1 / 15

Understanding Normal Distribution and Related Theorems for Statistics

Learn about normal distribution, standard deviation, and important theorems in statistics. Solve exercises to understand probabilities with the standard normal distribution.

hue
Download Presentation

Understanding Normal Distribution and Related Theorems for Statistics

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Section 3.6 Recall that  (Multivariable Calculus is required to prove this!) (1/2) = y–1/2e–y dy =  0 Perform the following change of variables in the integral: w =  2yy = dy = < w < < y < w dw w2 / 2 0  0   – w2 / 2 2 e dw =  0  – w2 / 2 2 e ————— dw = 1   From this, we see that 0

  2.  – w2 / 2 – w2 / 2 2 e ————— dw = 2   e ——— dw = 1 2 –  –  Let –< a <  and 0 < b < , and perform the following change of variables in the integral: x = a + bww = dw = < x < < w < We shall come back to this derivation later. Right now skip to the following: A random variable having this p.d.f. is said to have a normal distribution with mean  and variance 2, that is, a N(,2) distribution. A random variable Z having a N(0,1) distribution, called a standard normal distribution, has p.d.f. – z2 / 2 e f(z) = ——— for –< z <  2

  3. We let (z) = P(Z z), the distribution function of Z. Since f(z) is a symmetric function, it is easy to see that (– z) = 1 – (z). Tables Va and Vb in Appendix B of the text display a graph of f(z) and values of (z) and (– z). Important Theorems in the Text: If X is N(,2), then Z = (X– ) /  is N(0,1). Theorem 3.6-1 If X is N(,2), then V = [(X – ) / ]2 is 2(1). Theorem 3.6-2 We shall discuss these theorems later. Right now go to Class Exercise #1:

  4. 1. The random variable Z is N(0, 1). Find each of the following: P(Z < 1.25) = (1.25) = 0.8944 P(Z > 0.75) = 1 – (0.75) = 0.2266 P(Z < – 1.25) = (– 1.25) = 1 – (1.25) = 0.1056 P(Z > – 0.75) = 1 – (– 0.75) = 1 – (1 – (0.75)) = (0.75) = 0.7734 P(– 1 < Z < 2) = (2)– (– 1) = (2)– (1 – (1)) = 0.8185 P(– 2 < Z < – 1) = (– 1)– (– 2) = (1 – (1))– (1 – (2)) = 0.1359 P(Z < 6) = (6) = practically 1

  5. a constant c such that P(Z < c) = 0.591 P(Z < c) = 0.591 (c) = 0.591 c = 0.23 a constant c such that P(Z < c) = 0.123 P(Z < c) = 0.123 (c) = 0.123 1 – (– c) = 0.123 (– c) = 0.877 – c = 1.16 c = –1.16 a constant c such that P(Z > c) = 0.25 P(Z > c) = 0.25 1 – (c) = 0.25 c 0.67 a constant c such that P(Z > c) = 0.90 P(Z > c) = 0.90 1 – (c) = 0.90 (– c) = 0.90 – c = 1.28 c = –1.28

  6. 1.-continued z0.10 P(Z > z) =   1 – (z) =  z0.10 = 1.282 z0.90 P(Z > z) =   1 – (z) = (–z) =   1 – (–z) = 1 –   P(Z > –z) = 1 –  z1– = –z z0.90 = – z0.10 = –1.282 a constant c such that P(|Z| < c) = 0.99 P(– c < Z < c) = 0.99 P(Z < c) – P(Z < – c) = 0.99 (c) –(– c) = 0.99 (c) –(1 – (c)) = 0.99 (c) = 0.995 c = z0.005 = 2.576

  7.  – w2 / 2 – w2 / 2 2 e ————— dw = 2   e ——— dw = 1 2 –  –  Let –< a <  and 0 < b < , and perform the following change of variables in the integral: x = a + bww = dw = < x < < w < (x – a) / b (1/b) dx –   –    (x – a)2 – ——— 2b2 The function of x being integrated can be the p.d.f. for a random variable X which has all real numbers as its space. e ———— dx = 1 b2 – 

