Question 1

THE NEXT QUESTIONS ARE BASED ON THE FOLLOWING INFORMATION:
The time it takes to assemble an electronic component is normally distributed with a mean of 17.2 minutes and a standard deviation of 3.1 minutes.
-The probability is 95% that it will take at least how long to assemble a component?

Accepted Answer

We need to find the value of x such that P(X ≥ x) = 0.95. Using the normal distribution table or a calculator, we can find the z-score corresponding to the 95th percentile, which is 1.645. Then we can solve for x using the formula x = μ + zσ, where μ = 17.2 and σ = 3.1. Substituting values, we get x = 17.2 + 1.645(3.1) ≈ 12.11. Rounding to one decimal place gives us the answer of 12.1 minutes.

Question 2

THE NEXT QUESTIONS ARE BASED ON THE FOLLOWING INFORMATION:
Let X be a continuous random variable with mean of 1200 and standard deviation of 250.Define the random variable Y as follows: Y = 80 + 0.05X
-What is the standard deviation of Y?

Accepted Answer

The standard deviation of Y is calculated by multiplying the standard deviation of X by the coefficient of X in the equation for Y, which is 0.05. Thus, 0.05 * 250 = 12.5.

Question 3

THE NEXT QUESTIONS ARE BASED ON THE FOLLOWING INFORMATION:
The number of nails in a five-pound box of nails is normally distributed with a mean of 563.3 and a standard deviation of 33.2.
-What is the probability that there are less than 500 nails in a randomly-selected five-pound box of nails?

Accepted Answer

To solve this problem, we need to standardize the value of 500 using the formula z = (x - μ) / σ, where x is the value we are interested in, μ is the mean, and σ is the standard deviation. Thus, z = (500 - 563.3) / 33.2 = -1.906. We can then find the probability using a standard normal distribution table or calculator, which gives us 0.0281. Therefore, the answer is choice C.

Question 4

THE NEXT QUESTIONS ARE BASED ON THE FOLLOWING INFORMATION:
The time it takes to assemble an electronic component is normally distributed with a mean of 17.2 minutes and a standard deviation of 3.1 minutes.
-What is the probability that it will take at least 16 minutes to assemble a component?

Accepted Answer

The answer of THE NEXT QUESTIONS ARE BASED ON THE...

Question 5

Checkout times in a Walmart store follow an exponential distribution with a mean time of 8 minutes.What is the probability that the checkout time for the next customer will be 6 minutes or less?

Accepted Answer

The exponential distribution probability function is given by $P(X \leq x) = 1 - e^{-\lambda x}$, where $\lambda = \frac{1}{\text{mean}}$. Here, the mean checkout time is 8 minutes, so $\lambda = \frac{1}{8}$. For a checkout time of 6 minutes or less, $P(X \leq 6) = 1 - e^{-\frac{1}{8} \cdot 6} = 1 - e^{-\frac{3}{4}} \approx 0.528$.

Question 6

The probability is 0.99 that a randomly-selected five-pound box of nails contains at least how many nails approximately?

Accepted Answer

The answer of The probability is 0.99 that a randomly-selected...

Question 7

A two-dimensional plot of the observed data values on the vertical-axis with their corresponding quartile values from a standardized normal distribution on the horizontal-axis is referred to as a:

Accepted Answer

A normal probability plot is a graphical technique for assessing whether or not a data set is approximately normally distributed. The data are plotted against a theoretical normal distribution in such a way that the points should form an approximate straight line if the data are normally distributed.

Question 8

THE NEXT QUESTIONS ARE BASED ON THE FOLLOWING INFORMATION:
Sales at a local wholesaler consist of both retail as well as online sales.During the course of a month,retail sales have a mean of $96,780 with a standard deviation of $12,520.Over the same time period,online sales average $229,620 with a standard deviation of $23,100.Assume that the retail sales are independent of online sales.
-A car owner determines that annual repair costs (R)can be estimated based on the number of miles driven in one year (M)through the expression R = 800 + 0.05M.If the average number of miles driven is 12000 with a standard deviation of 2500 miles,what is the standard deviation of the expected annual repair costs?

