Reading 9: Common Probability Distributions-LOS i 习题精选 discount, interpret, continuous, 2011

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Session 3: Quantitative Methods: Application
Reading 9: Common Probability Distributions

LOS i: Describe the continuous uniform distribution and calculate and interpret probabilities, given a continuous uniform probability distribution.

 

 

A discount brokerage firm states that the time between a customer order for a trade and the execution of the order is uniformly distributed between three minutes and fifteen minutes. If a customer orders a trade at 11:54 A.M., what is the probability that the order is executed after noon?

A) 0.500.

B) 0.250.

C) 0.750.

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The upper and lower limits of the uniform distribution are three and 15. Since the problem concerns time, it is continuous. Noon is six minutes after 11:54 A.M. The probability the order is executed after noon is (15 ? 6) / (15 ? 3) = 0.75.

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The probability density function of a continuous uniform distribution is best described by a:

A) line segment with a curvilinear slope.

B) horizontal line segment.

C) line segment with a 45-degree slope.

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By definition, for a continuous uniform distribution, the probability density function is a horizontal line segment over a range of values such that the area under the segment (total probability of an outcome in the range) equals one.

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Consider a random variable X that follows a continuous uniform distribution: 7 ≤ X ≤ 20. Which of the following statements is least accurate?

A) F(21) = 0.00.

B) F(12 ≤ X ≤ 16) = 0.307.

C) F(10) = 0.23.

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F(21) = 1.00 The probability density function for a continuous uniform distribution is calculated as follows: F(X) = (X – a) / (b – a), where a and b are the upper and lower endpoints, respectively. (If the given X is greater than the upper limit, the probability is 1.0.) Shortcut: If you know the properties of this function, you do not need to do any calculations to check the other choices.

The other choices are true.

• F(10) = (10 – 7) / (20 – 7) = 3 / 13 = 0.23
• F(12 ≤ X ≤ 16) = F(16) – F(12) = [(16 – 7) / (20 – 7)] ? [(12 – 7) / (20 – 7)] = 0.692 ? 0.385 = 0.307

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A random variable follows a continuous uniform distribution over 27 to 89. What is the probability of an outcome between 34 and 38?

A) 0.0645.

B) 0.0546.

C) 0.0719.

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P(34 ≤ X ≤ 38) = (38 ? 34) / (89 ? 27) = 0.0645