Question 1

Service time for customers at a drive-through coffee shop has been shown to be uniformly distributed between 2 and 10 minutes. Customers will complain when service time exceeds 7.5 minutes. Based on this information, the probability of getting a complaint based on service time is 0.3125.

Accepted Answer

The uniform distribution is defined from 2 to 10 minutes, making the total range 8 minutes (10 - 2). The range of service times that would cause complaints is from 7.5 to 10 minutes, a range of 2.5 minutes (10 - 7.5). The probability of a complaint is the range of complaint times over the total range, or 2.5 / 8 = 0.3125.

Question 2

An electronics repair shop has determined that the time between failures for a particular electronic component part is exponentially distributed with a mean time between failures of 200 hours. Based on this information, the probability that a part will fail between 20 and 100 hours is approximately 0.30.

Accepted Answer

The exponential distribution with a mean of 200 hours has a rate parameter λ = 1/200. The probability of failure between 20 and 100 hours can be calculated using the cumulative distribution function (CDF) of the exponential distribution, which results in approximately 0.30.

Question 3

The Varden Packaging Company has a contract to fill 50-gallon barrels with gasoline for use by the U.S. Army. The machine that Varden uses has an adjustable device that allows the average fill per barrel to be adjusted as desired. However, the actual distribution of fill volume from the machine is known to be normally distributed with a standard deviation equal to 0.5 gallons. The contract that Varden has with the military calls for no more than 2 percent of all barrels to contain less than 49.2 gallons of gasoline. Suppose Varden managers are unwilling to set the mean fill at any level higher than 50 gallons. Given that, in order to meet the requirements, they will need to increase the standard deviation of fill volume.

Accepted Answer

To meet the requirement of no more than 2 percent of all barrels containing less than 49.2 gallons of gasoline, Varden managers would need to adjust the mean fill level closer to 50 gallons, not increase the standard deviation. Increasing the standard deviation would result in a wider spread of the fill volumes, potentially increasing the percentage of barrels with less than 49.2 gallons, which would not help in meeting the contract requirements.

Question 4

An assembly process takes between 20 and 40 minutes to complete with the distribution of time thought to be uniformly distributed. Based on this, the percentage of assemblies that require less than 25 minutes is 0.05.

Accepted Answer

In a uniformly distributed process, the percentage of assemblies requiring less than 25 minutes can be calculated by finding the difference between 25 minutes and the minimum time (20 minutes), then dividing by the total range (40 minutes - 20 minutes). This gives (25-20) / (40-20) = 5 / 20 = 0.25 or 25%.

Question 5

The Varden Packaging Company has a contract to fill 50 gallon barrels with gasoline for use by the U.S. Army. The machine that Varden uses has an adjustable device that allows the average fill per barrel to be adjusted as desired. However, the actual distribution of fill volume from the machine is known to be normally distributed with a standard deviation equal to 0.5 gallons. The contract that Varden has with the military calls for no more than 2 percent of all barrels to contain less than 49.2 gallons of gasoline. In order to meet this requirement, Varden should set the mean fill to approximately 49.92 gallons.

Accepted Answer

The answer of The Varden Packaging Company has a contract...

Question 6

Suppose the time it takes for a customer to be served at a fast-food chain business is thought to be uniformly distributed between 3 and 8 minutes, then the probability that it will take exactly 5 minutes is 0.20.

Accepted Answer

In a continuous uniform distribution, the probability of observing any exact value is 0 because there are infinitely many possible values within any interval.

Question 7

A seafood shop sells salmon fillets where the weight of each fillet is normally distributed with a mean of 1.6 pounds and a standard deviation of 0.3 pounds. Based on this information we can conclude that 90 percent of the fillets weight more than 1.0 pound.

Accepted Answer

In a normal distribution, 90% of the data does not fall above 1.0 pound when the mean is 1.6 pounds and the standard deviation is 0.3 pounds. The Z-score for 1.0 pound is approximately -2, which corresponds to about 2.28% of the data below 1.0 pound, meaning about 97.72% of the fillets weigh more than 1.0 pound, not 90%.

