Question 1

An experiment of a gender selection method includes a control group of 8 couples who are not given any treatment intended to influence the genders of their babies. Construct a table listing the possible values of the random variable x (which represents the number of girls among the 8 births) and the corresponding probabilities. Then find the mean and standard deviation for the number of girls in such groups of 10 and the maximum and minimum usual values for the number of girls. Round all results to four decimal places.

Accepted Answer

The answer of An experiment of a gender selection method...

Question 2

Suppose that in one town 10% of people are left handed. Suppose that you want to find the probability of getting exactly 2 left-handed people when 4 people are randomly selected.
Can the answer be found as follows: Use the multiplication rule to find the probability of getting two left handers followed by two right handers, which is (0.1)(0.1)(0.9)(0.9)? If not,
explain why not and show how the required probability can be found.

Accepted Answer

The answer of Suppose that in one town 10% of...

Question 3

Find the standard deviation, ?, for the binomial distribution which has the stated values of n and p. Round your answer to the nearest hundredth.
-$\mathrm { n } = 693 ; \mathrm { p } = 0.7$

A) $\sigma = 16.18$
B) $\sigma = 9.65$
C) $\sigma = 12.06$
D) $\sigma = 15.33$

Accepted Answer

The answer of Find the standard deviation, ?, for the...

Question 4

Determine whether the given procedure results in a binomial distribution. If not, state the reason why.
-Rolling a single "loaded" die 11 times, keeping track of the numbers that are rolled.
A) Not binomial: the trials are not independent.
B) Not binomial: there are more than two outcomes for each trial.
C) Procedure results in a binomial distribution.
D) Not binomial: there are too many trials.

Accepted Answer

A binomial distribution requires only two possible outcomes per trial, but rolling a die results in six possible outcomes.

Question 5

List the four requirements for a binomial distribution. Describe an experiment which is binomial and discuss how the experiment fits each of the four requirements.

Accepted Answer

The answer of List the four requirements for a binomial...

Question 6

Determine whether the following is a probability distribution. If not, identify the requirement that is not satisfied.
-If a person is randomly selected from a certain town, the probability distribution for the number, x, of siblings is as described in the accompanying table.
 $$\begin{array} { c | c } 
\mathrm { x } & \mathrm { P } ( \mathrm { x } ) \
\hline 0 & 0.27 \
1 & 0.28 \
2 & 0.23 \
3 & 0.10 \
4 & 0.06 \
5 & 0.02
\end{array}$$

Accepted Answer

The answer of Determine whether the following is a probability...

Question 7

Suppose a mathematician computed the expected value of winnings for a person playing each of seven different games in a casino. What would you expect to be true for all expected values for these seven games?

Accepted Answer

The answer of Suppose a mathematician computed the expected value...

Question 8

Focus groups of 11 people are randomly selected to discuss products of the Famous Company. It is determined that the mean number (per group) who recognize the Famous brand name is 5.7, and The standard deviation is 0.50. Would it be unusual to randomly select 11 people and find that Greater than 9 recognize the Famous brand name?
A) Yes
B) No

Accepted Answer

To determine if it is unusual, we calculate the z-score: z = (9 - 5.7) / (0.50/√11) = 22.08. A z-score this high is extremely rare, indicating it is unusual.

Question 9

Find the indicated probability. Round to three decimal places.
-An airline estimates that 91% of people booked on their flights actually show up. If the airline books 76 people on a flight for which the maximum number is 74, what is the probability that the Number of people who show up will exceed the capacity of the plane?
A) 0.028
B) 0.006
C) 0.007
D) 0.001

Accepted Answer

The problem can be modeled using a binomial distribution with n = 76 and p = 0.91. We need to find the probability that more than 74 people show up, i.e., P(X > 74). This can be calculated using the complement rule: P(X > 74) = 1 - P(X ≤ 74). Using a binomial calculator or statistical software, P(X ≤ 74) is approximately 0.993, so P(X > 74) is approximately 0.007.

Question 10

Assume that a procedure yields a binomial distribution with a trial repeated n times. Use the binomial probability
formula to find the probability of x successes given the probability p of success on a single trial. Round to three decimal
places.
-$\mathrm { n } = 4 , \mathrm { x } = 3 , \mathrm { p } = \frac { 1 } { 6 }$

A) 0.012
B) 0.015
C) 0.004
D) 0.023

Accepted Answer

The binomial probability formula is $ P(X = x) = \binom{n}{x} p^x (1-p)^{n-x} $. For $ n = 4 $, $ x = 3 $, and $ p = \frac{1}{6} $, the calculation is $ \binom{4}{3} \left(\frac{1}{6}\right)^3 \left(\frac{5}{6}\right)^1 = 4 \times \frac{1}{216} \times \frac{5}{6} = 0.015 $.

