Class 12. Daniel B. Rowe, Ph.D. Department of Mathematics, Statistics, and Computer Science. Marquette University MATH 1700

Transcription

1 Class 12 Daniel B. Rowe, Ph.D. Department of Mathematics, Statistics, and Computer Science Copyright 2017 by D.B. Rowe 1

2 Agenda: Recap Chapter Lecture Chapter Problem Solving Session. 2

3 Recap Chapter

4 6.1 Normal Probability Distributions The mathematical formula for the normal distribution is (p 269): f( x) e 1 x 2 2 where e = π = μ = population mean σ = population std. deviation 2 f(x) x, 0 We will not use this formula. Figure from Johnson & Kuby,

5 6.1 Normal Probability Distributions How are areas found in math? f(x) Aside: Don t need to know. a b A b 1 e 2 a 1 x 2 f( x) This can not be done analytically and can only be done through numerical integration (Calculus). 2 dx Figure modified from Johnson & Kuby,

6 6.2 The Standard Normal Probability Distributions Example: Here is a normal distribution with μ = 5 and σ 2 = 4. f(x) μ= σ=2 x But we can t do calculus in this class. Someone had the idea to convert normal distribution to the standard normal. Subtract μ and divide this by σ for every value of x. z = (x- μ)/σ. Area between x 1 and x 2 is the same as area between z 1 and z 2. 6

7 6.2 The Standard Normal Probability Distributions x Example: Here is a normal distribution with μ = 5 and σ 2 = 4. z f(x) f(z) x 1 = 2.28 x 2 = 9.28 σ=2 x x 1 z1 2 z2 σ=1 z 1 = z 2 = 2.14 We find z 1 = and z 2 = 2.14? Do we agree with my z's? x 1 = μ= x 2 = z 1 = μ= z 2 = x z 7

8 Appendix B, Table 3, Page P(-1.36<z<2.14) = P(z<2.14) - P(z<-1.36) = = - z Red Area= z 8 z

9 6.2 The Standard Normal Probability Distributions Example: Here is a normal distribution with μ = 5 and σ 2 = 4. f(x) f(z) x Area between x 1 and x 2 is same as the area between z 1 and z 2. z 9

10 Questions? Homework: Chapter 6 # 7, 9, 13, 17, 19, 29, 31, 33, 41, 45, 47, 53, 61, 75, 95, 99 Read Chapter 7. 10

11 Lecture Chapter

12 6.3 Applications of Normal Distributions You may recall that in Chapter 2 we discussed a Standard Score, or z-score. It was discussed then using and s. Now, we will be using μ and σ. Standard score, or z-score: The position a particular value of x has relative to the mean, measured in standard deviations. z i x x - mean of x x i std. dev. of x (6.3) 12

13 6.3 Applications of Normal Distributions Assume that IQ scores x are normally distributed with a mean μ of 100 and a standard deviation σ of 16. μ Figures from Johnson & Kuby,

14 6.3 Applications of Normal Distributions Example: Assume that IQ scores x are normally distributed with a mean μ of 100 and a standard deviation σ of 16. If a person is picked at random, what is the probability that his or her IQ is between 100 and 115? i.e. P(100 x115)? μ μ Figures from Johnson & Kuby,

15 6.3 Applications of Normal Distributions IQ scores normally distributed μ=100 and σ=16. P(100 x 115) P( z z z ) z z 1 x x 1 x1 100 x z 2 x 2 Figures from Johnson & Kuby,

16 6.3 Applications of Normal Distributions IQ scores normally distributed μ=100 and σ=16. P(100 x 115) P(0 z 0.94) Now we can use the table. = - Figure from Johnson & Kuby,

17 6.3 Applications of Normal Distributions Now we can use the table. = - P(0 z 0.94) Figures from Johnson & Kuby,

18 6.4 Notation We can use the table in reverse. Before we had a z value then looked up the probability (area) less than z. Now we will have a probability (area), call it α, and want to know the z value, call it z(α), that has a probability (area) of α larger than it. α=p(z>z(α)) z Figure from Johnson & Kuby,

19 6.4 Notation Example: Let α=0.05. Let s find z(0.05). P(z>z(0.05))= =P(z>z(0.05)) Figure from Johnson & Kuby,

20 6.4 Notation Example: Same as finding P(z<z(0.05))= Figures from Johnson & Kuby,

21 6.5 Normal Approximation of the Binomial Distribution In Chapter 5 we discussed the binomial distribution n! x P( x) p (1 p) x!( n x)! x = # of heads when we flip a coin n times n=2 p=1/2 nx x P( x) x 0,..., n x

22 6.5 Normal Approximation of the Binomial Distribution If we flip the coin a large number of times n! x P( x) p (1 p) x!( n x)! x = # of heads when we flip a coin n times It gets tedious to find the n=14 probabilities! nx n=14 p=1/2 x 0,..., n Figure from Johnson & Kuby,

23 6.5 Normal Approximation of the Binomial Distribution It gets tedious to find the n=14 probabilities! n=14 p=1/2 So what we can do is use a histogram representation, Figures from Johnson & Kuby,

24 6.5 Normal Approximation of the Binomial Distribution So what we can do is use a histogram representation, n=14 p=1/2 Then approximate binomial probabilities with normal areas. Figures from Johnson & Kuby,

25 6.5 Normal Approximation of the Binomial Distribution Approximate binomial probabilities with normal areas. 2 Use a normal with np, np(1 p) n=14 p=1/2 (14)(.5) 7 2 (14)(.5)(1.5) 3.5 Figures from Johnson & Kuby,

26 6.5 Normal Approximation of the Binomial Distribution n=14, p=1/2 We then approximate binomial probabilities with normal areas. Px ( 4) from the binomial formula is approximately P(3.5 x4.5) from the normal with 2 7, 3.5 the ±.5 is called a continuity correction Figures from Johnson & Kuby,

27 6.5 Normal Approximation of the Binomial Distribution From the binomial formula 14! P(4) (.5) (1.5) 4!(14 4)! Px ( 4) From the Normal Distribution P(3.5 x4.5) z z 1 2 x1 x2 n=14, p=1/2 2 7,

28 Questions? Homework: Chapter 6 # 7, 9, 13, 17, 19, 29, 31, 33, 41, 45, 47, 53, 61, 75, 95, 99 Read Chapter 7. 32