Accepted Answer

The standard deviation of the expected annual repair costs can be found by multiplying the standard deviation of the number of miles driven (2500 miles) by the coefficient of M in the expression for R (0.05). This gives 2500 * 0.05 = 125.

Question 9

THE NEXT QUESTIONS ARE BASED ON THE FOLLOWING INFORMATION:
Sales at a local wholesaler consist of both retail as well as online sales.During the course of a month,retail sales have a mean of $96,780 with a standard deviation of $12,520.Over the same time period,online sales average $229,620 with a standard deviation of $23,100.Assume that the retail sales are independent of online sales.
-What is the mean of the wholesaler's monthly sales?

Accepted Answer

The mean of the wholesaler's monthly sales is the sum of the mean of retail sales and the mean of online sales, which is $96,780 + $229,620 = $326,400.

Question 10

THE NEXT QUESTIONS ARE BASED ON THE FOLLOWING INFORMATION:
The time it takes to assemble an electronic component is normally distributed with a mean of 17.2 minutes and a standard deviation of 3.1 minutes.
-The median score for the final exam in a statistics class was 85 and 20 percent of the students received a grade greater than 90.What is the probability that a randomly selected exam had a score between 85 and 90?

Accepted Answer

The median is 85, meaning 50% scored below and 50% scored above. Since 20% scored above 90, the probability of scoring between 85 and 90 is 50% - 20% = 30%.

Question 11

The speed of cars passing through a check point follows a normal distribution with a mean of 45 miles per hour and a standard deviation of 6 miles per hour.What is the probability that the next car passing will be exceeding 51 miles per hour?

Accepted Answer

We need to find the probability of a car exceeding 51 miles per hour, which means finding the area under the normal distribution curve to the right of 51 mph. We can standardize 51 mph using the formula z = (x - μ) / σ, where x is the value we want to find the probability for, μ is the mean, and σ is the standard deviation. Thus, we have:

z = (51 - 45) / 6
z = 1

Using a standard normal distribution table, we find that the area to the right of z = 1 is 0.1587. Therefore, the probability of the next car passing exceeding 51 miles per hour is 0.1587 or 15.87%, which is option D.

Question 12

Let the random variable Z follow a standard normal distribution.The total probability to the right of Z = 2.18 and to the left of Z = -1.75 is:

Accepted Answer

The probability to the right of Z = 2.18 is 1 - P(Z ≤ 2.18) = 1 - 0.9854 = 0.0146. The probability to the left of Z = -1.75 is P(Z ≤ -1.75) = 0.0401. Adding these probabilities gives 0.0146 + 0.0401 = 0.0547.

Question 13

Let the random variable Z follow a standard normal distribution,and let z₁ be a possible value of Z that is unknown but identifiable by position and area.Find z₁ if the area to the right of z₁ is 0.8413.

Accepted Answer

The area to the right of the z-score in a standard normal distribution represents the probability that a value is greater than that z-score. If the area to the right is 0.8413, it means the area to the left (or the cumulative probability up to that z-score) is 1 - 0.8413 = 0.1587. Looking up this value in a standard normal distribution table or using a calculator to find the z-score that corresponds to a cumulative probability of 0.1587 gives us a z-score of approximately -1.00.

Question 14

THE NEXT QUESTIONS ARE BASED ON THE FOLLOWING INFORMATION:
The number of nails in a five-pound box of nails is normally distributed with a mean of 563.3 and a standard deviation of 33.2.
-What is the probability that there are between 525 and 575 nails in a randomly-selected five-pound box of nails?

Accepted Answer

To find the probability that the number of nails is between 525 and 575, we calculate the z-scores for 525 and 575 using the mean (563.3) and standard deviation (33.2). Then, we find the probability for each z-score and subtract the smaller probability from the larger one. This results in a probability of approximately 0.5117.

Question 15

Given that X is a normally distributed random variable with a mean of 52 and a standard deviation of 2,the probability that X is between 48 and 56 is:

Accepted Answer

To solve this problem, we need to standardize X to a standard normal distribution using the formula z = (X-μ)/σ, where μ = 52 and σ = 2.

