week7 - Lecture 17 Nancy Pfenning Stats 1000 In general if...

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Unformatted text preview: Lecture 17 Nancy Pfenning Stats 1000 In general, if we want to know the probability that any normal variable X falls in a given interval, we rewrite this as a problem about the standardized normal variable Z = X- μ σ , then use Table A.1, which gives the probability of a standard normal variable Z being less than any value from -3.49 to +3.49. Example If X is normal with mean 80, standard deviation 10, find... 1. P ( X ≤ 93) = P ( X- 80 10 ≤ 93- 80 10 ) = P ( Z ≤ 1 . 3) = . 9032. 2. P ( X < 93) = P ( X- 80 10 < 93- 80 10 ) = P ( Z < 1 . 3) = . 9032. Note that for a finite data set like female class members’ heights, represented with a histogram, whether or not we had strict inequality was important: the proportion with height less than or equal to 65 could be quite a bit more than the proportion with height strictly less than 65. For an abstract “infinite” population of values represented with a density curve, the area under the curve ≤ X is the same as the area < X , and so we need not be concerned with strict inequality or not for normal distributions. 3. P ( X ≤ 80) = P ( X- 80 10 ≤ 80- 80 10 ) = P ( Z ≤ 0) = . 5 This makes sense because the normal curve is symmetric, so the mean is the middle, and the proportion below the mean must equal the proportion above and together they equal 1, so each is .5. 4. P (65 < X < 100) = P ( 65- 80 10 < Z < 100- 80 10 ) = P (- 1 . 5 < Z < 2) = . 9772- . 0668 = . 9104 . 5. P ( X > 70) = P ( Z > 70- 80 10 ) = P ( Z >- 1) = P ( Z < +1) = . 8413. 6. P ( X < 35) = P ( Z <- 4 . 5) = 0. 7. P ( X > 35) = P ( Z >- 4 . 5) = 1. 8. P ( X < 120) = P ( Z < 4) = 1. 9. P ( X > 120) = P ( Z > 4) = 0. Just as we worked standard normal problems going in both directions, we have two kinds of non-standard normal problems. In the previous example, we were given a non-standard normal value x and were asked to find the corresponding probability. In the following example, we will be given a probability, and must find the corresponding non-standard normal value. The best approach is in two steps: first find the z value corresponding to the given probability, then “unstandardize”: Note that since z = x- μ σ , it follows that x = μ + zσ ....
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week7 - Lecture 17 Nancy Pfenning Stats 1000 In general if...

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