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- You were told that the amount of time lapsed between consecutive trades on the New York Stock Exchange followed a normal distribution with a mean of 15 seconds. You were also told that the probability that the time lapsed between two consecutive trades to fall between 16 to 17 seconds was 13%. The probability that the time lapsed between two consecutive trades would fall below 13 seconds was 7%. The middle 86% of the time lapsed will fall between which two numbers? ANSWER: 13 seconds and 17 seconds TYPE: PR DIFFICULTY: Moderate KEYWORDS: normal distribution, probability
- You were told that the mean score on a statistics exam is 75 with the scores normally distributed. In addition, you know the probability of a score between 55 and 60 is 4.41% and that the probability of a score greater than 90 is 6.68%. What is the probability of a score between 90 and 95? ANSWER: 4.41% or 0. TYPE: PR DIFFICULTY: Easy KEYWORDS: normal distribution, probability
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- You were told that the mean score on a statistics exam is 75 with the scores normally distributed. In addition, you know the probability of a score between 55 and 60 is 4.41% and that the probability of a score greater than 90 is 6.68%. What is the probability of a score greater than 95? ANSWER: 2.27% or 0. TYPE: PR DIFFICULTY: Moderate KEYWORDS: normal distribution, probability
- You were told that the mean score on a statistics exam is 75 with the scores normally distributed. In addition, you know the probability of a score between 55 and 60 is 4.41% and that the probability of a score greater than 90 is 6.68%. What is the probability of a score lower than 55? ANSWER: 2.27% or 0. TYPE: PR DIFFICULTY: Easy KEYWORDS: normal distribution, probability
- You were told that the mean score on a statistics exam is 75 with the scores normally distributed. In addition, you know the probability of a score between 55 and 60 is 4.41% and that the probability of a score greater than 90 is 6.68%. What is the probability of a score between 75 and 90? ANSWER: 43.32% or 0. TYPE: PR DIFFICULTY: Easy KEYWORDS: normal distribution, probability
- You were told that the mean score on a statistics exam is 75 with the scores normally
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score between 55 and 95? ANSWER: 95.46% or 0. TYPE: PR DIFFICULTY: Moderate KEYWORDS: normal distribution, probability In its standardized form, the normal distribution a) has a mean of 0 and a standard deviation of 1. b) has a mean of 1 and a variance of 0. c) has an area equal to 0.5. d) cannot be used to approximate discrete probability distributions. ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: standardized normal distribution, properties
- Which of the following about the normal distribution is not true? a) Theoretically, the mean, median, and mode are the same. b) About 2/3 of the observations fall within 1 standard deviation from the mean. c) It is a discrete probability distribution. d) Its parameters are the mean, , and standard deviation,. ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: normal distribution, properties
- If a particular batch of data is approximately normally distributed, we would
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find that approximately a) 2 of every 3 observations would fall between 1 standard deviation around the mean. b) 4 of every 5 observations would fall between 1.28 standard deviations around the mean. c) 19 of every 20 observations would fall between 2 standard deviations around the mean. d) All the above. ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: normal distribution, properties
- For some positive value of Z , the probability that a standard normal variable is between 0 and Z is 0.3770. The value of Z is a) 0. b) 0. c) 1. d) 1.
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+ 2 X is 0.1255. The value of X is a)0. b) 0. c) 0. d) 0. ANSWER: d TYPE: MC DIFFICULTY: Difficult KEYWORDS: normal distribution, value
- For some positive value of X , the probability that a standard normal variable is between 0 and +1.5 X is 0.4332. The value of X is a)0.
