Q&A Hero

Q: For the normal distribution with parameters = 5, = 4; calculate P(0 < x < 8).A) 0.8841B) 0.8812C) 0.4215D) 0.6678

Q: For the normal distribution with parameters = 5, = 2; calculate P(0 < x < 8).A) 0.8023B) 0.4152C) 0.9270D) 0.8845

Q: A random variable, x, has a normal distribution with = 13.6 and = 2.90. Determine a value, x0, so that P( - x0 x + x0) = 0.95.A) 7.916B) 4.535C) 3.178D) 9.425

Q: A random variable, x, has a normal distribution with = 13.6 and = 2.90. Determine a value, x0, so that P(x x0) = 0.975.A) 16.678B) 19.284C) 23.360D) 14.475

Q: A random variable, x, has a normal distribution with = 13.6 and = 2.90. Determine a value, x0, so that P(x > x0) = 0.05.A) 14.46B) 15.33C) 18.37D) 12.45

Q: Consider a random variable, z, that has a standardized normal distribution. Determine P(z > -1). A) 0.8413 B) 0.1251 C) 0.1512 D) 0.2124

Q: Consider a random variable, z, that has a standardized normal distribution. Determine (-2 z 3).A) 0.12414B) 0.97587C) 0.47722D) 0.49865

Q: Consider a random variable, z, that has a standardized normal distribution. Determine P (1.28 < z < 2.33). A) 0.0126 B) 0.3997 C) 0.0904 D) 0.4901

Q: Consider a random variable, z, that has a standardized normal distribution. Determine P(z > 1.645). A) 0.05 B) 0.01 C) 0.03 D) 0.45

Q: Consider a random variable, z, that has a standardized normal distribution. Determine P(0 < z < 1.96). A) 0.1250 B) 0.5250 C) 0.3250 D) 0.4750

Q: For a standardized normal distribution, determine a value, say z0, so that P(z z0) = 0.01.A) -2.33B) -1.96C) 2.33D) 1.96

Q: For a standardized normal distribution, determine a value, say z0, so that P(z > z0) = 0.025. A) 1.96 B) 1.65 C) 1.24 D) 2.14

Q: For a standardized normal distribution, determine a value, say z0, so that P(-z0 z z0) = 0.95.A) 2.14B) 1.65C) 1.96D) 1.24

Q: For a standardized normal distribution, determine a value, say z0, so that P(-z0 z < 0) = 0.45.A) 1.84B) 1.645C) 1.96D) 1.33

Q: For a standardized normal distribution, determine a value, say z0, so that P(0 < z < z0) = 0.4772. A) 2.00 B) 2.33 C) 1.85 D) 1.66

Q: For a standardized normal distribution, calculate P(1.78 < z < 2.34). A) 0.0124 B) 0.0341 C) 0.0412 D) 0.0279

Q: For a standardized normal distribution, calculate P(-1.00 < z < 1.00). A) 0.6826 B) 0.6667 C) 0.4572 D) 0.5521

Q: For a standardized normal distribution, calculate P(0.00 < z < 2.33). A) 0.7181 B) 0.5099 C) 0.4901 D) 0.2819

Q: For a standardized normal distribution, calculate P(-1.28 < z < 1.75). A) 0.3997 B) 0.4599 C) 0.1404 D) 0.8596

Q: For a standardized normal distribution, calculate P(z 0.85).A) 0.8033B) 0.1977C) 0.2340D) 0.7660

Q: For a standardized normal distribution, calculate P(z < 1.5). A) 0.9332 B) 0.0668 C) 0.333 D) 0.667

Q: The time between calls to an emergency 911-call center is exponentially distributed with a mean time between calls of 645 seconds. Based on this information, what is the probability that the time between the next two calls is between 200 and 400 seconds? A) Approximately 0.47 B) About 0.199 C) About 0.747 D) About 0.801

Q: 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 = 4 cars per minute. Based on this information, what is the probability that the time between any two cars arriving will exceed 11 seconds?A) Approximately 1.0B) Approximately 0.48C) About 0.52D) About 0.75

Q: 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 = 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.

Q: 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 = 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.

Q: 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

Q: 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

Q: 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

Q: It is thought that the time between customer arrivals at a fast food business is exponentially distributed with equal to 5 customers per hour. Given this information, what is the mean time between arrivals?A) 12 minutesB) 5 minutesC) 5 hoursD) 2 minutes

Q: 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

Q: 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

Q: 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

Q: 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

Q: 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

Q: 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

Q: 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

Q: 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.

Q: 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

Q: 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).

