Question 1

A binomial distribution has 50 trials (n = 50)with a probability of success of 0.50 (π = 0.50). To use 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 use the following formulas:

µ = nπ
σ = √(nπ(1-π))

Plugging in the values, we get:
µ = 50 * 0.50 = 25
σ = √(50 * 0.50 * 0.50) = 3.5355

Therefore, the best choice is A.

Question 2

The center of a normal probability distribution is ________.

Accepted Answer

The center of a normal probability distribution is the mean of the distribution. This is where the highest point of the curve occurs, and therefore serves as the measure of central tendency for the distribution. The standard deviation is a measure of the spread or variability of the distribution, and may or may not be equal to the mean. The center of the distribution is not always equal to zero, and can be positive or negative depending on the mean and shape of the distribution.

Question 3

Consider a continuous random variable x,which is uniformly distributed between 65 and 85. The probability of x taking on a value between 75 to 90 is ________.

Accepted Answer

Since the continuous random variable is uniformly distributed between 65 to 85, the probability density function (PDF) is a constant value over this range. Therefore, the area under the PDF between any two values of x will give the probability of x taking on a value in that range.

To find the probability of x taking on a value between 75 to 90, we need to find the area under the PDF between these two values. However, the PDF is only defined between 65 to 85, so any values outside this range will have a PDF of 0.

Therefore, we need to find the area under the PDF between 75 to 85 and add the area under the PDF between 85 to 90, which is:

(85-75)/(85-65) + 0 = 0.50

So the probability of x taking on a value between 75 to 90 is 0.50 or 50%.

Question 4

A binomial distribution has 50 trials (n = 50)with a probability of success of 0.50 (π = 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

The z-score is calculated using the formula: z = (X - 0.5 - μ) / σ, where X is 34, μ = nπ = 25, and σ = √(nπ(1-π)) = √12.5. This gives z = (34 - 0.5 - 25) / √12.5 = 2.404.

Question 5

Customers of the Key Refining Company charge an average of $70 per month. The distribution of amounts charged 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 Customers of the Key Refining Company charge...

Question 6

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

Accepted Answer

If 12 customers are served per hour, then the time per customer is 60 minutes divided by 12 customers, which equals 5 minutes per customer.

Question 7

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 8

The z-value for a standard normal distribution ________.

Accepted Answer

The z-value for a standard normal distribution can be either positive or negative, depending on how far the observed value is from the mean. It is calculated as the difference between the observed value and the mean, divided by the standard deviation. A positive z-value indicates that the observed value is above the mean, while a negative z-value indicates that it is below the mean. Therefore, the z-value can take on any value within the range of negative infinity to positive infinity.

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 test score separates the top 25% of the students from the lower 75% of students?

Accepted Answer

We can use the z-score formula to find the test score that separates the top 25% from the lower 75%:

z = (x - μ) / σ

where x is the test score we want to find, μ is the mean (70), and σ is the standard deviation (7).

To find the z-score that corresponds to the top 25%, we look up the z-score that corresponds to a cumulative area of 0.75 in a standard normal table. This is approximately 0.67.

So we can set up the equation:

0.67 = (x - 70) / 7

Solving for x, we get:

x = 74.69

Therefore, the test score that separates the top 25% of the students from the lower 75% is approximately 74.69.

Question 10

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 11

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

Accepted Answer

First, we need to convert the given average into a rate of arrival per minute.

Rate of arrival per minute = 12 customers per hour / 60 minutes per hour = 0.2 customers per minute

Now we can use the Poisson distribution formula to calculate the probability of the next customer arriving in 3 minutes or less:

P(x ≤ 1) = e^(-λ) * λ^(x=0 or 1) / x!

P(x ≤ 1) = e^(-0.6) * 0.6^0/0! + e^(-0.6) * 0.6^1/1!

P(x ≤ 1) = 0.5488 ≈ 0.55

Therefore, the probability that the next customer will arrive in 3 minutes or less is approximately 0.55 or 55%, which corresponds to choice C.

Question 12

A binomial distribution has 100 trials (n = 100)with a probability of success of 0.25 (π = 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

When approximating a binomial distribution with a normal distribution and looking for the probability of "75 or more" successes, we apply the continuity correction factor by subtracting 0.5 from the lower bound of the desired outcome. Thus, for 75 or more successes, we use 74.5 as the value to calculate the z-score.

Question 13

Consider a standard normal random variable z. What is the value of z if the area to the right of z is 0.2643?

Accepted Answer

The answer of Consider a standard normal random variable z....

Question 14

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% of the tires will have to be replaced. What guaranteed mileage should the manufacturer announce?

Accepted Answer

To ensure no more than 5% of the tires will have to be replaced, the manufacturer should set the guaranteed mileage at the 5th percentile of the normal distribution. Using the Z-score formula, Z = (X - μ) / σ, where Z is the Z-score corresponding to the 5th percentile, X is the mileage, μ is the mean, and σ is the standard deviation. The Z-score for the 5th percentile is approximately -1.645. Plugging in the values: -1.645 = (X - 47,900) / 2,050. Solving for X gives X ≈ 44,518 miles.

Question 15

Management is considering a bonus system to increase production. One suggestion is to pay a bonus on the highest 5% of production based on past experience. Past records indicate that an average of 4,000 units of a small assembly is 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% of production,the bonus will be paid on how many units or more?

Accepted Answer

To find the number of units for the upper 5% of production, we use the z-score formula for a normal distribution. The z-score corresponding to the upper 5% (95th percentile) is approximately 1.645. The formula to calculate the value is: X = μ + (Z * σ), where μ is the mean, Z is the z-score, and σ is the standard deviation. Plugging in the values: X = 4000 + (1.645 * 60) = 4000 + 98.7 ≈ 4099. Therefore, the bonus would be paid on 4099 units or more.

Question 16

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 17

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 the 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) = 81.515, which rounds to 81.55.

Question 18

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

Accepted Answer

First, we need to convert the given average of 12 customers per hour to the average number of customers per minute. 
12 customers/hour = 12/60 customers/minute = 0.2 customers/minute.

Now, we can use the Poisson distribution formula to calculate the probability of the next customer arriving in 6 minutes or less:

P(X ≤ 1) = e-λ(λ^0/0!) + e-λ(λ^1/1!) 
where 
X = number of customers arriving in 6 minutes 
λ = average number of customers per minute = 0.2

P(X ≤ 1) = e^-0.2(0.2^0/0!) + e^-0.2(0.2^1/1!) 
= 0.6703

Therefore, the probability of the next customer arriving in 6 minutes or less is 0.6703 or approximately 0.70. 
Hence, the best answer choice is C.

Question 19

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

Accepted Answer

To approximate the binomial distribution with a normal distribution, we need to use the mean (µ) and standard deviation (σ) of the normal distribution. The mean of the binomial distribution is µ = nπ = 100(0.25) = 25. The variance of the binomial distribution is σ^2 = nπ(1-π) = 100(0.25)(0.75) = 18.75, so the standard deviation is σ = √(σ^2) = √(18.75) ≈ 4.33. Therefore, the best choice is C, with µ = 25 and σ = 4.33.

Question 20

A binomial distribution has 100 trials (n = 100)with a probability of success of 0.25 (π = 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...

Quiz 7: Continuous Probability Distributions

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