Question 1

Determine if the outcome is unusual. Consider as unusual any result that differs from the mean by more than 2 standard deviations. That is, unusual values are either less than $\mu - 2 \sigma$ or greater than $\mu + 2 \sigma$.
-A survey for brand recognition is done and it is determined that 68% of consumers have heard of Dull Computer Company. A survey of 800 randomly selected consumers is to be conducted. For such groups of 800, would it be unusual to get 687 consumers who recognize the Dull Computer Company name?
A)Yes
B)No

Accepted Answer

The answer of Determine if the outcome is unusual. Consider...

Question 2

Solve the problem.
-In a certain town, 60 percent of voters are in favor of a given ballot measure and 40 percent are opposed. For groups of 360 voters, find the mean for the number who oppose the measure.
A)60
B)144
C)40
D)216

Accepted Answer

The mean number of voters who oppose the measure can be found by multiplying the total number of voters in a group (360) by the percentage that opposes the measure (40%). So, 360 * 0.40 = 144.

Question 3

Solve the problem.
-On a multiple choice test with 24 questions, each question has four possible answers, one of which is correct. For students who guess at all answers, find the variance for the number of correct answers.
A)4.32
B)4.313
C)4.14
D)4.5

Accepted Answer

The answer of Solve the problem.
-On a multiple choice test...

Question 4

Solve the problem.
-The probability of winning a certain lottery is 1/56,728. For people who play 517 times, find the standard deviation for the number of wins.
A)0.0955
B)0.1046
C)0.0091
D)2.1706

Accepted Answer

The standard deviation for a binomial distribution is calculated using the formula √(n * p * (1-p)), where n is the number of trials and p is the probability of success. Here, n = 517 and p = 1/56,728. Plugging these values into the formula gives approximately 0.0955.

Question 5

Solve the problem.
-The probability that a radish seed will germinate is 0.7. A gardener plants seeds in batches of 13. Find the standard deviation for the number of seeds germinating in each batch.
A)1.652
B)1.568
C)1.587
D)1.632

Accepted Answer

The standard deviation for a binomial distribution is calculated using the formula √(n * p * (1-p)), where n is the number of trials (13), and p is the probability of success (0.7). Plugging in these values gives √(13 * 0.7 * 0.3) = √(2.73) ≈ 1.652.

Question 6

Determine if the outcome is unusual. Consider as unusual any result that differs from the mean by more than 2 standard deviations. That is, unusual values are either less than $\mu - 2 \sigma$ or greater than $\mu + 2 \sigma$.
-A survey for brand recognition is done and it is determined that 68% of consumers have heard of Dull Computer Company. A survey of 800 randomly selected consumers is to be conducted. For such groups of 800, would it be unusual to get 481 consumers who recognize the Dull Computer Company name?
A)No
B)Yes

Accepted Answer

The answer of Determine if the outcome is unusual. Consider...

Question 7

Solve the problem.
-The probability of winning a certain lottery is 1/66,857. For people who play 618 times, find the mean number of wins.
A)0.0016
B)108.2
C)0.000015
D)0.0092

Accepted Answer

The mean number of wins can be found by multiplying the probability of winning (1/66,857) by the number of times played (618), which equals approximately 0.0092.

Question 8

Determine if the outcome is unusual. Consider as unusual any result that differs from the mean by more than 2 standard deviations. That is, unusual values are either less than $\mu - 2 \sigma$ or greater than $\mu + 2 \sigma$.
-A survey for brand recognition is done and it is determined that 68% of consumers have heard of Dull Computer Company. A survey of 800 randomly selected consumers is to be conducted. For such groups of 800, would it be unusual to get 541 consumers who recognize the Dull Computer Company name?
A)Yes
B)No

Accepted Answer

To determine if the outcome is unusual, we calculate the expected mean and standard deviation for the survey results. The mean ($\mu$) is given by $n \times p = 800 \times 0.68 = 544$, where $n = 800$ is the number of trials and $p = 0.68$ is the probability of success (recognizing the company). The standard deviation ($\sigma$) is given by $\sqrt{n \times p \times (1-p)} = \sqrt{800 \times 0.68 \times (1-0.68)} \approx 13.86$. The range for usual outcomes is $\mu \pm 2\sigma = 544 \pm 2 \times 13.86$, which is approximately $544 \pm 27.72$. Since 541 falls within this range, it is not considered unusual.