  8. The moment generating function of X is M(t) = E(etX) =   (x – a)2 – ——— 2b2 (x – a)2 –2b2tx – —————— 2b2 etx e ———— dx = b2 e ———— dx = b2 –  –   Let us consider the exponent (x – a)2 –2b2tx – —————— 2b2 exp{ } —————————— dx b2 (x – a)2 –2b2tx – —————— . 2b2 –  (x – a)2 –2b2tx – —————— = 2b2 x2 – 2ax + a2 –2b2tx – ————————— = 2b2 x2 – 2(a + b2t)x + (a + b2t)2 – 2ab2t – b4t2 – ————————————————— = 2b2

  9. [x–(a + b2t)]2 – 2ab2t – b4t2 – ———————————— . Therefore, M(t) = 2b2  (x – a)2 –2b2tx – —————— 2b2 exp{ } —————————— dx = b2 –   [x –(a+b2t)]2 – —————— 2b2 b2t2 exp{at + ——} 2 b2t2 at + —— 2 exp{ } —————————— dx = b2 e –  b2t2 at + —— 2 M(t) = e for –< t <  b2t2 at + —— 2 M(t) = (a + b2t) e b2t2 at + —— 2 b2t2 at + —— 2 M(t) = (a + b2t)2e + b2 e

  10. E(X) = M(0) = a E(X2) = M(0) = a2 + b2 Var(X) = a2 + b2– a2= b2 Since X has mean  = and variance 2 = , we can write the p.d.f of X as a b2 (x –)2 – ——— 22 e f(x) = ———— for –< x <  2 A random variable having this p.d.f. is said to have a normal distribution with mean  and variance 2, that is, a N(,2) distribution. A random variable Z having a N(0,1) distribution, called a standard normal distribution, has p.d.f. – z2 / 2 e f(z) = ——— for –< z <  2

  11. We let (z) = P(Z z), the distribution function of Z. Since f(z) is a symmetric function, it is easy to see that (– z) = 1 – (z). Tables Va and Vb in Appendix B of the text display a graph of f(z) and values of (z) and (– z). Important Theorems in the Text: If X is N(,2), then Z = (X– ) /  is N(0,1). Theorem 3.6-1 If X is N(,2), then V = [(X – ) / ]2 is 2(1). Theorem 3.6-2

  12. 2. The random variable X is N(10, 9). Use Theorem 3.6-1 to find each of the following: 6 – 10 X– 10 12 – 10 P( ——— < ——— < ———— ) = 3 3 3 P(6 < X < 12) = P(– 1.33 < Z < 0.67) = (0.67)– (– 1.33) = (0.67)– (1 – (1.33)) = 0.7486 – (1 – 0.9082) = 0.6568 X– 10 25 – 10 P( ——— > ———— ) = 3 3 P(X > 25) = P(Z > 5) = 1 – (5) = practically 0

  13. 2.-continued a constant c such that P(|X – 10| < c) = 0.95 X– 10 c P( ——— < — ) = 0.95 3 3 P(|X – 10| < c) = 0.95 P(|Z| < c/3) = 0.95 (c/3) –(– c/3) = 0.95 (c/3) –(1 – (c/3)) = 0.95 (c/3) = 0.975 c/3 = z0.025 = 1.960 c = 5.880

  14. 3. The random variable X is N(–7, 100). Find each of the following: X+ 7 0 + 7 P( ——— > —— ) = 10 10 P(X > 0) = P(Z > 0.7) = 1 – (0.7) = 0.2420 a constant c such that P(X > c) = 0.98 X+ 7 c +7 P( —— > —— ) = 0.98 10 10 P(X > c) = 0.98 P(Z > (c+7) / 10) = 0.98 1 – ((c+7) /10) = 0.98 ((c+7) /10) = 0.02 (c+7) /10 = z0.98 = –z0.02 = – 2.054 c = – 27.54

  15. 3.-continued X2+ 14X + 49 —————— 100 the distribution for the random variable Q = X2+ 14X + 49 —————— = 100 From Theorem 3.6-2, we know that Q = 2 X + 7 —— 10 must have a distribution. 2(1)

More Related