Question 1

How is the standard deviation of a uniform distribution computed? _________

Accepted Answer

The answer of How is the standard deviation of a...

Question 2

The mean amount of gasoline and services charged by Key Refining Company credit customers is $70 per month. The distribution of amounts spent is approximately normal with a standard deviation of $10. What is the probability of selecting a credit card customer at random and finding the customer charged between $70 and $83?

Accepted Answer

The answer of The mean amount of gasoline and services...

Question 3

A binomial distribution has 50 trials (n = 50) with a probability of success of 0.50 ($\pi$=0.50). To use the normal distribution to approximate the binomial, what are the mean and standard deviation?

Accepted Answer

According to the normal approximation to the binomial distribution, the mean of the normal distribution is equal to the mean of the binomial distribution (µ = np = 50 × 0.50 = 25) and the standard deviation of the normal distribution is equal to the square root of the product of the number of trials, the probability of success, and the probability of failure (σ = sqrt(np(1-p)) = sqrt(50 × 0.50 × 0.50) = 3.5355). Therefore, option A is the best choice.

Question 4

The average score of 100 students taking a statistics final was 70 with a standard deviation of 7. Assuming a normal distribution, what is the probability that a student scored 90 or higher?

Accepted Answer

To find the probability of a student scoring 90 or higher, we calculate the z-score for 90 using the formula z = (X - μ) / σ, where X is the score (90), μ is the mean (70), and σ is the standard deviation (7). This gives us a z-score of (90-70)/7 = 20/7 ≈ 2.86. Looking up this z-score in a standard normal distribution table or using a calculator, we find the probability of a z-score being above 2.86 is approximately 0.0021, indicating a very low probability of a student scoring 90 or higher.

Question 5

If an average of 12 customers is served per hour, what is the probability that the next customer will arrive in 6 minutes or less?

Accepted Answer

To solve this problem, we can use the Poisson distribution, which models the number of events occurring in a certain amount of time, given an average rate of occurrence. In this case, the average rate of customers being served is 12 per hour, or λ = 12/60 = 0.2 per minute.

Let X be the number of customers served in 6 minutes. Then X has a Poisson distribution with parameter λ' = 0.2*6 = 1.2.

We want to find P(X ≤ 1), which represents the probability that at most one customer arrives in 6 minutes or less. Using the Poisson distribution formula, we have:

P(X ≤ 1) = e^(-λ')*(λ'^0/0!) + e^(-λ')*(λ'^1/1!) = e^(-1.2)*(1 + 1.2) = 0.70

Therefore, the probability that the next customer will arrive in 6 minutes or less is 0.70, or approximately 70%. The best choice is C.

Question 6

Suppose a tire manufacturer wants to set a mileage guarantee on its new XB 70 tire. Tests revealed that the tire's mileage is normally distributed with a mean of 47,900 miles and a standard deviation of 2,050 miles. The manufacturer wants to set the guaranteed mileage so that no more than 5 percent of the tires will have to be replaced. What guaranteed mileage should the manufacturer announce?

Accepted Answer

To find the guaranteed mileage, we need to determine the mileage at the 5th percentile of the normal distribution. Using the z-score for the 5th percentile, which is approximately -1.645, we calculate: 47,900 + (-1.645 * 2,050) = 44,528 miles.

Question 7

The shape of the uniform distribution is ____________.

Accepted Answer

The answer of The shape of the uniform distribution is...

Question 8

Management is considering adopting a bonus system to increase production. One suggestion is to pay a bonus on the highest 5 percent of production based on past experience. Past records indicate that, on the average, 4,000 units of a small assembly are produced during a week. The distribution of the weekly production is approximately normally distributed with a standard deviation of 60 units. If the bonus is paid on the upper 5 percent of production, the bonus will be paid on how many units or more?

Accepted Answer

To find the number of units for the top 5% of production, we need to find the z-score that corresponds to the 95th percentile of a standard normal distribution, which is approximately 1.645. Using the formula for a z-score: z = (X - μ) / σ, where μ = 4000 and σ = 60, we solve for X: 1.645 = (X - 4000) / 60. Solving for X gives X ≈ 4099.

Question 9

The average score of 100 students taking a statistics final was 70 with a standard deviation of 7. Assuming a normal distribution, what is the probability that a student scored greater than 65?

Accepted Answer

Using the Z-score formula, Z = (X - μ) / σ, where X is the score (65), μ is the mean (70), and σ is the standard deviation (7), we find Z = (65 - 70) / 7 = -0.714. Looking up this Z-score in a standard normal distribution table or using a calculator, we find the probability of scoring less than 65 is approximately 0.2389. Therefore, the probability of scoring greater than 65 is 1 - 0.2389 = 0.7611.

Question 10

A binomial distribution has 100 trials (n = 100) with a probability of success of 0.25 ($\pi$=0.25). We would like to find the probability of 34 or more successes using the normal distribution to approximate the binomial. Applying the continuity correction factor, what z-score should be used?

Accepted Answer

The answer of A binomial distribution has 100 trials (n...