Question 9

Solve the problem.
-In a certain town, 44% of voters favor a given ballot measure. For groups of 31 voters, find the variance for the number who favor the measure.
A)2.76
B)7.64
C)13.64
D)58.35

Accepted Answer

The variance for a binomial distribution is calculated using the formula: variance = n * p * (1 - p), where n is the number of trials (31 voters), and p is the probability of success (0.44). So, variance = 31 * 0.44 * (1 - 0.44) = 7.64.

Question 10

Solve the problem.
-A company manufactures batteries in batches of 23 and there is a 3% rate of defects. Find the variance for the number of defects per batch.
A)0.69
B)0.665
C)0.64
D)0.669

Accepted Answer

The variance for a binomial distribution is given by the formula $ \sigma^2 = np(1-p) $, where $ n $ is the number of trials (in this case, batteries per batch, which is 23), and $ p $ is the probability of success (in this case, the defect rate, which is 3% or 0.03). Plugging in the values, we get $ \sigma^2 = 23 \times 0.03 \times (1 - 0.03) = 23 \times 0.03 \times 0.97 = 0.6691 $. Rounding to three decimal places gives 0.669.

Question 11

Determine if the outcome is unusual. Consider as unusual any result that differs from the mean by more than 2 standard deviations. That is, unusual values are either less than $\mu - 2 \sigma$ or greater than $\mu + 2 \sigma$.
-The Acme Candy Company claims that 60% of the jawbreakers it produces weigh more than .4 ounces. Suppose that 800 jawbreakers are selected at random from the production lines. Would it be unusual for this sample of 800 to contain 418 jawbreakers that weigh more than .4 ounces?
A)Yes
B)No

Accepted Answer

The answer of Determine if the outcome is unusual. Consider...

Question 12

Determine if the outcome is unusual. Consider as unusual any result that differs from the mean by more than 2 standard deviations. That is, unusual values are either less than $\mu - 2 \sigma$ or greater than $\mu + 2 \sigma$.
-According to AccuData Media Research, 36% of televisions within the Chicago city limits are tuned to "Eyewitness News" at 5:00 pm on Sunday nights. At 5:00 pm on a given Sunday, 2500 such televisions are randomly selected and checked to determine what is being watched. Would it be unusual to find that 872 of the 2500 televisions are tuned to "Eyewitness News"?
A)No
B)Yes

Accepted Answer

The mean number of televisions tuned to "Eyewitness News" is 0.36 * 2500 = 900. The standard deviation is $\sqrt{0.36 * 0.64 * 2500} = 12$. So, the unusual values are less than 900 - 2 * 12 = 876 or greater than 900 + 2 * 12 = 924. Since 872 is less than 876, it is not unusual.

Question 13

Determine if the outcome is unusual. Consider as unusual any result that differs from the mean by more than 2 standard deviations. That is, unusual values are either less than $\mu - 2 \sigma$ or greater than $\mu + 2 \sigma$.
-The Acme Candy Company claims that 8% of the jawbreakers it produces actually result in a broken jaw. Suppose 9571 persons are selected at random from those who have eaten a jawbreaker produced at Acme Candy Company. Would it be unusual for this sample of 9571 to contain 792 persons with broken jaws?
A)No
B)Yes

Accepted Answer

The expected number of broken jaws, given the 8% rate, is $0.08 \times 9571 = 765.68$. The standard deviation for a binomial distribution is $\sqrt{n p (1-p)} = \sqrt{9571 \times 0.08 \times (1-0.08)} \approx 25.5$. The range for usual outcomes is $765.68 \pm 2 \times 25.5$, or approximately $714.68$ to $816.68$. Since 792 is within this range, it is not unusual.