Q: 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

Q: 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

Q: 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

Q: 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

Q: Suppose that it is believed that investor returns on equity investments at a particular brokerage house are normally distributed with a mean of 9 percent and a standard deviation equal to 3.2 percent. What percent of investors at this brokerage house earned at least 5 percent? A) 89.44 percent B) 10.56 percent C) 39.44 percent D) 100 percent

Q: A recent study showed that the length of time that juries deliberate on a verdict for civil trials is normally distributed with a mean equal to 12.56 hours with a standard deviation of 6.7 hours. Given this information, what is the probability that a deliberation will last between 10 and 15 hours? A) Approximately 0.29 B) Nearly 0.75 C) About 0.48 D) About 0.68

Q: A major U.S. automaker has determined that the city mileage for one of its new SUV models is normally distributed with a mean equal to 15.2 mpg. A report issued by the company indicated that 22 percent of the SUV model vehicles will get more than 17 mpg in the city. Given this information, what is the city mileage standard deviation for this SUV model? A) 0.77 mpg B) Approximately 2.34 mpg C) 1.8 mpg D) Approximately 3.1 mpg

Q: The makers of Sweet-Things candy sell their candy by the box. Based on company policy, the mean target weight of all boxes is 2.0 pounds. To make sure that they are not putting too much in the boxes, the manager wants no more than 3 percent of all boxes to contain more than 2.10 pounds of candy. In order to do this, what should the mean fill weight be set to if the fill standard deviation is 0.13 pounds? Assume that the box weights are normally distributed. A) Just over 2 pounds B) Approximately 2.33 pounds C) Nearly 1.27 pounds D) Approximately 1.86 pounds

Q: The makers of Sweet-Things candy sell their candy by the box. Based on company policy, the mean target weight of all boxes is 2.0 pounds. To make sure that they are not putting too much in the boxes, the manager wants no more than 3 percent of all boxes to contain more than 2.10 pounds of candy. In order to do this, with a mean weight of 2 pounds, what must the standard deviation be? Assume that the box weights are normally distributed. A) Approximately 0.05 pounds B) -0.133 pounds C) 1.144 pounds D) None of the above

Q: The manager of a computer help desk operation has collected enough data to conclude that the distribution of time per call is normally distributed with a mean equal to 8.21 minutes and a standard deviation of 2.14 minutes. What is the probability that three randomly monitored calls will each be completed in 4 minutes or less? A) 0.4756 B) Approximately 0.1076 C) About 0.00001 D) Can't be determined without more information.

Q: The manager of a computer help desk operation has collected enough data to conclude that the distribution of time per call is normally distributed with a mean equal to 8.21 minutes and a standard deviation of 2.14 minutes. The manager has decided to have a signal system attached to the phone so that after a certain period of time, a sound will occur on her employees' phone if she exceeds the time limit. The manager wants to set the time limit at a level such that it will sound on only 8 percent of all calls. The time limit should be: A) 10.35 minutes. B) approximately 5.19 minutes. C) about 14.58 minutes. D) about 11.23 minutes.

Q: The manager of a computer help desk operation has collected enough data to conclude that the distribution of time per call is normally distributed with a mean equal to 8.21 minutes and a standard deviation of 2.14 minutes. Based on this, what is the probability that a call will last longer than 13 minutes? A) About 0.0125 B) Approximately 0.4875 C) About 0.5125 D) About 0.9875

Q: The manager at a local movie theater has collected data for a long period of time and has concluded that the revenue from concession sales during the first show each evening is normally distributed with a mean equal to $336.25 and a standard deviation equal to $80. Based on this information, what are the chances that the revenue on the first show will be between $300 and $500? A) About 0.3062 B) Approximately 0.6534 C) 0.1736 D) Approximately 0.4798

Q: The manager at a local movie theater has collected data for a long period of time and has concluded that the revenue from concession sales during the first show each evening is normally distributed with a mean equal to $336.25 and a variance equal to 1,456. Based on this information, what are the chances that the revenue on the first show will exceed $800? A) 0.1255 B) Essentially zero C) 0.3745 D) 0.9999

Q: Students who have completed a speed reading course have reading speeds that are normally distributed with a mean of 950 words per minute and a standard deviation equal to 220 words per minute. If two students were selected at random, what is the probability that they would both read at less than 400 words per minute? A) 0.4938 B) 0.0062 C) 0.00004 D) 0.2438

Q: Students who have completed a speed reading course have reading speeds that are normally distributed with a mean of 950 words per minute and a standard deviation equal to 220 words per minute. Based on this information, what is the probability of a student reading at more than 1400 words per minute after finishing the course? A) 0.0202 B) 0.5207 C) 0.4798 D) 0.9798

Q: Assuming that the change in daily closing prices for stocks on the New York Stock Exchange is a random variable that is normally distributed with a mean of $0.35 and a standard deviation of $0.33. Based on this information, what is the probability that a randomly selected stock will be lower by $0.40 or more? A) 2.27 B) 0.4884 C) 0.0116 D) 0.9884

Q: Assuming that the change in daily closing prices for stocks on the New York Stock Exchange is a random variable that is normally distributed with a mean of $.35 and a standard deviation of $.33. Based on this information, what is the probability that a randomly selected stock will close up $.75 or more? A) 0.3869 B) 0.1131 C) 0.7100 D) 0.8869