Question 11

A binomial distribution has 100 trials (n = 100) with a probability of success of 0.25 ($\pi$=0.25). To apply the normal distribution to approximate the binomial, what are the mean and standard deviation?

Accepted Answer

To approximate a binomial distribution using normal distribution, we need to calculate the mean (µ) and standard deviation (σ) using the following formulas:
- µ = n * p
- σ = sqrt(n * p * (1 - p))

In this case, n = 100 and p = 0.25, so:
- µ = 100 * 0.25 = 25
- σ = sqrt(100 * 0.25 * (1 - 0.25)) = 4.33 (rounded to 2 decimal places)

Therefore, the best choice is C, which has µ = 25 and σ = 4.33.

Question 12

The average score of 100 students taking a statistics final was 70 with a standard deviation of 7. Assuming a normal distribution, what test score separates the top 5% of the students from the lower 95% of students?

Accepted Answer

To find the score that separates the top 5% from the lower 95%, we need to find the z-score that corresponds to the 95th percentile, which is approximately 1.645. Using the formula: score = mean + (z-score * standard deviation), we get 70 + (1.645 * 7) = 81.48.

Question 13

If an average of 60 customers are served per hour, then one customer arrives every

Accepted Answer

If 60 customers are served per hour, dividing 60 minutes by 60 customers gives 1 minute per customer.

Question 14

How is the expected value of a uniform distribution computed? __________

Accepted Answer

The answer of How is the expected value of a...

Question 15

The average score of 100 students taking a statistics final was 70 with a standard deviation of 7. Assuming a normal distribution, what test score separates the top 25% of the students from the lower 75% of students?

Accepted Answer

We need to find the test score that separates the top 25% from the lower 75%. 
The area under the normal curve corresponding to the top 25% is equal to 0.25. 
Using a standard normal distribution table or calculator, we can find the z-score that corresponds to this area, which is approximately 0.67. 
We can then use the formula z = (x - mu)/sigma, where x is the test score we want to find, mu is the mean score (70), and sigma is the standard deviation (7). 
Solving for x, we get x = z*sigma + mu = 0.67*7 + 70 = 74.69. 
Therefore, the test score that separates the top 25% from the lower 75% is approximately 74.69.

Question 16

If an average of 12 customers are served per hour, then one customer arrives every

Accepted Answer

If an average of 12 customers are served per hour, then one customer is served every 5 minutes (60 minutes divided by 12 customers equals 5 minutes per customer). Therefore, one customer arrives every 5 minutes.

Question 17

A binomial distribution has 50 trials (n = 50) with a probability of success of 0.50 ($\pi$=0.50). We would like to find the probability of 34 or more successes using the normal distribution to approximate the binomial. Applying the continuity correction factor, what z-score should be used?

Accepted Answer

To approximate the binomial distribution using the normal distribution, apply the continuity correction factor. For 34 or more successes, use 33.5 as the cutoff. The mean is $n\pi = 25$ and the standard deviation is $\sqrt{n\pi(1-\pi)} = 3.54$. The z-score is $(33.5 - 25) / 3.54 = 2.404$.

Question 18

If an average of 12 customers is served per hour, what is the probability that the next customer will arrive in 3 minutes or less?

Accepted Answer

To solve this problem, we need to first find the average time between customers.

Average time between customers (in minutes) = 60 minutes/hour ÷ 12 customers/hour = 5 minutes/customer

Next, we need to find the probability that the next customer will arrive in 3 minutes or less.

To do this, we will use the exponential distribution with a rate parameter of 1/5 (since the average time between customers is 5 minutes):

P(X ≤ 3) = 1 - e^(-1/5 * 3) = 0.4493

Therefore, the probability that the next customer will arrive in 3 minutes or less is approximately 0.45.

The best choice is C.

Question 19

A binomial distribution has 100 trials (n = 100) with a probability of success of 0.25 ($\pi$=0.25). We would like to find the probability of 75 or more successes using the normal distribution to approximate the binomial. Applying the continuity correction factor, what value would be used to calculate a z-score?

Accepted Answer

The continuity correction factor is applied to ensure that the normal distribution is a good approximation for the binomial distribution. The correction factor is 0.5, which is added to the number of successes and subtracted from the number of trials. In this case, the continuity correction factor would be applied as follows:75 + 0.5 = 75.5100 - 75.5 = 24.5Therefore, the value used to calculate the z-score would be 75.5.

Question 20

If an average of 12 customers is served per hour, what is the probability that the next customer will arrive in 3 minutes or more?

Accepted Answer

The average rate is 12 customers per hour, which is 1 customer every 5 minutes. The probability that the next customer arrives in 3 minutes or more is the complement of the probability that the customer arrives in less than 3 minutes. Using the exponential distribution, P(X ≥ 3) = 1 - P(X < 3) = 1 - (1 - e^(-3/5)) = e^(-3/5) ≈ 0.45.

