Question 1

At an airport, 77% of recent flights have arrived on time. A sample of 11 flights is studied. What is the mean number of flights that are on time?
A) 1.95
B) 8.47
C) 2.53
D) 2.91

Accepted Answer

The mean number of flights that are on time can be found by multiplying the total number of flights in the sample (11) by the probability of a flight arriving on time (77% or 0.77). Thus, the mean is 11 * 0.77 = 8.47.

Question 2

It is estimated that 30% of households own a riding lawn mower. A sample of 18 households is studied. What is the probability that no more than 3 of these own a riding lawn mower?
A) 0.94
B) 0.1646
C) 0.8354
D) 0.0600

Accepted Answer

This is a binomial probability problem with parameters n = 18 and p = 0.30. We want to find the probability that no more than 3 households own a riding lawn mower, so we can use the cumulative probability function of the binomial distribution:

P(X ≤ 3) = Σ P(X = i), i from 0 to 3

Using a calculator or software, we can find that:

P(X ≤ 3) = 0.1646

Therefore, the best choice is B.

Question 3

It is estimated that 45% of households own a riding lawn mower. A sample of 15 households is studied. What is the probability that exactly 7 of these own a riding lawn mower?
A) 0.8353
B) 0.1647
C) 0.7987
D) 0.2013

Accepted Answer

The answer of It is estimated that 45% of households...

Question 4

At an airport, 83% of recent flights have arrived on time. A sample of 12 flights is studied. What is the standard deviation of the number of flights that are on time?
A) 1.69
B) 9.96
C) 1.30
D) 0.38

Accepted Answer

The standard deviation of a binomial distribution can be calculated using the formula σ = sqrt(npq), where n is the number of trials, p is the probability of success, and q is the probability of failure (q = 1 - p). 
In this case, n = 12 and p = 0.83 (since 83% of recent flights have arrived on time, the probability of success is 0.83). Therefore, q = 1 - 0.83 = 0.17. 
Using the formula, we get: 
σ = sqrt(12 x 0.83 x 0.17) ≈ 1.30 
Therefore, the standard deviation of the number of flights that are on time is approximately 1.30, which corresponds to choice C.

Question 5

In a poll conducted by a survey firm, 75% of respondents said that their jobs were sometimes or always stressful. Nine workers are chosen at random. What is the probability that less than 4 of Them find their jobs stressful?
A) 0.0100
B) 0.9900
C) 0.0489
D) 0.0389

Accepted Answer

The probability of a worker finding their job stressful is 0.75. The probability of a worker not finding their job stressful is 0.25. The probability of less than 4 workers finding their jobs stressful is the sum of the probabilities of 0, 1, 2, and 3 workers finding their jobs stressful. This can be calculated using the binomial distribution:P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)= (9 choose 0) * (0.75)^0 * (0.25)^9 + (9 choose 1) * (0.75)^1 * (0.25)^8 + (9 choose 2) * (0.75)^2 * (0.25)^7 + (9 choose 3) * (0.75)^3 * (0.25)^6= 0.0019 + 0.0146 + 0.0547 + 0.1367= 0.2079Therefore, the probability of less than 4 workers finding their jobs stressful is 0.2079, which is closest to choice A, 0.0100.

Question 6

In a poll conducted by a survey firm, 73% of respondents said that their jobs were sometimes or always stressful. Eight workers are chosen at random. What is the mean number of workers who Find their job stressful?
A) 5.84
B) 1.58
C) 1.26
D) 2.16

Accepted Answer

The mean number of workers who find their job stressful can be found by multiplying the total number of workers chosen by the percentage who find their job stressful. Thus, $8 \times 0.73 = 5.84$.

Question 7

In a poll conducted by a survey firm, 82% of respondents said that their jobs were sometimes or always stressful. Eleven workers are chosen at random. What is the probability that more than 9 of Them find their jobs stressful?
A) 0.3164
B) 0.3849
C) 0.6836
D) 0.6151

Accepted Answer

The probability that more than 9 out of 11 workers find their jobs stressful can be calculated using the binomial probability formula. Since the probability of a single worker finding their job stressful is 0.82, we need to find the probability of exactly 10 and exactly 11 workers finding their jobs stressful, and then sum these probabilities.For exactly 10 workers: $P(X=10) = \binom{11}{10}(0.82)^{10}(0.18)^{1}$For exactly 11 workers: $P(X=11) = \binom{11}{11}(0.82)^{11}(0.18)^{0}$Summing these probabilities gives the total probability of more than 9 out of 11 workers finding their jobs stressful, which is approximately 0.3849.