Question 11

Find the mean of the given probability distribution.
-$\begin{array}{c|c}
\mathrm{x} & \mathrm{P}(\mathrm{x}) \
\hline 0 & 0.23 \
1 & 0.20 \
2 & 0.37 \
3 & 0.06 \
4 & 0.14
\end{array}$

A) $\mu = 1.81$
B) $\mu = 1.68$
C) $\mu = 1.58$
D) $\mu = 1.91$

Accepted Answer

The mean of a probability distribution is calculated as $\mu = \sum [x \cdot P(x)]$. Here, $\mu = (0 \cdot 0.23) + (1 \cdot 0.20) + (2 \cdot 0.37) + (3 \cdot 0.06) + (4 \cdot 0.14) = 1.68$.

Question 12

If a procedure meets all the conditions of a binomial distribution except that the number of trials is not fixed, then the geometric distribution can be used. The probability of getting the first success on The xth trial is given by $P ( x ) = p ( 1 - p ) ^ { x - 1 }$ , where p is the probability of success on any one trial.Assume that the probability of choosing a yellow piece of candy in a bag of hard candy is 0.240.
Find the probability that the first yellow candy is found in the fourth inspected. Round your answer to the nearest thousandth.
A) 0.08
B) 0.061
C) 0.439
D) 0.105

Accepted Answer

The answer of If a procedure meets all the conditions...

Question 13

List the three methods for finding binomial probabilities in the table below, and then complete the table to discuss the advantages and disadvantages of each.
 $$\begin{array} { | l | l | l | } 
\hline \text { Methods } & \text { Advantage } & \text { Disadvantage } \
\hline & & \
\hline & & \
\hline & & \
\hline & & \
\hline
\end{array}$$

Accepted Answer

The answer of List the three methods for finding binomial...

Question 14

Find the mean of the given probability distribution.
-The probabilities that a batch of 4 computers will contain 0, 1, 2, 3, and 4 defective computers are 0.6561, 0.2916, 0.0486, 0.0036, and 0.0001, respectively. Round answer to the nearest hundredth. 
A) $\mu = 0.30$
B) $\mu = 2.00$
C) $\mu = 1.06$
D) $\mu = 0.40$

Accepted Answer

The answer of Find the mean of the given probability...

Question 15

Use the given values of n and p to find the minimum usual value $\mu - 2 \sigma$ and the maximum usual value $\mu + 2 \sigma$ . Round your answer to the nearest hundredth unless otherwise noted.
-$n = 460 , p = \frac { 2 } { 7 }$
A) Minimum: 112.05; maximum: 150.81
B) Minimum: 121.74; maximum: 141.12
C) Minimum: 150.81; maximum: 112.05
D) Minimum: 117.73; maximum: 145.13

Accepted Answer

The answer of Use the given values of n and...

Question 16

Find the indicated probability. Round to three decimal places.
-A car insurance company has determined that 8% of all drivers were involved in a car accident last year. Among the 14 drivers living on one particular street, 3 were involved in a car accident last year. If 14 drivers are randomly selected, what is the probability of getting 3 or more who were involved in a car accident last year?
A) 0.074
B) 0.926
C) 0.407
D) 0.096

Accepted Answer

The probability of getting 3 or more drivers involved in an accident can be calculated using the binomial probability formula. The probability of 3 or more is 1 minus the probability of 0, 1, or 2, which results in approximately 0.096.

Question 17

A multiple choice test has 7 questions each of which has 5 possible answers, only one of which is correct. If Judy, who forgot to study for the test, guesses on all questions, what is the probability That she will answer exactly 3 questions correctly?

A) 0.115
B) 0.885
C) 0.275
D) 0.00800

Accepted Answer

The probability of guessing a question correctly is 1/5, and incorrectly is 4/5. Using the binomial probability formula for exactly 3 correct answers out of 7, we calculate: C(7,3) * (1/5)^3 * (4/5)^4 = 0.115.

Question 18

In a certain town, 90 percent of voters are in favor of a given ballot measure and 10 percent are opposed. For groups of 260 voters, find the mean for the number who oppose the measure.
A) 26
B) 90
C) 10
D) 234

Accepted Answer

The mean number of voters who oppose the measure in a group of 260 is calculated by multiplying the total number of voters (260) by the percentage who oppose (10%), which equals 26.

Question 19

A 28-year-old man pays $118 for a one-year life insurance policy with coverage of $140,000. If the probability that he will live through the year is 0.9993, what is the expected value for the insurance
Policy?
A) $139,902.00
B) -$20.00
C) -$117.92
D) $98.00

Accepted Answer

The answer of A 28-year-old man pays $118 for a...

Question 20

The brand name of a certain chain of coffee shops has a 50% recognition rate in the town of Coffleton. An executive from the company wants to verify the recognition rate as the company is Interested in opening a coffee shop in the town. He selects a random sample of 9 Coffleton Residents. Find the probability that the number that recognize the brand name is not 4.
A) 0.754
B) 0.164
C) 0.246
D) 0.00195

Accepted Answer

The probability that exactly 4 out of 9 recognize the brand is calculated using the binomial probability formula. The probability of not getting exactly 4 is 1 minus this probability, which is approximately 0.754.

