Chapter 7: Theoretical Probability Distributions Variable - Measured/Categorized characteristic

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1 BSTT523: Pagano & Gavreau, Chapter 7 1 Chapter 7: Theoretical Probability Distributions Variable - Measured/Categorized characteristic Random Variable (R.V.) X Assumes values (x) by chance Discrete R.V. Can assume a finite number of values Continuous R.V. Can assume any value within an interval 1. Discrete Random Variables p.2 2. Some Discrete Distributions Bernoulli distribution p.5 Binomial distribution p.7 Poisson distribution p Continuous Random Variables p The Normal/Gaussian Distribution p.18

2 BSTT523: Pagano & Gavreau, Chapter Discrete Random Variables: Definition: The probability distribution of a discrete r.v. X: A table, graph, formula, or other device that specifies all possible values of X and their respective probabilities Example 7.1 Table form: Birth order of children in the U.S. x: Birth Order P(X = x) Total 1.000

3 BSTT523: Pagano & Gavreau, Chapter 7 3 Discrete Probability Density Function (Discrete PDF): f(x) = P(X = x) Properties of the discrete PDF: i. 0 f(x i ) 1 for all x i Non-negative ii. f(x i ) {all x i } = 1 Exhaustive iii. P X = x i X = x j = f(x i ) + f(x j ) Additive Cumulative Distribution Function (CDF): F(x) = P(X x) = x i x P(X = x i ) = x i x f(x i ) Note: i. PDF f(x i ) = P(X = x i ) = F(x i ) F(x i 1 ) ii. P(X < x i ) = F(x i 1 ) iii. P(a < X b) = F(b) F(a)

4 BSTT523: Pagano & Gavreau, Chapter 7 4 Example 7.1: Birth Order x: Birth Order PDF f(x) = P(X = x) CDF F(x) = P(X x) Q1. Prob. that a child picked at random was mother s 1 st or 2 nd child? Q2. Prob. that a child picked at random was of birth order fewer than 4? Q3. Prob. that a child picked at random was of order 5 or more? Q4. Prob. that a child picked at random was of order between 3 and 5?

5 BSTT523: Pagano & Gavreau, Chapter Some Discrete Distributions Bernoulli Distribution Bernoulli Variable: Binary Variable 1, success X = 0, failure Bernoulli Trial: One performance of experiment with 0/1 outcome Denote p = P(X = 1) q = P(X = 0) = 1 p The PDF of the Bernoulli distribution is p if X = 1 f(x) = q if X = 0 = p x q 1 x, x = 0,1 = p x (1 p) 1 x, x = 0,1 The Bernoulli distribution has one parameter = p

6 BSTT523: Pagano & Gavreau, Chapter 7 6 If X follows a Bernoulli distribution, then Mean: Variance: μ = E(X) = p σ 2 = Var(X) = pq = p(1 p) Examples of Bernoulli variables: 1, Heads Ex. 1: flip a coin X = 0, Tails Ex. 2: roll a die, interested in 3 s X = 1, die falls on 3 0, otherwise

7 BSTT523: Pagano & Gavreau, Chapter 7 7 Binomial Distribution Perform n independent Bernoulli trials. X = number of successes (1 s) p = probability of success in each trial X~BIN(n, p) q = 1 p Q: What is the PDF f(x), x = 0,1,, n of X~BIN(n, p)? i.e., what is the probability of x successes in n Bernoulli trials? Q1. 5 Bernoulli trials, X~BIN(5, p) P(result is 10010)=? Solution: pqqpq = p 2 q 3 Q2. Other results with 2 successes out of 5? Number Sequence There are 10 ways to get 2 successes out of 5 The probability of each sequence is p 2 q 3 P(Sequence 1 or 2 or or 10) = 10p 2 q 3

8 BSTT523: Pagano & Gavreau, Chapter 7 8 Definition: A Combination of n subjects taken x at a time = Number of unordered subsets of x ( n choose x ) = ncx = n! x!(n x)! where x! = x(x-1)(x-2) (2)(1) and define 0!=1 Example: 5 choose 2 how many subsets of 2 out of 5? 5C2 = 5! = 5 4 = 10 2!(5 2)! 2 1 Back to binomial distribution question: X~BIN(5, p); f(2)=? f(5)=? Ans: f(2) = 5C2p 2 q 3 = 10p 2 q 3 f(5) = 5C5p 5 q 0 = 5! 5!0! p5 q 0 = 1p 5 1= p 5

9 BSTT523: Pagano & Gavreau, Chapter 7 9 Binomial PDF f(x) : X~BIN(n, p) P(x successes in n Bernoulli trials) f(x) = ncx p x q n-x, x = 0, 1,, n = 0, otherwise Number of Successes x Probability f(x) 0 nc0 q n 1 nc1 pq n x ncx p x q n-x n-1 ncn-1 p n-1 q n ncn p n Total 1 Important Binomial distribution features: Mean: Variance: μ = E(X) = np σ 2 = Var(X) = npq

10 BSTT523: Pagano & Gavreau, Chapter 7 10 Example 7.2 Smoking in the U.S.: 29% are smokers, or p =.29 Select a random sample of size 10. Q1. What is P(4 smokers in the sample)? X = number of smokers out of n = 10 X~BIN(10,.29) Solution 1. f(4) = 10C4 (0.29) 4 (0.71) 6 = 10! (.00707)(.1281) = !6! Solution 2. Table A.1 (P.A1): Binomial PDF p = 0.05 to 0.5, n = 2 to 20 f(4) : n = 10, p.30 f(4).2001 Solution 3. SAS: PROBBNML(p, n, m) CDF PDF( BINOMIAL, x, p, n) PDF CDF( BINOMIAL, x, p, n) - CDF Q2. P(6 or more smokers in the sample)=? P(X 6) = 1 F(5) = 1 (. 9596) =.0404