Question 8

If the time it takes for a customer to be served at a fast-food chain business is thought to be uniformly distributed between 3 and 8 minutes, then the probability that the time it takes for a randomly selected customer to be served will be less than 5 minutes is 0.40.

Accepted Answer

In a uniform distribution, the probability is calculated as the length of the interval of interest divided by the total length of the distribution. Here, the interval of interest is 3 to 5 minutes, and the total interval is 3 to 8 minutes. The probability is (5-3)/(8-3) = 2/5 = 0.40.

Question 9

Suppose the time it takes for a customer to be served at a fast-food chain business is thought to be uniformly distributed between 3 and 8 minutes, then the probability that a customer is served in less than 3 minutes is 0.

Accepted Answer

In a uniform distribution between 3 and 8 minutes, the minimum time is 3 minutes, so the probability of being served in less than 3 minutes is 0.

Question 10

It has been determined the weight of bricks made by the Dillenger Stone Company is uniformly distributed between 1 and 1.5 pounds. Based on this information, the probability that two randomly selected bricks will each weigh more than 1.3 pounds is 0.16.

Accepted Answer

Since the weight of bricks is uniformly distributed between 1 and 1.5 pounds, the range of possible weights is 0.5 pounds. The probability that one brick weighs more than 1.3 pounds is the length of the interval from 1.3 to 1.5 pounds divided by the total interval length (1.5 - 1 = 0.5), which is (1.5 - 1.3) / 0.5 = 0.4. The probability that two bricks independently each weigh more than 1.3 pounds is 0.4 * 0.4 = 0.16.

Question 11

For a normal distribution, the probability of a value being between a positive z-value and its population mean is the same as that of a value being between a negative z-value and its population mean.

Accepted Answer

In a normal distribution, the curve is symmetrical around the mean. Therefore, the probability of a value falling between any positive z-value and the mean is the same as the probability of a value falling between its corresponding negative z-value and the mean.

Question 12

One of the basic differences between a uniform probability distribution and a normal probability distribution is that the uniform is symmetrical but the normal is skewed depending on the value of the standard deviation.

Accepted Answer

Both uniform and normal probability distributions can be symmetrical. The uniform distribution is symmetrical around its center, and a normal distribution is also symmetrical, not skewed, around its mean. The standard deviation affects the spread of a normal distribution, not its symmetry.

Question 13

The miles per gallon for hybrid vehicles on city streets have been determined to be normally distributed with a mean of 33.2 mpg and a variance of 16. Based on this information, the probability that if three randomly selected vehicles are monitored and that two of the three will exceed the 35 mpg is slightly greater than 0.18.

Accepted Answer

Given the mean (μ) = 33.2 mpg and variance (σ^2) = 16, the standard deviation (σ) = √16 = 4 mpg. To find the probability of a single vehicle exceeding 35 mpg, we first calculate the Z-score for 35 mpg, which is (35 - 33.2) / 4 = 0.45. The probability associated with this Z-score can be found using a standard normal distribution table or calculator, which gives a probability of a single vehicle exceeding 35 mpg. However, the question asks for the probability that exactly two out of three vehicles exceed 35 mpg, which involves binomial probability calculations, not directly provided by the information given. The statement's conclusion about the probability being slightly greater than 0.18 without showing these calculations or the exact probability for a single vehicle exceeding 35 mpg is not directly inferable from the given data.

Question 14

An electronics repair shop has determined that the time between failures for a particular electronic component part is exponentially distributed with a mean time between failures of 200 hours. Based on this information, the probability that a part will fail in the first 20 hours is approximately 0.095.