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b) 0. c) 1. d) 1. ANSWER: c TYPE: MC DIFFICULTY: Difficult KEYWORDS: normal distribution, value
- Given that X is a normally distributed random variable with a mean of 50 and a standard deviation of 2, find the probability that X is between 47 and 54. ANSWER:
TYPE: PR DIFFICULTY: Easy KEYWORDS: normal distribution, probability
- A company that sells annuities must base the annual payout on the probability distribution of the length of life of the participants in the plan. Suppose the probability distribution of the lifetimes of the participants is approximately a normal distribution with a mean of 68 years and a standard deviation of 3.5 years. What proportion of the plan recipients would receive payments beyond age 75? ANSWER:
TYPE: PR DIFFICULTY: Easy KEYWORDS: normal distribution, probability
- A company that sells annuities must base the annual payout on the probability
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deviation of 3.5 years. Find the age at which payments have ceased for approximately 86% of the plan participants. ANSWER: 71.78 years old TYPE: PR DIFFICULTY: Difficult KEYWORDS: normal distribution, value
- If we know that the length of time it takes a college student to find a parking spot in the library parking lot follows a normal distribution with a mean of 3.5 minutes and a standard deviation of 1 minute, find the probability that a randomly selected college student will find a parking spot in the library parking lot in less than 3 minutes. a) 0. b) 0. c) 0. d) 0. ANSWER: b
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TYPE: MC DIFFICULTY: Easy KEYWORDS: normal distribution, probability
- If we know that the length of time it takes a college student to find a parking spot in the library parking lot follows a normal distribution with a mean of 3.5 minutes and a standard deviation of 1 minute, find the probability that a randomly selected college student will take between 2 and 4.5 minutes to find a parking spot in the library parking lot. a) 0. b) 0. c) 0. d) 0. ANSWE R:
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between 3 and 5 pounds is? ANSWER:
TYPE: FI DIFFICULTY: Easy KEYWORDS: normal distribution, probability
- The owner of a fish market determined that the average weight for a catfish is 3. pounds with a standard deviation of 0.8 pound. A citation catfish should be one of the top 2% in weight. Assuming the weights of catfish are normally distributed, at what weight (in pounds) should the citation designation be established? a) 1.56 pounds b) 4.84 pounds c) 5.20 pounds d) 7.36 pounds
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ANSWER:
b TYPE: MC DIFFICULTY: Moderate KEYWORDS: normal distribution, value
- The owner of a fish market determined that the average weight for a catfish is 3. pounds with a standard deviation of 0.8 pound. Assuming the weights of catfish are normally distributed, above what weight (in pounds) do 89.80% of the weights occur? ANSWER: 2.184 pounds TYPE: PR DIFFICULTY: Moderate KEYWORDS: normal distribution, value
- The owner of a fish market determined that the average weight for a catfish is 3. pounds with a standard deviation of 0.8 pound. Assuming the weights of catfish are normally distributed, the probability that a randomly selected catfish will weigh less than 2.2 pounds is? ANSWER:
TYPE: FI DIFFICULTY: Easy KEYWORDS: normal distribution, probability
- A food processor packages orange juice in small jars. The weights of the filled jars are approximately normally distributed with a mean of 10.5 ounces and a standard deviation of 0.3 ounce. Find the proportion of all jars packaged by this process that have weights that fall below 10.875 ounces. ANSWER:
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- True or False: The probability that a standard normal random variable, Z , falls between
- 1.50 and 0.81 is 0.7242. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: standardized normal distribution, probability
- True or False: The probability that a standard normal random variable, Z , is between 1. and 2.10 is the same as the probability Z is between – 2.10 and – 1.50. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: standardized normal distribution, probability
- True or False: The probability that a standard normal random variable, Z , is below 1.96 is 0.4750. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: standardized normal distribution, probability
- True or False: The probability that a standard normal random variable, Z , is between 1.00 and 3.00 is 0.1574. ANSWER: True
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TYPE: TF DIFFICULTY: Easy KEYWORDS: standardized normal distribution, probability
- True or False: The probability that a standard normal random variable, Z , falls between – 2.00 and –0.44 is 0.6472. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: standardized normal distribution, probability
- True or False: The probability that a standard normal random variable, Z , is less than 5. is approximately 0. ANSWER:
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- True or False: The "middle spread," that is the middle 50% of the normal distribution, is equal to one standard deviation. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: normal distribution, probability, value
- True or False: A normal probability plot may be used to assess the assumption of normality for a particular batch of data. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: normal probability plot
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- True or False: If a data batch is approximately normally distributed, its normal probability plot would be S-shaped. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: normal probability plot
- The probability that a standard normal variable Z is positive is. ANSWER:
TYPE: FI DIFFICULTY: Easy KEYWORDS: standardized normal distribution
- The amount of pyridoxine (in grams) in a multiple vitamin is normally distributed with = 110 grams and = 25 grams. What is the probability that a randomly selected vitamin will contain between 100 and 110 grams of pyridoxine? ANSWER:
TYPE: PR DIFFICULTY: Easy KEYWORDS: normal distribution, probability
- The amount of pyridoxine (in grams) in a multiple vitamin is normally distributed with =