Question 14

Solve the problem.
-A company manufactures batteries in batches of 28 and there is a 3% rate of defects. Find the standard deviation for the number of defects per batch.
A)0.903
B)0.917
C)0.9
D)0.886

Accepted Answer

The standard deviation for a binomial distribution is calculated using the formula: sqrt(n * p * (1-p)), where n is the number of trials (28) and p is the probability of success (0.03). Plugging in the values gives sqrt(28 * 0.03 * 0.97) ≈ 0.903.

Question 15

Determine if the outcome is unusual. Consider as unusual any result that differs from the mean by more than 2 standard deviations. That is, unusual values are either less than $\mu - 2 \sigma$ or greater than $\mu + 2 \sigma$.
-The Acme Candy Company claims that 60% of the jawbreakers it produces weigh more than .4 ounces. Suppose that 800 jawbreakers are selected at random from the production lines. Would it be unusual for this sample of 800 to contain 463 jawbreakers that weigh more than .4 ounces?
A)Yes
B)No

Accepted Answer

To determine if the outcome is unusual, we calculate the mean and standard deviation for the distribution of the sample proportion. The mean of the sample proportion ($\mu_{\hat{p}}$) is equal to the population proportion, which is 0.60. The standard deviation of the sample proportion ($\sigma_{\hat{p}}$) is calculated using the formula $\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}}$, where $p = 0.60$ and $n = 800$. Thus, $\sigma_{\hat{p}} = \sqrt{\frac{(0.60)(0.40)}{800}} = \sqrt{\frac{0.24}{800}} = \sqrt{0.0003} = 0.0173$.To find if 463 out of 800 is unusual, we convert 463 to a proportion: $\frac{463}{800} = 0.57875$. We then compare this to the mean ($0.60$) in terms of standard deviations. The difference from the mean is $0.60 - 0.57875 = 0.02125$, which is less than 2 standard deviations ($2 \times 0.0173 = 0.0346$).Since the difference from the mean is less than 2 standard deviations, the result is not considered unusual.

Question 16

Determine if the outcome is unusual. Consider as unusual any result that differs from the mean by more than 2 standard deviations. That is, unusual values are either less than $\mu - 2 \sigma$ or greater than $\mu + 2 \sigma$.
-According to AccuData Media Research, 36% of televisions within the Chicago city limits are tuned to "Eyewitness News" at 5:00 pm on Sunday nights. At 5:00 pm on a given Sunday, 2500 such televisions are randomly selected and checked to determine what is being watched. Would it be unusual to find that 986 of the 2500 televisions are tuned to "Eyewitness News"?
A)Yes
B)No

Accepted Answer

Given that 36% of televisions are tuned to "Eyewitness News," the expected number of televisions tuned in out of 2500 is $2500 \times 0.36 = 900$. The standard deviation ($\sigma$) for a binomial distribution is calculated using the formula $\sigma = \sqrt{n \cdot p \cdot (1-p)}$, where $n$ is the number of trials and $p$ is the probability of success. Substituting the given values, $\sigma = \sqrt{2500 \cdot 0.36 \cdot (1-0.36)} = \sqrt{2500 \cdot 0.36 \cdot 0.64} \approx 24$. The range for usual outcomes is $\mu - 2\sigma$ to $\mu + 2\sigma$, or $900 - 2 \cdot 24$ to $900 + 2 \cdot 24$, which simplifies to $852$ to $948$. Since 986 is outside this range, it is considered unusual.

Question 17

Determine if the outcome is unusual. Consider as unusual any result that differs from the mean by more than 2 standard deviations. That is, unusual values are either less than $\mu - 2 \sigma$ or greater than $\mu + 2 \sigma$.
-The Acme Candy Company claims that 60% of the jawbreakers it produces weigh more than .4 ounces. Suppose that 800 jawbreakers are selected at random from the production lines. Would it be unusual for this sample of 800 to contain 528 jawbreakers that weigh more than .4 ounces?
A)No
B)Yes

Accepted Answer

The expected number of jawbreakers weighing more than 0.4 ounces is $0.6 \times 800 = 480$. The standard deviation is $\sqrt{800 \times 0.6 \times 0.4} = \sqrt{192} \approx 13.86$. The range for usual values is $480 \pm 2 \times 13.86$, which is approximately $452.28$ to $507.72$. Since 528 is outside this range, it is unusual.

