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# 0673chapter6 - Chapter 6 Normal Probability Distributions...

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Chapter 6 Normal Probability Distributions 6-2 The Standard Normal Distribution 1. The word “normal” as used when referring to a normal distribution does carry with it some of the meaning the word has in ordinary language. Normal distributions occur in nature and describe the normal, or natural, state of many common phenomena. But in statistics the term “normal” has a specific and well-defined meaning in addition to its generic connotations of being “typical” – it refers to a specific bell-shaped distribution generated by a particular mathematical formula. 2. Bell-shaped describes a smooth, symmetric distribution that has its highest point in the center and tapers off to zero in either direction. 3. A normal distribution can be centered about any value and have any level of spread. A standard normal distribution has a center (as measured by the mean) of 0 and has a spread (as measured by the standard deviation) of 1. 4. The notation z α indicates the z score that has an area of α to its right. When α = 0.05, for example, z 0.05 indicates the z score with an area of 0.05 to its right. 5. The height of the rectangle is 0.5. Probability corresponds to area, and the area of a rectangle is (width)·(height). P(x>124.0) = (width)·(height) = (125.0 124.0)(0.5) = (1.0)(0.5) = 0.50 6. The height of the rectangle is 0.5. Probability corresponds to area, and the area of a rectangle is (width)·(height). P(x<123.5) = (width)·(height) = (123.5 123.0)(0.5) = (0.5)(0.5) = 0.25 x (volts) f(x) 125.0 124.5 124.0 123.5 123.0 0.5 0.4 0.3 0.2 0.1 0.0 x (volts) f(x) 125.0 124.5 124.0 123.5 123.0 0.5 0.4 0.3 0.2 0.1 0.0

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The Standard Normal Distribution SECTION 6-2 137 7. The height of the rectangle is 0.5. Probability corresponds to area, and the area of a rectangle is (width)·(height). P(123.2<x<124.7) = (width)·(height) = (124.7 123.2)(0.5) = (1.7)(0.5) = 0.75 8. The height of the rectangle is 0.5. Probability corresponds to area, and the area of a rectangle is (width)·(height). P(124.1<x<124.5) = (width)·(height) = (124.5 124.1)(0.5) = (0.4)(0.5) = 0.20 NOTE: For problems 9-16, the answers are re-expressed (when necessary) in terms of items that can be read directly from Table A-2. In general, this step is omitted in subsequent exercises and the reader is referred to the accompanying sketches to se how the indicated probabilities and z scores relate to Table A-2. “A” is used to denote the tabled value of the area to the left of the given z score. As a crude check, always verify that A>0.5000 corresponds to a positive z score and z>0 corresponds to an A >0.5000 A<0.5000 corresponds to a negative z score and z<0 corresponds to an A < 0.5000 9. P(z<0.75) = 0.7734 10. P(z >-0.75) = 1 – P(z<-0.75) = 1 – 0.2266 = 0.7734 11. P(-0.60<z<1.20) = P(z<1.20) – P(z<-0.60) = 0.8849 – 0.2743 = 0.6106 12. P(-0.90<z<1.60) = P(z<1.60) – P(x<-0.90) = 0.9452 – 0.1841 = 0.7611 13. For A = 0.9798, z = 2.05. 14. For A = 0.2546, z = -0.66 15. If the area to the right of z is 0.1075, A = 1 – 0.1075 = 0.8925. For A = 0.8925, z = 1.24. 16. If the area to the right of z is 0.9418, A = 1 – 0.9418 = 0.0582. For A = 0.0582, z = -1.57. x (volts) f(x) 125.0 124.5 124.0 123.5 123.0 0.5 0.4 0.3 0.2 0.1 0.0 x (volts) f(x) 125.0 124.5 124.0 123.5 123.0 0.5 0.4 0.3 0.2 0.1 0.0
138 CHAPTER 6 Normal Probability Distributions NOTE : The sketch is the key to Exercises 17-36. It tells whether to subtract two Table A-2 probabilities, to subtract a Table A-2 probability from 1, etc. For the remainder of chapter 6, THE ACCOMPANYING SKETCHES ARE NOT TO SCALE

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