Question 1

The transportation manager for the State of New Jersey has determined that the time between arrivals at a toll booth on the state's turnpike is exponentially distributed with λ = 4 cars per minute.Based on this,the average time between arrivals is:
A)15 seconds.
B)12 seconds.
C)25 seconds.
D)4 minutes.

Accepted Answer

The average time between arrivals for an exponential distribution is given by 1/λ. Given λ = 4 cars per minute, the average time between arrivals is 1/4 minute per car, which is 0.25 minutes per car. Since there are 60 seconds in a minute, this converts to 0.25 * 60 = 15 seconds.

Question 2

It is assumed that the time customers spend in a record store is uniformly distributed between 3 and 12 minutes.Based on this information,what is the probability that a customer will spend more than 9 minutes in the record store?
A)0.33
B)0.1111
C)0.67
D)0.25

Accepted Answer

The total range of time is 12 - 3 = 9 minutes. The time spent over 9 minutes ranges from 9 to 12, which is 3 minutes. The probability is 3/9 = 1/3 or 0.33.

Question 3

It is thought that the time between customer arrivals at a fast food business is exponentially distributed with λ equal to 5 customers per hour.Given this information,what is the mean time between arrivals?
A)12 minutes
B)5 minutes
C)5 hours
D)2 minutes

Accepted Answer

The mean time between arrivals for an exponential distribution is given by 1/λ. Given λ = 5 customers per hour, the mean time is 1/5 hour per customer, which is 12 minutes (since 1 hour = 60 minutes, 60/5 = 12 minutes).

Question 4

In a standard normal distribution,the probability that z is greater than 0 is:
A)0.5
B)equal to 1
C)at least 0.5
D)1.96

Accepted Answer

In a standard normal distribution, the mean is 0, and the distribution is symmetric around the mean. Therefore, the probability that z is greater than 0 is exactly 0.5, as half of the distribution lies above 0 and half below it.

Question 5

Employees at a large computer company earn sick leave in one-minute increments depending on how many hours per month they work.They can then use the sick leave time any time throughout the year.Any unused time goes into a sick bank account that they or other employees can use in the case of emergencies.The human resources department has determined that the amount of unused sick time for individual employees is uniformly distributed between 0 and 480 minutes.The company has decided to give a cash payment to any employee that returns over 400 minutes of sick leave at the end of the year.What percentage of employees could expect a cash payment?
A)16.67 percent
B)0.1667 percent
C)Just over 43 percent
D)80 percent

Accepted Answer

The range of unused sick time is from 0 to 480 minutes. The portion of this range that qualifies for a cash payment is from 400 to 480 minutes, which is 80 minutes. To find the percentage of employees who could expect a cash payment, divide the qualifying range (80 minutes) by the total range (480 minutes) and multiply by 100: (80/480) * 100 = 16.67 percent.

Question 6

A store sells 6 different models of cell phones and have found that they sell an equal number of each model.The probability distribution that would describe this random variable is called:
A)uniform distribution.
B)Poisson distribution.
C)continuous distribution.
D)relative frequency distribution.

Accepted Answer

A uniform distribution is a probability distribution where all outcomes are equally likely. In this case, since the store sells an equal number of each of the 6 different models of cell phones, the probability of selling each model is equal. Hence, the probability distribution that would describe this random variable is a uniform distribution. The other options are not appropriate in this case as a Poisson distribution is used to model rare events, a continuous distribution is used for continuous variables, and a relative frequency distribution is used to describe the frequency of observed outcomes in a sample.

Question 7

A professor noted that the grades of his students were normally distributed with a mean of 75.07 and a standard deviation of 11.65.If only 10 percent of the students received grades of A,what is the minimum score needed to receive an A?
A)80.00
B)85.00
C)90.00
D)95.00

Accepted Answer

To find the minimum score needed to receive an A (top 10% of the grades), we use the concept of z-scores in a normal distribution. The z-score corresponding to the top 10% (90th percentile) is approximately 1.28. Using the formula for a z-score: $z = \frac{(X - \mu)}{\sigma}$, where $X$ is the score, $\mu$ is the mean, and $\sigma$ is the standard deviation, we can solve for $X$. Plugging in the values: $1.28 = \frac{(X - 75.07)}{11.65}$, solving for $X$ gives a score slightly above 89, which means the minimum score needed to receive an A is closest to 90.

Question 8

In a standard normal distribution,the probability P(-1.00< z < 1.20)is the same as:
A)P(1< z < 1.20)- P(0 < z < 1.00).
B)P(1< z < 1.20)- 2*P(0 < z < 1.00).
C)2 ∗ P(1< z < 1.20)- P(0 < z <1.00).
D)P(1 < z < 1.20)+ 2 ∗ P(0 < z <1.00).

Accepted Answer

The answer of In a standard normal distribution,the probability P(-1.00<...