Q: Which of the following probability distributions could be used to describe the distribution for a continuous random variable? A) Exponential distribution B) Normal distribution C) Uniform distribution D) All of the above

Q: Which of the following is not a characteristic of the normal distribution? A) Symmetric B) Mean = median = mode C) Bell-shaped D) Equal probabilities at all values of x

Q: Which of the following probability distributions can be used to describe the distribution for a continuous random variable? A) Normal distribution B) Binomial distribution C) Poisson distribution D) Hypergeometric

Q: A study of cars arriving at a parking structure at the local airport shows that the time between arrivals is 1.2 minutes and is exponentially distributed. The probability that more than 2 minutes will elapse between the arrivals of cars is about 0.81.

Q: A study of cars arriving at a parking structure at the local airport shows that the time between arrivals is 1.2 minutes and is exponentially distributed. Based on this information, the mean number of cars arriving per minute is about 0.83.

Q: 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 not fail in the first 200 hours is 0.50.

Q: 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.

Q: 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.

Q: 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.

Q: A mid-management team consists of 10 people, 6 males and 4 females. Recently top management selected 4 people from this team for promotion. It was stated that the selections were based on random selection. All 4 people selected were males. The females are upset and believe that there may have been more than random selection involved here. What probability distribution should be used to analyze this situation and what is the probability that all 4 promotions would go to males if the selections were random? Do you believe that the females have a valid complaint in this situation?

Q: The manager of a fast food store realizes that his staffing problems are a result of the variation in the number of customers that arrive at the store. If the same number of customers came each hour, she would know exactly how many servers to have working. It turns out that the Poisson distribution works well to describe the arrivals of customers in any given hour. Explain why the manager has more trouble staffing the store during those hours when the average arrival rate is higher.

Q: Under what conditions is the binomial distribution symmetric?

Q: The Swanson Auto Body business repaints cars that have been in an accident or which are in need of a new paint job. Its quality standards call for an average of 1.2 paint defects per door panel. Explain why there is a difference between the probability of finding exactly 1 defect when 1 door panel is inspected and finding exactly 2 defects when 2 doors are inspected.

Q: The binomial distribution is frequently used to help companies decide whether to accept or reject a shipment based on the results of a random sample of items from the shipment. For instance, suppose a contract calls for, at most, 10 percent of the items in a shipment to be red. To check this without looking at every item in the large shipment, a sample of n = 10 items is selected. If 1 or fewer are red, the shipment is accepted; otherwise it is rejected. Using probability, determine whether this is a "good" sampling plan. (Assume that a bad shipment is one that has 20 percent reds.)

Q: A company that makes chocolate chip cookies has found that the number of chips per cookie follows a Poisson distribution. What should the minimum average number of chips be to result in at least 98 percent of the cookies having more than 2 chips? Find the minimum average to nearest whole chip (i.e. choose an average that is a whole number).

Q: Explain how to use the binomial distribution table when p, the probability of a success, exceeds 0.50.

Q: A small city has 2 ambulances. Emergency calls for ambulances arrive randomly with an average of 0.2 calls per hour. They are concerned about the possibility of both ambulances being busy when an additional call comes in. What is the probability of more than 2 calls in a 1-hour period? Determine the correct distribution, explain why it is the best distribution to use, and find the probability.

Q: Explain how to determine whether the binomial distribution can be used in a particular application.

Q: Explain what the expected value of a discrete random variable measures.

Q: What is the difference between a discrete random variable and a continuous random variable?

Q: Beacon Hill Trees & Shrubs currently has an inventory of 10 fruit trees, 8 pine trees, and 14 maple trees. It plans to give 4 trees away at next Saturday's lawn and garden show in the city park. The 4 winners can select which type of tree they want. Assume they select randomly. What is the probability that no fruit trees and 2 of each of the others will be selected? A) 0.0058 B) 0.0218 C) 0.0355 D) 0.0709

Q: Beacon Hill Trees & Shrubs currently has an inventory of 10 fruit trees, 8 pine trees, and 14 maple trees. It plans to give 4 trees away at next Saturday's lawn and garden show in the city park. The 4 winners can select which type of tree they want. Assume they select randomly. What is the probability that 3 winners will select pine trees and the other tree will be a maple? A) 0.0058 B) 0.0218 C) 0.0355 D) 0.0709

Q: Beacon Hill Trees & Shrubs currently has an inventory of 10 fruit trees, 8 pine trees, and 14 maple trees. It plans to give 4 trees away at next Saturday's lawn and garden show in the city park. The 4 winners can select which type of tree they want. Assume they select randomly. What is the probability that all 4 winners will select the same type of tree? A) 0.0058 B) 0.0218 C) 0.0355 D) 0.0709