Question 8

The Australian sheep dog is a breed renowned for its intelligence and work ethic. It is estimated that 45% of adult Australian sheep dogs weigh 65 pounds or more. A sample of 16 adult dogs is studied.
What is the probability that no more than 3 of them weigh 65 lb or more?
A) 0.9719
B) 0.9934
C) 0.0066
D) 0.0281

Accepted Answer

This is a binomial probability problem, where:

n = 16 (sample size)
p = 0.45 (probability of success, i.e. a dog weighing 65 lb or more)

We want to find the probability of no more than 3 successes (dogs weighing 65 lb or more), which can be written as:

P(X ≤ 3)

Using a binomial probability table, we can find this probability to be 0.0281.

Therefore, the answer is D.

Question 9

The Australian sheep dog is a breed renowned for its intelligence and work ethic. It is estimated that 25% of adult Australian sheep dogs weigh 65 pounds or more. A sample of 11 adult dogs is studied.
What is the mean number of dogs who weigh 65 lb or more?
A) 11
B) 2.0625
C) 2.75
D) 1.4361

Accepted Answer

To find the mean number of dogs who weigh 65 pounds or more, we need to first calculate the total number of dogs who weigh 65 pounds or more in the sample.

25% of 11 dogs is (25/100) x 11 = 2.75 dogs.

Therefore, the mean number of dogs who weigh 65 pounds or more is 2.75.

The best choice is C.

Question 10

It is estimated that 40% of households own a riding lawn mower. A sample of 13 households is studied. What is the mean number of households who own a riding mower?
A) 3.12
B) 13
C) 1.7664
D) 5.2

Accepted Answer

The mean number of households who own a riding mower can be calculated by multiplying the total number of households by the proportion that own a riding mower. 
In this case: 
Mean = Total households x Proportion 
Mean = 13 x 0.4 
Mean = 5.2

Therefore, the best choice is D (5.2).

Question 11

The Australian sheep dog is a breed renowned for its intelligence and work ethic. It is estimated that 45% of adult Australian sheep dogs weigh 65 pounds or more. A sample of 12 adult dogs is studied.
What is the probability that more than 9 of them weigh 65 lb or more?
A) 0.0079
B) 0.9644
C) 0.0356
D) 0.0011

Accepted Answer

The probability of a dog weighing 65 lb or more is 0.45. The probability of a dog weighing less than 65 lb is 0.55. The probability of more than 9 of the 12 dogs weighing 65 lb or more is 0.0079.

Question 12

At an airport, 78% of recent flights have arrived on time. A sample of 11 flights is studied. Find the probability that exactly 7 of them were on time.
A) 0.1358
B) 0.0014
C) 0.8642
D) 0.0030

Accepted Answer

This is a binomial probability problem, where n = 11 (number of trials), k = 7 (number of successes), p = 0.78 (probability of success on a single trial), and q = 1 - p = 0.22 (probability of failure). The binomial probability formula is P(X = k) = (nCk) * p^k * q^(n-k). Plugging in the values, we get P(X = 7) ≈ 0.1358.

Question 13

Determine the indicated probability for a Poisson random variable with the given values of $\lambda \text { and } t$ $\lambda = 0.8 , t = 3 , P ( 2 )$
A) 0.0431
B) 0.2613
C) 0.9569
D) 0.3840

Accepted Answer

The probability of observing exactly $k$ events in a Poisson distribution is given by the formula $P(k; \lambda t) = \frac{e^{-\lambda t}(\lambda t)^k}{k!}$. For this problem, $\lambda = 0.8$, $t = 3$, and $k = 2$. Plugging these values into the formula gives $P(2) = \frac{e^{-2.4}(2.4)^2}{2!} = 0.2613$.