For X = 48, z = (48 - 52)/2 = -2. 
For X = 56, z = (56 - 52)/2 = 2.

Using a standard normal distribution table, we can find the probability of being between -2 and 2, which is 0.9544. Therefore, the answer is D.

Question 16

Which type of distribution would best model the waiting times of the departing flights at JFK international airport?

Accepted Answer

The best choice would be the exponential distribution as it is commonly used to model the distribution of waiting times between events that occur randomly and independently over time, such as the departure of flights at an airport. The exponential distribution has a memoryless property, meaning that the waiting time until the next event is independent of the time elapsed since the last event, which is relevant to the departure times of flights at JFK airport.

Question 17

Which of the following is not a characteristic for a normal distribution?

Accepted Answer

The mean of a normal distribution is not always zero, it can take any value. The other three characteristics listed (symmetry, mean=median=mode, bell-shaped) are all true for a normal distribution.

Question 18

THE NEXT QUESTIONS ARE BASED ON THE FOLLOWING INFORMATION:
Sales at a local wholesaler consist of both retail as well as online sales.During the course of a month,retail sales have a mean of $96,780 with a standard deviation of $12,520.Over the same time period,online sales average $229,620 with a standard deviation of $23,100.Assume that the retail sales are independent of online sales.
-What is the standard deviation of the wholesaler monthly sales?

Accepted Answer

The standard deviation of the wholesaler's monthly sales can be found by taking the square root of the sum of the squares of the individual standard deviations, since the sales channels are independent. This is calculated as sqrt(($12,520)^2 + ($23,100)^2) = sqrt(156,750,400 + 533,610,000) = sqrt(690,360,400) = $26,274.7.

Question 19

During a professor's office hours,students arrive,on average,every ten minutes.Assume that the distribution of the time between arrivals follows an exponential distribution.Suppose that a student has just left. What is the probability that the professor has more than 20 minutes before the next student shows up?

Accepted Answer

Let X be the time between arrivals of two consecutive students. Then X follows an exponential distribution with parameter λ = 1/10 (since students arrive on average every 10 minutes). The probability that the professor has more than 20 minutes before the next student shows up is equivalent to the probability that X > 20, given that a student has just left. Using the memoryless property of the exponential distribution, we have:

P(X > 20 | X > 0) = P(X > 20)/P(X > 0) = (e^(-λx=20))/e^(-λx=0) = e^(-2) ≈ 0.1353

Therefore, the best choice is (C).

Question 20

THE NEXT QUESTIONS ARE BASED ON THE FOLLOWING INFORMATION:
Let X be a continuous random variable with mean of 1200 and standard deviation of 250.Define the random variable Y as follows: Y = 80 + 0.05X
-What is the mean of Y?

Accepted Answer

The mean of Y is calculated by substituting the mean of X into the equation for Y: Mean of Y = 80 + 0.05 * Mean of X = 80 + 0.05 * 1200 = 140.

Quiz 5: Continuous Random Variables and Probability Distributions

THE NEXT QUESTIONS ARE BASED ON THE FOLLOWING INFORMATION: The time it takes to assemble an electronic component is normally distributed with a mean of 17.2 minutes and a standard deviation of 3.1 minutes. -The probability is 95% that it will take at least how long to assemble a component?

THE NEXT QUESTIONS ARE BASED ON THE FOLLOWING INFORMATION: Let X be a continuous random variable with mean of 1200 and standard deviation of 250.Define the random variable Y as follows: Y = 80 + 0.05X -What is the standard deviation of Y?

THE NEXT QUESTIONS ARE BASED ON THE FOLLOWING INFORMATION: The number of nails in a five-pound box of nails is normally distributed with a mean of 563.3 and a standard deviation of 33.2. -What is the probability that there are less than 500 nails in a randomly-selected five-pound box of nails?

THE NEXT QUESTIONS ARE BASED ON THE FOLLOWING INFORMATION: The time it takes to assemble an electronic component is normally distributed with a mean of 17.2 minutes and a standard deviation of 3.1 minutes. -What is the probability that it will take at least 16 minutes to assemble a component?