Quiz 7: Continuous Probability Distributions

How is the standard deviation of a uniform distribution computed? _________

A binomial distribution has 50 trials (n = 50) with a probability of success of 0.50 ( $\pi$ =0.50) . To use the normal distribution to approximate the binomial, what are the mean and standard deviation?

The average score of 100 students taking a statistics final was 70 with a standard deviation of 7. Assuming a normal distribution, what is the probability that a student scored 90 or higher?

If an average of 12 customers is served per hour, what is the probability that the next customer will arrive in 6 minutes or less?

The shape of the uniform distribution is ____________.

The average score of 100 students taking a statistics final was 70 with a standard deviation of 7. Assuming a normal distribution, what is the probability that a student scored greater than 65?

A binomial distribution has 100 trials (n = 100) with a probability of success of 0.25 ( $\pi$ =0.25) . We would like to find the probability of 34 or more successes using the normal distribution to approximate the binomial. Applying the continuity correction factor, what z-score should be used?

A binomial distribution has 100 trials (n = 100) with a probability of success of 0.25 ( $\pi$ =0.25) . To apply the normal distribution to approximate the binomial, what are the mean and standard deviation?

The average score of 100 students taking a statistics final was 70 with a standard deviation of 7. Assuming a normal distribution, what test score separates the top 5% of the students from the lower 95% of students?

If an average of 60 customers are served per hour, then one customer arrives every

How is the expected value of a uniform distribution computed? __________

The average score of 100 students taking a statistics final was 70 with a standard deviation of 7. Assuming a normal distribution, what test score separates the top 25% of the students from the lower 75% of students?

If an average of 12 customers are served per hour, then one customer arrives every

A binomial distribution has 50 trials (n = 50) with a probability of success of 0.50 ( $\pi$ =0.50) . We would like to find the probability of 34 or more successes using the normal distribution to approximate the binomial. Applying the continuity correction factor, what z-score should be used?

If an average of 12 customers is served per hour, what is the probability that the next customer will arrive in 3 minutes or less?

If an average of 12 customers is served per hour, what is the probability that the next customer will arrive in 3 minutes or more?

Quiz 7: Continuous Probability Distributions

How is the standard deviation of a uniform distribution computed? _________

A binomial distribution has 50 trials (n = 50) with a probability of success of 0.50 ( π\piπ =0.50) . To use the normal distribution to approximate the binomial, what are the mean and standard deviation?

The average score of 100 students taking a statistics final was 70 with a standard deviation of 7. Assuming a normal distribution, what is the probability that a student scored 90 or higher?

If an average of 12 customers is served per hour, what is the probability that the next customer will arrive in 6 minutes or less?

The shape of the uniform distribution is ____________.

The average score of 100 students taking a statistics final was 70 with a standard deviation of 7. Assuming a normal distribution, what is the probability that a student scored greater than 65?

A binomial distribution has 100 trials (n = 100) with a probability of success of 0.25 ( π\piπ =0.25) . We would like to find the probability of 34 or more successes using the normal distribution to approximate the binomial. Applying the continuity correction factor, what z-score should be used?

A binomial distribution has 100 trials (n = 100) with a probability of success of 0.25 ( π\piπ =0.25) . To apply the normal distribution to approximate the binomial, what are the mean and standard deviation?

The average score of 100 students taking a statistics final was 70 with a standard deviation of 7. Assuming a normal distribution, what test score separates the top 5% of the students from the lower 95% of students?

If an average of 60 customers are served per hour, then one customer arrives every

How is the expected value of a uniform distribution computed? __________

The average score of 100 students taking a statistics final was 70 with a standard deviation of 7. Assuming a normal distribution, what test score separates the top 25% of the students from the lower 75% of students?

If an average of 12 customers are served per hour, then one customer arrives every

A binomial distribution has 50 trials (n = 50) with a probability of success of 0.50 ( π\piπ =0.50) . We would like to find the probability of 34 or more successes using the normal distribution to approximate the binomial. Applying the continuity correction factor, what z-score should be used?

If an average of 12 customers is served per hour, what is the probability that the next customer will arrive in 3 minutes or less?

If an average of 12 customers is served per hour, what is the probability that the next customer will arrive in 3 minutes or more?

A binomial distribution has 50 trials (n = 50) with a probability of success of 0.50 ( $\pi$ =0.50) . To use the normal distribution to approximate the binomial, what are the mean and standard deviation?

A binomial distribution has 100 trials (n = 100) with a probability of success of 0.25 ( $\pi$ =0.25) . We would like to find the probability of 34 or more successes using the normal distribution to approximate the binomial. Applying the continuity correction factor, what z-score should be used?

A binomial distribution has 100 trials (n = 100) with a probability of success of 0.25 ( $\pi$ =0.25) . To apply the normal distribution to approximate the binomial, what are the mean and standard deviation?

A binomial distribution has 50 trials (n = 50) with a probability of success of 0.50 ( $\pi$ =0.50) . We would like to find the probability of 34 or more successes using the normal distribution to approximate the binomial. Applying the continuity correction factor, what z-score should be used?