60% of female students
B)the random sample of 8 students
C)the number of female students out of 8
D)the student body size

Accepted Answer

The random variable in this experiment is the number of female students out of 8, as it is the value of interest that we are measuring and analyzing.

Question 11

Exhibit 5-10
The probability that Pete will catch fish on a particular day when he goes fishing is 0.8. Pete is going fishing 3 days next week.
-Refer to Exhibit 5-10. What is the random variable in this experiment?
A)the 0.8 probability of catching fish
B)the 3 days
C)the number of days out of 3 that Pete catches fish
D)the number of fish in the body of water

Accepted Answer

The random variable in this experiment is the number of days out of 3 that Pete catches fish. This is because it is the outcome we are interested in measuring and it can take on different values (0, 1, 2, or 3). The probability of catching fish is simply the probability of a certain outcome occurring, not the random variable itself. The number of days Pete is going fishing (3) is not a random variable, it is a given quantity. The number of fish in the body of water is not relevant to the question being asked.

Question 12

Exhibit 5-8
The student body of a large university consists of 60% female students. A random sample of 8 students is selected.
-Refer to Exhibit 5-8. 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 5-8
The student body of a large...

Question 13

Exhibit 5-9
Forty percent of all registered voters in a national election are female. A random sample of 5 voters is selected.
-Refer to Exhibit 5-9. 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 is a binomial distribution problem, with n=5 and p=0.4 (probability of a female voter). 
The formula for probability of x successes out of n trials is: 
P(x) = (nCx) * p^x * (1-p)^(n-x), where nCx is the combination of n things taken x at a time. 
So, for P(2), we have: 
P(2) = (5C2) * 0.4^2 * 0.6^3 = 10 * 0.16 * 0.216 = 0.3456 
Therefore, the answer is D.

Question 14

Exhibit 5-7
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 5-7. 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 5-7
A sample of 2,500 people was...

Question 15

Exhibit 5-7
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 5-7. The expected number of cups of coffee is
A)1
B)1.2
C)1.5
D)1.7

Accepted Answer

The expected number of cups of coffee is calculated by taking the sum of (frequency x value) for each value and dividing by the total frequency. 
Expected number of cups of coffee = (400x0 + 500x1 + 600x2 + 800x3 + 200x4)/2500 
Expected number of cups of coffee = 3200/2500 = 1.28 
Therefore, the best choice is B – 1.2 cups of coffee.

Question 16

Exhibit 5-6
Probability Distribution
 
-Refer to Exhibit 5-6. The variance of x equals
A)9.165
B)84
C)85
D)93.33

Accepted Answer

The answer of Exhibit 5-6
Probability Distribution
 
-Refer to Exhibit 5-6....

Question 17

Exhibit 5-10
The probability that Pete will catch fish on a particular day when he goes fishing is 0.8. Pete is going fishing 3 days next week.
-Refer to Exhibit 5-10. The variance of the number of days Pete will catch fish is
A).16
B).48
C).8
D)2.4

Accepted Answer

The variance of a binomial distribution can be calculated using the formula npq, where n is the number of trials, p is the probability of success, and q is the probability of failure (1-p). In this case, n=3, p=0.8, and q=0.2. Therefore, the variance is (3)(0.8)(0.2) = 0.48.

Question 18

Exhibit 5-11
The random variable x is the number of occurrences of an event over an interval of ten minutes. It can be assumed that the probability of an occurrence is the same in any two time periods of an equal length. It is known that the mean number of occurrences in ten minutes is 5.3.
-Refer to Exhibit 5-11. The probability that there are less than 3 occurrences is
A).0659
B).0948
C).1016
D).1239

Accepted Answer

The answer of Exhibit 5-11
The random variable x is the...

Question 19

Exhibit 5-10
The probability that Pete will catch fish on a particular day when he goes fishing is 0.8. Pete is going fishing 3 days next week.
-Refer to Exhibit 5-10. The expected number of days Pete will catch fish is
A).6
B).8
C)2.4
D)3

Accepted Answer

To find the expected number of days Pete will catch fish, we multiply the probability of catching fish (0.8) by the number of days he is going fishing (3). Expected number of days = 0.8 x 3 = 2.4. Therefore, the best choice is C.

Question 20

Exhibit 5-9
Forty percent of all registered voters in a national election are female. A random sample of 5 voters is selected.
-Refer to Exhibit 5-9. 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 of selecting a female voter in the sample is 0.4, and the probability of selecting a male voter is 0.6. The probability of selecting no females in the sample is therefore (0.6)^5, which is approximately 0.0778.

Quiz 5: Discrete Probability Distributions

Exhibit 5-10 The probability that Pete will catch fish on a particular day when he goes fishing is 0.8. Pete is going fishing 3 days next week. -Refer to Exhibit 5-10. The probability that Pete will catch fish on one day or less is

Exhibit 5-9 Forty percent of all registered voters in a national election are female. A random sample of 5 voters is selected. -Refer to Exhibit 5-9. What is the random variable in this experiment?

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

Exhibit 5-10 The probability that Pete will catch fish on a particular day when he goes fishing is 0.8. Pete is going fishing 3 days next week. -Refer to Exhibit 5-10. The probability that Pete will catch fish on exactly one day is

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

Exhibit 5-8 The student body of a large university consists of 60% female students. A random sample of 8 students is selected. -Refer to Exhibit 5-8. What is the random variable in this experiment?

Exhibit 5-10 The probability that Pete will catch fish on a particular day when he goes fishing is 0.8. Pete is going fishing 3 days next week. -Refer to Exhibit 5-10. What is the random variable in this experiment?

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

Exhibit 5-9 Forty percent of all registered voters in a national election are female. A random sample of 5 voters is selected. -Refer to Exhibit 5-9. The probability that the sample contains 2 female voters is

Exhibit 5-7 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 5-7. The variance of the number of cups of coffee is

Exhibit 5-7 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 5-7. The expected number of cups of coffee is

Exhibit 5-6 Probability Distribution -Refer to Exhibit 5-6. The variance of x equals

Exhibit 5-10 The probability that Pete will catch fish on a particular day when he goes fishing is 0.8. Pete is going fishing 3 days next week. -Refer to Exhibit 5-10. The variance of the number of days Pete will catch fish is

Exhibit 5-10 The probability that Pete will catch fish on a particular day when he goes fishing is 0.8. Pete is going fishing 3 days next week. -Refer to Exhibit 5-10. The expected number of days Pete will catch fish is

Exhibit 5-9 Forty percent of all registered voters in a national election are female. A random sample of 5 voters is selected. -Refer to Exhibit 5-9. The probability that there are no females in the sample is