Question 18

Solve the problem.
-On a multiple choice test with 11 questions, each question has four possible answers, one of which is correct. For students who guess at all answers, find the standard deviation for the number of correct answers.
A)1.407
B)1.342
C)1.369
D)1.436

Accepted Answer

The standard deviation for a binomial distribution is given by the formula $\sqrt{n \cdot p \cdot (1-p)}$, where $n$ is the number of trials (in this case, 11 questions), and $p$ is the probability of success on each trial (since there is one correct answer out of four choices, $p = 0.25$). Plugging these values in, we get $\sqrt{11 \cdot 0.25 \cdot (1-0.25)} = \sqrt{11 \cdot 0.25 \cdot 0.75} = \sqrt{2.0625} \approx 1.436$.

Question 19

Solve the problem.
-According to a college survey, 22% of all students work full time. Find the standard deviation for the number of students who work full time in samples of size 16.
A)3.52
B)2.75
C)1.88
D)1.66

Accepted Answer

The standard deviation for the number of students who work full time in samples of size $n$ can be found using the formula for the standard deviation of a proportion in a sample: $\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}}$, where $p$ is the proportion of students who work full time (0.22) and $n$ is the sample size (16). Plugging in the values, we get $\sigma_{\hat{p}} = \sqrt{\frac{0.22(1-0.22)}{16}} = \sqrt{\frac{0.22(0.78)}{16}} = \sqrt{\frac{0.1716}{16}} = \sqrt{0.010725} = 0.10357$. This is the standard deviation of the proportion, but to find the standard deviation of the number of students, we multiply by the sample size: $0.10357 \times 16 = 1.65712$, which rounds to 1.66.

Question 20

Solve the problem.
-A die is rolled 15 times and the number of twos that come up is tallied. If this experiment is repeated many times, find the standard deviation for the number of twos.
A)1.394
B)1.449
C)1.94
D)1.443

Accepted Answer

The standard deviation for the number of twos is the square root of the variance, which is the product of the probability of getting a two and the probability of not getting a two, multiplied by the number of rolls. The probability of getting a two is 1/6, and the probability of not getting a two is 5/6. Therefore, the variance is (1/6) * (5/6) * 15 = 1.443. The standard deviation is the square root of 1.443, which is 1.443.

Quiz 5: Probability Distributions

Solve the problem. -In a certain town, 60 percent of voters are in favor of a given ballot measure and 40 percent are opposed. For groups of 360 voters, find the mean for the number who oppose the measure.

Solve the problem. -On a multiple choice test with 24 questions, each question has four possible answers, one of which is correct. For students who guess at all answers, find the variance for the number of correct answers.

Solve the problem. -The probability of winning a certain lottery is 1/56,728. For people who play 517 times, find the standard deviation for the number of wins.

Solve the problem. -The probability that a radish seed will germinate is 0.7. A gardener plants seeds in batches of 13. Find the standard deviation for the number of seeds germinating in each batch.

Solve the problem. -The probability of winning a certain lottery is 1/66,857. For people who play 618 times, find the mean number of wins.

Solve the problem. -In a certain town, 44% of voters favor a given ballot measure. For groups of 31 voters, find the variance for the number who favor the measure.

Solve the problem. -A company manufactures batteries in batches of 23 and there is a 3% rate of defects. Find the variance for the number of defects per batch.

Solve the problem. -A company manufactures batteries in batches of 28 and there is a 3% rate of defects. Find the standard deviation for the number of defects per batch.

Solve the problem. -On a multiple choice test with 11 questions, each question has four possible answers, one of which is correct. For students who guess at all answers, find the standard deviation for the number of correct answers.

Solve the problem. -According to a college survey, 22% of all students work full time. Find the standard deviation for the number of students who work full time in samples of size 16.

Solve the problem. -A die is rolled 15 times and the number of twos that come up is tallied. If this experiment is repeated many times, find the standard deviation for the number of twos.