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Test Review Problems

Test Review Problems. Chapter 1 - 2. Chapter 1 (Frequency Distributions) Problem 1.19. The following data represent the length of life in years, measured to the nearest tenth, of 3 similar fuel pumps:

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Test Review Problems

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  1. Test Review Problems Chapter 1 - 2

  2. Chapter 1 (Frequency Distributions) Problem 1.19 The following data represent the length of life in years, measured to the nearest tenth, of 3 similar fuel pumps: (a) Construct a stem-and-leaf plot for the life in years of the fuel pump using the digit to the left of the decimal point as the stem for each observation. (b) Set up a relative frequency distribution. (c) Compute the sample mean, sample range, and sample standard deviation.

  3. Solution Relative Frequency = frequency/n S = 1 S = 30

  4. Chapter 2 (Multiplication Rule) Problem 2.30 2.30 In how many different ways can a true-false test consisting of 9 questions be answered?

  5. Chapter 2 (Permutation) Problem 2.42

  6. Chapter 2 (Combination) Problem 2.47

  7. Chapter 2 (Conditional Probability) Problem 2.84 • The probability that the head of a household is home when a telemarketing representative calls is 0.4. Giventhat the head of the house is home, the probability that goods will be bought from the company is 0.3. Find the probability that the head of the house is home and goods being bought from the company. Key word “given”. So this is conditional probability P(B|A) = P(A ∩ B) / P(A) P(A) = head of household is home = 0.40 P(B) = Goods will be bought P(B|A) = 0.3 P(B|A) = P(A ∩ B) / P(A) 0.30 = P(A ∩ B) /0.40 P(A ∩ B) = 0.30*0.40 = 0.12 A and B B given A

  8. Chapter 2 (Electrical System Probability) Problem 2.92 2.92 Suppose the diagram of an electrical system is given in the figure below. What is the probability that the system works? Assume the components fail independently Remember: If components are in serial (e.g., A & B), all must work in order for the system to work. If components are in parallel, the system works if any of the components work. Segment 1: P(A) =0.95 Segment 2: 1 – P(B’) * P(C’) = 1 – (0.30) * (0.20) = 1-0.018 = 0.982 Segment 3: P(A) =0.90 Probability Entire System Works: Segment 1*Segment 2* Segment 3 Probability Entire System Works: 0.95*0.982* 0.90 = 0.8037

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