Question 9

Employees at a large computer company earn sick leave in one-minute increments depending on how many hours per month they work.They can then use the sick leave time any time throughout the year.Any unused time goes into a sick bank account that they or other employees can use in the case of emergencies.The human resources department has determined that the amount of unused sick time for individual employees is uniformly distributed between 0 and 480 minutes.Based on this information,what is the probability that an employee will have less than 20 minutes of unused sick time?
A)0.002
B)0.966
C)0.063
D)0.042

Accepted Answer

The answer of Employees at a large computer company earn...

Question 10

Employees at a large computer company earn sick leave in one-minute increments depending on how many hours per month they work.They can then use the sick leave time any time throughout the year.Any unused time goes into a sick bank account that they or other employees can use in the case of emergencies.The human resources department has determined that the amount of unused sick time for individual employees is uniformly distributed between 0 and 480 minutes.The company has decided to give a cash payment to any employee that returns over a specified amount of sick leave minutes.Assuming that the company wishes no more than 5 percent of all employees to get a cash payment,what should the required number of minutes be?
A)24 minutes
B)419 minutes
C)456 minutes
D)470 minutes

Accepted Answer

The answer of Employees at a large computer company earn...

Question 11

It is assumed that the time failures for an electronic component are exponentially distributed with a mean of 50 hours between consecutive failures.Based on this information,what is the probability that a randomly selected part will fail in less than 10 hours?
A)About 0.82
B)About 0.20
C)About 0.33
D)About 0.18

Accepted Answer

The answer of It is assumed that the time failures...

Question 12

It is assumed that the time failures for an electronic component are exponentially distributed with a mean of 50 hours between consecutive failures.If one extra component is installed as a backup,what is the probability of at least one of the two components working for at least 60 hours?
A)About 0.51
B)About 0.09
C)About 0.06
D)About 0.70

Accepted Answer

The answer of It is assumed that the time failures...

Question 13

The transportation manager for the State of New Jersey has determined that the time between arrivals at a toll booth on the state's turnpike is exponentially distributed with λ = 4 cars per minute.Based on this information,the standard deviation for the time between arrivals is:
A)25 seconds.
B)3.87 seconds.
C)15 seconds.
D)2 minutes.

Accepted Answer

The standard deviation for an exponential distribution is given by 1/λ, where λ is the rate parameter (4 cars per minute in this case). Since the rate is in cars per minute, the time between arrivals is in minutes per car. Therefore, the standard deviation is 1/4 minute per car, which is 0.25 minutes per car. Converting 0.25 minutes to seconds (0.25 * 60), we get 15 seconds.

Question 14

A major cell phone service provider has determined that the number of minutes that its customers use their phone per month is normally distributed with a mean equal to 445.5 minutes with a standard deviation equal to 177.8 minutes.As a promotion,the company plans to hold a drawing to give away one free vacation to Hawaii for a customer who uses between 400 and 402 minutes during a particular month.Based on the information provided,what proportion of the company's customers would be eligible for the drawing?
A)Approximately 0.1026
B)About 0.004
C)Approximately 0.2013
D)About 0.02

Accepted Answer

To calculate the proportion of customers who use between 400 and 402 minutes, we need to standardize the values using the formula z=(x-μ)/σ, where x is the value of interest, μ is the mean, and σ is the standard deviation.

For 400 minutes: 
z = (400 - 445.5) / 177.8 = -0.256

For 402 minutes: 
z = (402 - 445.5) / 177.8 = -0.245

Using a standard normal distribution table, we can find the probability of a value being between these two z-scores:

P(-0.256 < Z < -0.245) = P(Z < -0.245) - P(Z < -0.256) = 0.3990 - 0.3980 = 0.0010

Therefore, only about 0.004 (0.1% * 0.399) of the company's customers would be eligible for the drawing.

Question 15

Which of the following probability distributions would most likely be used to describe the time between failures for electronic components?
A)Binomial distribution
B)Exponential distribution
C)Uniform distribution
D)Normal distribution

Accepted Answer

The exponential distribution is commonly used to model the time between consecutive events in which the probability of an event occurring remains constant over time. Therefore, it is a suitable choice for describing the time between failures of electronic components, as the likelihood of a component failure remains relatively constant over time. The binomial distribution, uniform distribution, and normal distribution are not appropriate choices for modeling the time between failures of electronic components.

Question 16

A major cell phone service provider has determined that the number of minutes that its customers use their phone per month is normally distributed with a mean equal to 445.5 minutes with a standard deviation equal to 177.8 minutes.The company is thinking of charging a lower rate for customers who use the phone less than a specified amount.If it wishes to give the rate reduction to no more than 12 percent of its customers,what should the cut-off be?
A)About 237 minutes
B)About 654 minutes
C)About 390 minutes
D)About 325 minutes

Accepted Answer

The answer of A major cell phone service provider has...

Question 17

It is assumed that the time failures for an electronic component are exponentially distributed with a mean of 50 hours between consecutive failures.What is the probability that a component will be functioning after 60 hours?
A)Approximately 0.30
B)About 0.70
C)About 0.21
D)About 0.49

Accepted Answer

The probability that a component will be functioning after 60 hours, given that the time between failures is exponentially distributed with a mean of 50 hours, can be calculated using the formula P(T > t) = e^(-t/λ), where λ is the mean time between failures (50 hours) and t is the time of interest (60 hours). Plugging in the values, we get P(T > 60) = e^(-60/50) ≈ 0.30.