Quiz 5: Discrete Probability Distributions

Find the standard deviation, ?, for the binomial distribution which has the stated values of n and p. Round your answer to the nearest hundredth. - $\mathrm { n } = 693 ; \mathrm { p } = 0.7$

Determine whether the given procedure results in a binomial distribution. If not, state the reason why. -Rolling a single "loaded" die 11 times, keeping track of the numbers that are rolled.

List the four requirements for a binomial distribution. Describe an experiment which is binomial and discuss how the experiment fits each of the four requirements.

Suppose a mathematician computed the expected value of winnings for a person playing each of seven different games in a casino. What would you expect to be true for all expected values for these seven games?

Find the mean of the given probability distribution. - $\begin{array}{c|c}\mathrm{x} & \mathrm{P}(\mathrm{x}) \\\hline 0 & 0.23 \\1 & 0.20 \\2 & 0.37 \\3 & 0.06 \\4 & 0.14\end{array}$

Find the mean of the given probability distribution. -The probabilities that a batch of 4 computers will contain 0, 1, 2, 3, and 4 defective computers are 0.6561, 0.2916, 0.0486, 0.0036, and 0.0001, respectively. Round answer to the nearest hundredth.

Use the given values of n and p to find the minimum usual value $\mu - 2 \sigma$ and the maximum usual value $\mu + 2 \sigma$ . Round your answer to the nearest hundredth unless otherwise noted. - $n = 460 , p = \frac { 2 } { 7 }$

A multiple choice test has 7 questions each of which has 5 possible answers, only one of which is correct. If Judy, who forgot to study for the test, guesses on all questions, what is the probability That she will answer exactly 3 questions correctly?

In a certain town, 90 percent of voters are in favor of a given ballot measure and 10 percent are opposed. For groups of 260 voters, find the mean for the number who oppose the measure.

A 28-year-old man pays $118 for a one-year life insurance policy with coverage of $140,000. If the probability that he will live through the year is 0.9993, what is the expected value for the insurance Policy?

Quiz 5: Discrete Probability Distributions

Find the standard deviation, ?, for the binomial distribution which has the stated values of n and p. Round your answer to the nearest hundredth. - n=693;p=0.7\mathrm { n } = 693 ; \mathrm { p } = 0.7n=693;p=0.7

Determine whether the given procedure results in a binomial distribution. If not, state the reason why. -Rolling a single "loaded" die 11 times, keeping track of the numbers that are rolled.

List the four requirements for a binomial distribution. Describe an experiment which is binomial and discuss how the experiment fits each of the four requirements.

Suppose a mathematician computed the expected value of winnings for a person playing each of seven different games in a casino. What would you expect to be true for all expected values for these seven games?

Find the mean of the given probability distribution. - xP(x) 00.2310.2020.3730.0640.14\begin{array}{c|c}\mathrm{x} & \mathrm{P}(\mathrm{x}) \\\hline 0 & 0.23 \\1 & 0.20 \\2 & 0.37 \\3 & 0.06 \\4 & 0.14\end{array}x01234​P(x) 0.230.200.370.060.14​​

Find the mean of the given probability distribution. -The probabilities that a batch of 4 computers will contain 0, 1, 2, 3, and 4 defective computers are 0.6561, 0.2916, 0.0486, 0.0036, and 0.0001, respectively. Round answer to the nearest hundredth.

Use the given values of n and p to find the minimum usual value μ−2σ\mu - 2 \sigmaμ−2σ and the maximum usual value μ+2σ\mu + 2 \sigmaμ+2σ . Round your answer to the nearest hundredth unless otherwise noted. - n=460,p=27n = 460 , p = \frac { 2 } { 7 }n=460,p=72​

A multiple choice test has 7 questions each of which has 5 possible answers, only one of which is correct. If Judy, who forgot to study for the test, guesses on all questions, what is the probability That she will answer exactly 3 questions correctly?

In a certain town, 90 percent of voters are in favor of a given ballot measure and 10 percent are opposed. For groups of 260 voters, find the mean for the number who oppose the measure.

A 28-year-old man pays $118 for a one-year life insurance policy with coverage of $140,000. If the probability that he will live through the year is 0.9993, what is the expected value for the insurance Policy?

Find the standard deviation, ?, for the binomial distribution which has the stated values of n and p. Round your answer to the nearest hundredth. - $\mathrm { n } = 693 ; \mathrm { p } = 0.7$

Find the mean of the given probability distribution. - $\begin{array}{c|c}\mathrm{x} & \mathrm{P}(\mathrm{x}) \\\hline 0 & 0.23 \\1 & 0.20 \\2 & 0.37 \\3 & 0.06 \\4 & 0.14\end{array}$

Use the given values of n and p to find the minimum usual value $\mu - 2 \sigma$ and the maximum usual value $\mu + 2 \sigma$ . Round your answer to the nearest hundredth unless otherwise noted. - $n = 460 , p = \frac { 2 } { 7 }$