Accepted Answer

The exponential distribution is given by the formula $P(T \leq t) = 1 - e^{-\lambda t}$, where $\lambda = \frac{1}{\text{mean time between failures}}$. For a mean time of 200 hours, $\lambda = \frac{1}{200}$. Plugging in $t = 20$ hours, we get $P(T \leq 20) = 1 - e^{-\frac{20}{200}} \approx 0.095$, which matches the given probability.

Question 15

The amount of drying time for the paint applied to a plastic component part is thought to be uniformly distributed between 30 and 60 minutes. Currently, the automated process selects the part from the drying bin after the part has been there for 50 minutes. Based on this, the probability that a part selected will not be dry is approximately 0.33.

Accepted Answer

The drying time is uniformly distributed between 30 and 60 minutes. The probability that a part is not dry after 50 minutes is the probability that the drying time is greater than 50 minutes, which is (60-50)/(60-30) = 10/30 = 0.33.

Question 16

If a uniform distribution and normal distribution both have the same mean and the same range, the normal distribution will have a larger standard deviation than the uniform distribution

Accepted Answer

For distributions with the same range, the uniform distribution maximizes the standard deviation. A normal distribution with the same mean and range will have a smaller standard deviation because its values are more concentrated around the mean.

Question 17

A seafood shop sells salmon fillets where the weight of each fillet is normally distributed with a mean of 1.6 pounds and a standard deviation of 0.3 pounds. They want to classify the largest fillets as extra large and charge a higher price for them. If they want the largest 15 percent of the fillets to be classified as extra large, the minimum weight for an extra large fillet should be 1.91 pounds.

Accepted Answer

To find the minimum weight that classifies the largest 15% of the salmon fillets as extra large, we use the Z-score table or a calculator for the normal distribution. The Z-score corresponding to the top 15% (or the 85th percentile, since the normal distribution is symmetric) is approximately 1.036. Using the Z-score formula $Z = \frac{X - \mu}{\sigma}$, where $X$ is the value we seek, $\mu = 1.6$ pounds is the mean, and $\sigma = 0.3$ pounds is the standard deviation, we solve for $X$: $1.036 = \frac{X - 1.6}{0.3}$. Solving this equation gives $X \approx 1.91$ pounds, which matches the statement.

Question 18

Any normal distribution can be converted to a standard normal distribution.

Accepted Answer

Any normal distribution can be converted to a standard normal distribution by using the Z-score formula, which standardizes the values by subtracting the mean and dividing by the standard deviation.

Question 19

The Varden Packaging Company has a contract to fill 50-gallon barrels with gasoline for use by the U.S. Army. The machine that Varden uses has an adjustable device that allows the average fill per barrel to be adjusted as desired. However, the actual distribution of fill volume from the machine is known to be normally distributed with a standard deviation equal to 0.5 gallons. The contract that Varden has with the military calls for no more than 2 percent of all barrels to contain less than 49.2 gallons of gasoline. In order to meet this requirement, Varden should set the mean fill to approximately 50.225 gallons.

Accepted Answer

To ensure that no more than 2% of barrels contain less than 49.2 gallons, Varden needs to set the mean fill such that 49.2 gallons is at the 2nd percentile of the normal distribution. Using the z-score for the 2nd percentile (-2.05) and the standard deviation (0.5 gallons), the mean fill should be approximately 50.225 gallons.

Question 20

The amount of drying time for the paint applied to a plastic component part is thought to be uniformly distributed between 30 and 60 minutes. Currently, the automated process selects the part from the drying bin after the part has been there for 50 minutes. The probability that none of three parts picked are still wet when they are selected is approximately 0.04.

Accepted Answer

The probability that a part is not wet when selected after 50 minutes is calculated based on the uniform distribution from 30 to 60 minutes. This means the drying time has a range of 30 minutes, and selecting after 50 minutes gives a 20-minute window where the part is dry. The probability for one part being dry is 20/30 or 2/3. For three parts, the probability that all are dry is (2/3)^3 = 8/27, which is approximately 0.296, not 0.04.

Quiz 6: Introduction to Continuous Probability Distributions