Question 14

In a poll conducted by a survey firm, 76% of respondents said that their jobs were sometimes or always stressful. Eight workers are chosen at random. What is the standard deviation of the Number of workers who find their jobs stressful?
A) 1.21
B) 0.43
C) 1.46
D) 2.47

Accepted Answer

The standard deviation is the square root of the variance. The variance is the product of the probability of success, the probability of failure, and the number of trials. In this case, the probability of success is 0.76, the probability of failure is 0.24, and the number of trials is 8. So the variance is 0.76 * 0.24 * 8 = 1.46. The standard deviation is the square root of 1.46, which is 1.21.

Question 15

The Australian sheep dog is a breed renowned for its intelligence and work ethic. It is estimated that 25% of adult Australian sheep dogs weigh 65 pounds or more. A sample of 14 adult dogs is studied.
What is the standard deviation of the number of dogs who weigh 65 lb or more?
A) 3.5
B) 14
C) 1.6202
D) 2.625

Accepted Answer

We know that p = 0.25 (proportion of adult Australian sheep dogs weighing 65 pounds or more)
We also know that n = 14 (sample size)
Using the formula for the standard deviation of a binomial distribution, we get:

σ = sqrt(np(1-p))
σ = sqrt(14*0.25*0.75)
σ = 1.6202

Therefore, the standard deviation of the number of dogs who weigh 65 lb or more is 1.6202. Option C is the correct answer.

Question 16

At an airport, 78% of recent flights have arrived on time. A sample of 14 flights is studied. Find the probability that more than 12 of them were on time.
A) 0.3761
B) 0.6239
C) 0.1527
D) 0.8473

Accepted Answer

The probability of more than 12 flights arriving on time out of 14 can be found by calculating the probability of exactly 13 and exactly 14 flights arriving on time and adding those probabilities together. This is a binomial probability problem where n = 14, p = 0.78 (probability of success), and q = 0.22 (probability of failure). The formula for binomial probability is P(X = k) = (nCk)(p^k)(q^(n-k)), where nCk is the combination of n items taken k at a time. Calculate P(13) and P(14) and add them together to find the total probability.

Question 17

At an airport, 77% of recent flights have arrived on time. A sample of 11 flights is studied. Find the probability that no more than 4 of them were on time.
A) 0.0039
B) 0.0046
C) 0.9954
D) 0.9961

Accepted Answer

The problem can be solved using the binomial probability formula. The probability of a flight arriving on time is 0.77, and we want the probability of 4 or fewer flights arriving on time out of 11. Using the binomial distribution, the cumulative probability for 0 to 4 flights arriving on time is approximately 0.0046.

Question 18

It is estimated that 35% of households own a riding lawn mower. A sample of 10 households is studied. What is the standard deviation of the number of households who own a riding lawn mower?
A) 3.5
B) 1.5083
C) 2.275
D) 10

Accepted Answer

We can use the formula for standard deviation of a binomial distribution: 
SD = √[np(1-p)] 
where n is the sample size and p is the probability of success (in this case, owning a riding lawn mower). 
Plugging in the values, we get: 
SD = √[10(0.35)(0.65)] 
SD ≈ 1.5083 
Therefore, the standard deviation of the number of households who own a riding lawn mower is approximately 1.5083.

Question 19

In a poll conducted by a survey firm, 84% of respondents said that their jobs were sometimes or always stressful. Ten workers are chosen at random. What is the probability that exactly 6 of them Find their jobs stressful?
A) 0.0018
B) 0.0483
C) 0.9517
D) 0.0012

Accepted Answer

This is a binomial probability problem with n=10 and p=0.84.
The probability of exactly 6 workers finding their jobs stressful is:
P(X=6) = (10 choose 6) * 0.84^6 * 0.16^4
Using a calculator or software, we get:
P(X=6) = 0.0483
Therefore, the best choice is B.

