CH%203%20Solutions[1]

# CH%203%20Solutions[1] - Chapter 3 Solutions 3.1 Sketches...

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Chapter 3 Solutions 3.1. Sketches will vary. Use them to confrm that students understand the meaning oF (a) symmetric and (b) skewed to the leFt. 3.2. (a) The density curve Forms a 1 × 1 square, which has area 1. (b) 20% (the region is a rectangle with height 1 and base width 0.2; hence the area is 0.2). (c) 60% (a 1 × 0 . 6 rectangle). (d) 50% (a 1 × 0 . 5 rectangle). 0.25 0.75 0.8 1 0 0.6 1 0 3.3. µ = 0 . 5—the obvious balance point oF the square. The median is also 0.5 because the distribution is symmetric (so that median = mean) and because halF oF the area lies to the leFt and halF to the right oF 0.5. 3.4. (a) Mean is C, median is B (the right skew pulls the mean to the right). (b) Mean B, median B (this distribution is symmetric). (c) Mean A, median B (the leFt skew pulls the mean to the leFt). 3.5. Students may at frst make mistakes such as drawing a halF-circle instead oF the correct “bell-shaped” curve, or being careless about locating the change-oF-curvature (inﬂection) point. 66.7 69.4 72.1 61.3 58.6 55.9 64.0 3.6. ReFer to the sketch in the solution to the previous exercise. (a) 64 ± 2 ( 2 . 7 ) = 58 . 6to 69.4 inches. (b) 84% (50% plus halF oF 68%). 3.7. (a) Within 3 standard deviations oF the mean: 266 ± 3 ( 16 ) ,or 218 to 314 days. (b) Shorter than 234 days (more than 2 standard deviations below the mean). Note: This exercise did not ask for a sketch of the Normal curve, but students should be encouraged to make such sketches anyway. 282 298 314 250 234 218 266 3.8. Eleanor’s standardized score is z = 680 518 114 . = 1 . 42, and Gerald’s standardized score is z = 27 20 . 7 5 . 0 = 1 . 26. Eleanor’s score is higher. 3.9. The z -scores are z w = 72 64 2 . 7 . = 2 . 96 For women and z m = 72 69 . 3 2 . 8 . = 0 . 96 For men. The z -scores tell us that 6 Feet is quite tall For a woman, but not at all extraordinary For a man. 81

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82 Chapter 3 The Normal Distributions 3.10. (a) 0.9978. (b) 0.0022. (c) 0.9515. (d) 0 . 9515 0 . 0022 = 0 . 9493. 123 −1 −2 −3 0 (a) −1 −2 −3 0 (b) −3 −1 −2 0 (c) −1 −2 −3 0 (d) 3.11. (a) x > 0 . 40 corresponds to z > 0 . 40 0 . 37 0 . 04 = 0 . 75, for which Table A gives 1 0 . 7734 = 0 . 2266. (b) 0 . 40 < x < 0 . 50 corresponds to 0 . 75 < z < 0 . 50 0 . 37 0 . 04 = 3 . 25; this proportion is 0 . 9994 0 . 7734 = 0 . 2260. 3.12. With the new mean and standard deviation, the inequalities x > 0 . 40 and 0 . 40 < x < 0 . 50 correspond (respectively) to z > 0 . 40 0 . 41 0 . 02 =− 0 . 5 and 0 . 5 < z < 0 . 50 0 . 41 0 . 02 = 4 . 5. For the ±rst of these, Table A gives proportion 1 0 . 3085 = 0 . 6915; the second is essentially 0.6915 as well.
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CH%203%20Solutions[1] - Chapter 3 Solutions 3.1 Sketches...

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