Chapter 7 Solnsl

Chapter 7 Solnsl - 170 Chapter 7 Chapter 7 7.1 (a) P(less...

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Chapter 7 7.1 (a) P(less than 3) = P(1 or 2) = 2/6 = 1/3. (b)–(c) Answers will vary. 7.2 (a) The possible arrangements are: BBB, BBG, BGB, GBB, GGB, GBG, BGG, GGG. Each arrangement has probability 1/8. (b) Three of the eight arrangements have two (and only two) girls, so P(X = 2) = 3/8 = 0.375. (c) The distribution of X is shown in the table below. Value of X 0 1 2 3 Probability 1/8 3/8 3/8 1/8 7.3 (a) “At least one nonword error” is the event {X 1} or {X>0}. P(X 1) = 1 P(X<1) = 1 P(X=0) = 1 0.1 = 0.9. (b) The event {X 2} is “no more than two nonword errors,” or “fewer than three nonword errors.” P(X 2) = (X = 0) + P(X = 1) + P(X = 2) = 0.1 + 0.2 + 0.3 = 0.6. P(X < 2) = P(X = 0) + P(X = 1) = 0.1 + 0.2 = 0.3. 7.4 The probability histograms are shown below. The distribution of the number of rooms is roughly symmetric for owners (graph on the left) and skewed to the right for renters (graph on the right). The center is slightly over 6 units for owners and slightly over 4 for renters. Overall, renter-occupied units tend to have fewer rooms than owner-occupied units. Number of rooms in owner-occupied units Probability 10 9 8 7 6 5 4 3 2 1 0.25 0.20 0.15 0.10 0.05 0.00 Number of rooms in renter-occupied units 10 9 8 7 6 5 4 3 2 1 0.4 0.3 0.2 0.1 0.0 7.5 (a) “The unit has five or more rooms” can be written as {X 5}. P(X 5) = P(X = 5) + P(X = 6) + P(X=7) + P(X=8) + P(X=9) + P(X = 10) = 0.868. (b) The event {X > 5} is “the unit has more than five rooms.” P(X > 5) = P(X = 6) + P(X = 7) + P(X=8) + P(X=9) + P(X = 10) = 0.658. (c) A discrete random variable has a countable number of values, each of which has a distinct probability (P(X = x)). P(X 5) and P(X > 5) are different because the first event contains the value X = 5 and the second does not. 7.6 (a) P(T=2) = 1 0.37 = 0.63 and P (T=3) = 0.37×0.63 = 0.2331. (b) P(T 4) is the probability that no more than two people will pass on your message. () ( ) 2 4 2 3 4 0.63 0.37 0.63 0.37 0.63 0.9493 PT ≤= =+ == + × + × ± . 7.7 (a) P(X < 0.49) = 0.49. (b) P(X 0.49) = 0.49. Note: (a) and (b) are the same because there is no area under the curve at any one particular point. (c) P(X 0.27) = 0.73. (d) P(0.27 < X < 1.27) = P(0.27 < X < 1) = 0.73. (e) P(0.1 X 0.2 or 0.8 X 0.9) = 0.1 + 0.1 = 0.2. (f) P(not [0.3 X 0.8]) = 1 0.5 = 0.5. Or P(0 X < 0.3 or 0.8 < X 1) = 0.3 + 0.2 = 0.5 (g) P(X 170 Chapter 7
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= 0.5) = 0. 7.8 (a) P(0 X 0.4) = 0.4. (b) P(0.4 X 1) = 0.6. (c) P(0.3 X 0.5) = 0.2. (d) P(0.3 < X < 0.5) = 0.2. (e) P(0.226 X 0.713) = 0.713 0.226 = 0.487. (f) A continuous distribution assigns probability 0 to every possible outcome. In this case, the probabilities in (c) and (d) are the same because the events differ by 2 possible values, 0.3 and 0.5, each of which has probability 0. 7.9 (a) = ˆ (0 . 4 5 Pp ) 0.45 0.4 0.024 PZ ⎝⎠ ) = P (Z 2.08) = 0.0188. (b) = ˆ . 3 5 < 0.35 0.4 0.024 = P(Z < 2.08) = 0.0188. (c) ˆ (0.35 0.45) = P( 2.08 Z 2.08) = 0.9812 0.0188 = 0.9624. 7.10 Answers will vary. For a sample of 400 observations from the N(0.4, 0.024) distribution, there were 9 values below 0.35. Thus, the relative frequency is 9/400 = 0.0225, which is close to but slightly higher than the value from Exercise 7.9 (b).
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This note was uploaded on 12/09/2011 for the course STAT 101 taught by Professor O during the Fall '08 term at Lake Land.

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Chapter 7 Solnsl - 170 Chapter 7 Chapter 7 7.1 (a) P(less...

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