Question 20

It is estimated that 40% of households own a riding lawn mower. A sample of 13 households is studied. What is the probability that more than 10 of these own a riding lawn mower?
A) 0.0078
B) 0.9922
C) 0.0013
D) 0.0001

Accepted Answer

The probability of more than 10 households (i.e., 11 or 12 or 13 households) owning a riding lawn mower can be calculated using the binomial probability formula. Given the probability (p) of a single household owning a riding lawn mower is 0.40, and the number of trials (n) is 13, we calculate the probability for exactly 11, 12, and 13 households owning a mower and sum these probabilities. The calculations yield a very small probability, with the correct answer being approximately 0.0013, which matches option C.

Quiz 6: Discrete Probability Distributions

At an airport, 77% of recent flights have arrived on time. A sample of 11 flights is studied. What is the mean number of flights that are on time?

It is estimated that 30% of households own a riding lawn mower. A sample of 18 households is studied. What is the probability that no more than 3 of these own a riding lawn mower?

It is estimated that 45% of households own a riding lawn mower. A sample of 15 households is studied. What is the probability that exactly 7 of these own a riding lawn mower?

At an airport, 83% of recent flights have arrived on time. A sample of 12 flights is studied. What is the standard deviation of the number of flights that are on time?

In a poll conducted by a survey firm, 75% of respondents said that their jobs were sometimes or always stressful. Nine workers are chosen at random. What is the probability that less than 4 of Them find their jobs stressful?

In a poll conducted by a survey firm, 73% of respondents said that their jobs were sometimes or always stressful. Eight workers are chosen at random. What is the mean number of workers who Find their job stressful?

In a poll conducted by a survey firm, 82% of respondents said that their jobs were sometimes or always stressful. Eleven workers are chosen at random. What is the probability that more than 9 of Them find their jobs stressful?

The Australian sheep dog is a breed renowned for its intelligence and work ethic. It is estimated that 45% of adult Australian sheep dogs weigh 65 pounds or more. A sample of 16 adult dogs is studied. What is the probability that no more than 3 of them weigh 65 lb or more?

The Australian sheep dog is a breed renowned for its intelligence and work ethic. It is estimated that 25% of adult Australian sheep dogs weigh 65 pounds or more. A sample of 11 adult dogs is studied. What is the mean number of dogs who weigh 65 lb or more?

It is estimated that 40% of households own a riding lawn mower. A sample of 13 households is studied. What is the mean number of households who own a riding mower?

The Australian sheep dog is a breed renowned for its intelligence and work ethic. It is estimated that 45% of adult Australian sheep dogs weigh 65 pounds or more. A sample of 12 adult dogs is studied. What is the probability that more than 9 of them weigh 65 lb or more?

At an airport, 78% of recent flights have arrived on time. A sample of 11 flights is studied. Find the probability that exactly 7 of them were on time.

Determine the indicated probability for a Poisson random variable with the given values of λ and t\lambda \text { and } tλ and t λ=0.8,t=3,P(2) \lambda = 0.8 , t = 3 , P ( 2 ) λ=0.8,t=3,P(2)

In a poll conducted by a survey firm, 76% of respondents said that their jobs were sometimes or always stressful. Eight workers are chosen at random. What is the standard deviation of the Number of workers who find their jobs stressful?

The Australian sheep dog is a breed renowned for its intelligence and work ethic. It is estimated that 25% of adult Australian sheep dogs weigh 65 pounds or more. A sample of 14 adult dogs is studied. What is the standard deviation of the number of dogs who weigh 65 lb or more?

At an airport, 78% of recent flights have arrived on time. A sample of 14 flights is studied. Find the probability that more than 12 of them were on time.

At an airport, 77% of recent flights have arrived on time. A sample of 11 flights is studied. Find the probability that no more than 4 of them were on time.

It is estimated that 35% of households own a riding lawn mower. A sample of 10 households is studied. What is the standard deviation of the number of households who own a riding lawn mower?

In a poll conducted by a survey firm, 84% of respondents said that their jobs were sometimes or always stressful. Ten workers are chosen at random. What is the probability that exactly 6 of them Find their jobs stressful?

It is estimated that 40% of households own a riding lawn mower. A sample of 13 households is studied. What is the probability that more than 10 of these own a riding lawn mower?

Determine the indicated probability for a Poisson random variable with the given values of $\lambda \text { and } t$ $\lambda = 0.8 , t = 3 , P ( 2 )$