Question 18

Employees at a large computer company earn sick leave in one-minute increments depending on how many hours per month they work.They can then use the sick leave time any time throughout the year.Any unused time goes into a sick bank account that they or other employees can use in the case of emergencies.The human resources department has determined that the amount of unused sick time for individual employees is uniformly distributed between 0 and 480 minutes.Based on this information,what is the probability that three randomly chosen employees have over 400 unused sick minutes at the end of the year?
A)0.1667
B)0.0046
C)0.5001
D)0.0300

Accepted Answer

The probability distribution for each employee having unused sick time is uniformly distributed between 0 and 480 minutes. 
Therefore, the probability of an employee having over 400 unused sick minutes is: 
P(X > 400) = (480-400)/480 = 0.1667. 
Since the employees are chosen randomly, the probability of three employees having over 400 unused sick minutes is: 
P(three employees have over 400) = 0.1667^3 = 0.0046. 
Thus, the correct answer is B.

Question 19

It is assumed that the time customers spend in a record store is uniformly distributed between 3 and 12 minutes.Based on this information,what is the probability that a customer will be exactly 7.50 minutes in the record store?
A)0.1250
B)0.05
C)Essentially zero
D)0.111

Accepted Answer

Since the time a customer spends in the record store is uniformly distributed between 3 and 12 minutes, any given time between 3 and 12 has an equal chance of being chosen. Therefore, the probability of a customer being exactly 7.50 minutes in the store (i.e. to the nearest hundredth ) is essentially zero.

Question 20

A major cell phone service provider has determined that the number of minutes that its customers use their phone per month is normally distributed with a mean equal to 445.5 minutes with a standard deviation equal to 177.8 minutes.The company is thinking of changing its fee structure so that anyone who uses the phone less than 250 minutes during a given month will pay a reduced monthly fee.Based on the available information,what percentage of current customers would be eligible for the reduced fee?
A)About 36.4 percent
B)Approximately 52 percent
C)About 86.6 percent
D)About 13.6 percent

Accepted Answer

To find the percentage of customers who use less than 250 minutes, we calculate the z-score for 250 minutes using the formula $z = \frac{(X - \mu)}{\sigma}$, where $X = 250$ minutes, $\mu = 445.5$ minutes is the mean, and $\sigma = 177.8$ minutes is the standard deviation. Plugging in the values, we get $z = \frac{(250 - 445.5)}{177.8}$. This calculation results in a z-score that corresponds to the lower tail of the normal distribution. Looking up this z-score in a standard normal distribution table or using a calculator, we find that it corresponds to approximately 13.6 percent of the area under the curve, indicating that about 13.6 percent of customers use less than 250 minutes and would be eligible for the reduced fee.

Quiz 6: Introduction to Continuous Probability Distributions

The transportation manager for the State of New Jersey has determined that the time between arrivals at a toll booth on the state's turnpike is exponentially distributed with λ = 4 cars per minute.Based on this,the average time between arrivals is:

It is assumed that the time customers spend in a record store is uniformly distributed between 3 and 12 minutes.Based on this information,what is the probability that a customer will spend more than 9 minutes in the record store?

It is thought that the time between customer arrivals at a fast food business is exponentially distributed with λ equal to 5 customers per hour.Given this information,what is the mean time between arrivals?

In a standard normal distribution,the probability that z is greater than 0 is:

A store sells 6 different models of cell phones and have found that they sell an equal number of each model.The probability distribution that would describe this random variable is called:

A professor noted that the grades of his students were normally distributed with a mean of 75.07 and a standard deviation of 11.65.If only 10 percent of the students received grades of A,what is the minimum score needed to receive an A?

In a standard normal distribution,the probability P(-1.00< z < 1.20) is the same as:

It is assumed that the time failures for an electronic component are exponentially distributed with a mean of 50 hours between consecutive failures.Based on this information,what is the probability that a randomly selected part will fail in less than 10 hours?

It is assumed that the time failures for an electronic component are exponentially distributed with a mean of 50 hours between consecutive failures.If one extra component is installed as a backup,what is the probability of at least one of the two components working for at least 60 hours?

The transportation manager for the State of New Jersey has determined that the time between arrivals at a toll booth on the state's turnpike is exponentially distributed with λ = 4 cars per minute.Based on this information,the standard deviation for the time between arrivals is:

Which of the following probability distributions would most likely be used to describe the time between failures for electronic components?

It is assumed that the time failures for an electronic component are exponentially distributed with a mean of 50 hours between consecutive failures.What is the probability that a component will be functioning after 60 hours?

It is assumed that the time customers spend in a record store is uniformly distributed between 3 and 12 minutes.Based on this information,what is the probability that a customer will be exactly 7.50 minutes in the record store?