11 BSTT523: Pagano & Gavreau, Chapter 7 11 Q3. Among the 10 individuals chosen, what is the expected number of smokers? E(X) = np = = 2.9 Variance and SD: Var(X) = npq = 10 (. 29) (. 71) = SD = npq = = 1.43 Note: Using Table A.1, what if p>0.5? f(x, n, p) = ncx p x (1-p) n-x f(n x, n, 1 p) = ncn-x (1-p) n-x (p) x ncx= n! x!(n x)! = n! (n x)!x! = ncn-x f(x, n, p) = f(n x, n, 1 p) i.e. if p>0.5 then treat X C as success. P(X x), X~BIN(n, p) = P(X C n x), X C ~BIN(n, 1 p)

12 BSTT523: Pagano & Gavreau, Chapter 7 12 Example 7.3 What do you think about the problem of childhood obesity? Poll in 2003: 55% of residents think it is serious. Randomly select n=12 residents. Q1. P(8 people think it is serious )? X~BIN(12,.55) f(8) =.1700 Same as P(4 out of 12 do not think serious ); X~BIN(12,.45) f(4) =.1700 Q2. P(5 or fewer think serious ) =? P(X 5 n = 12, p =.55) = P(X 7 n = 12, p =.45) = 1 P(X 6 n = 12, p =.45) = =.2607 Q3. Among the sample of 12, what is the expected number of people who think childhood obesity is serious? E(X) = np = = 6.6 Q4. What is the variance of the number who think childhood obesity is serious? Var(X) = npq = 12 (. 55) (. 45) = 2.97

13 BSTT523: Pagano & Gavreau, Chapter 7 13 Poisson Distribution X = number of event occurrences in a given interval of time/space/volume etc. i.e. Count Data Probability that x events will occur: f(x) = e λ λ x x!, x=0, 1, 2,... X~POI(λ) Important Poisson features: Mean: E(X) = λ Variance: Var(X) = λ When λ is small, the distribution is right-skewed; when λ increases (λ 10), the distribution becomes symmetric.

14 BSTT523: Pagano & Gavreau, Chapter 7 14 Example 7.4 Allergic reaction to anesthesia (Laake and Rottingen) Occurrences of reaction Poisson, about 12 incidents per year expected Q1. In the next year, what is the probability of seeing 3 incidents? Solution: X~POI(12) f(3) = e ! = Q2. What is the probability that at least 3 will have a reaction in the next year? Solution 1: P(X 3) = 1 P(X 2) = 1 F(2) = 1 {f(0) + f(1) + f(2)} = 1 e ! + e ! + e ! = =

15 BSTT523: Pagano & Gavreau, Chapter 7 15 Solution 2: Table A.2 (P.A-6): POISSON PDF P(X 3) = 1 F(2) = 1 ( ) =.9995 Solution 3: SAS: POISSON(λ, x) CDF PDF( POISSON, x, λ) PDF CDF( POISSON, x, λ) CDF

16 BSTT523: Pagano & Gavreau, Chapter Continuous Random Variables Continuous X can assume any value within its range. Within any interval, there are theoretically an infinite number of values. Subareas of histograms represent frequency of occurrence of values within class intervals Total frequency of values between a and b: add all subareas for intervals a through b. If width of class intervals is very small, then connecting midpoints (creating a frequency polygon) creates a smooth curve. If probability is shown on the y-axis and we have a smooth curve: probability density function (PDF) f(x) P(a < X b) = total area under f(x) between a and b, b a or f(t)dt.

17 BSTT523: Pagano & Gavreau, Chapter 7 17 Cumulative density function (CDF) of X: F(x) = x f(t)dt Note: Total area under f(x) = 1, i.e., + f(t)dt = 1 and f(x) = d F(x) = F (x) dx

18 BSTT523: Pagano & Gavreau, Chapter A special continuous distribution: the Normal or Gaussian Normal PDF: f(x) = 1 (x μ) 2 2πσ e 2σ 2, < x < + X~N(μ, σ 2 ) Characteristics: Distribution is symmetric around μ Mean = Median = Mode = μ Total area under the curve = 1, i.e., (x μ) πσ e 2σ 2 Area under the curve between σ and +σ.68 Area under the curve between 2σ and +2σ.95 Area under the curve between 3σ and +3σ.997 = 1 E(X) = μ Var(X) = σ 2 location parameter scale parameter Standard Normal Distribution: Z~N(0,1) has PDF φ(z) = 1 z 2 2π e 2, < z < +

19 BSTT523: Pagano & Gavreau, Chapter 7 19 Table A.3: Standard Normal Upper Tail Cumulative Probabilities P(Z z 0 ) = 1 Φ(z 0 ), z 0 0 where Φ(z) = z φ(t)dt is the CDF for Z for z 0 < 0, Φ(z 0 ) = P(Z z 0 ) = P(Z ( z 0 )), z 0 0 Example 7.5 Given a variable that follows the standard normal distribution, i.e. Z~N(0,1), what is P(z 1) and P(z 1)? Solution: by Table A.3, P(z 1)=0.159 and P(z 1) = P(z 1) = Example 7.6 Randomly pick a value z from the standard normal distribution. P(z has a value between -2 and +2) =? Solution: Note that for a continuous distribution P(X = x) = 0. P( 2 z +2) = P( 2 < z < +2) = 1 P(z 2) P(z 2) = 1 2 P(z 2) = 1 2 (. 023) = 0.954

20 BSTT523: Pagano & Gavreau, Chapter 7 20 How is the N(0,1) distribution related to N(μ, σ 2 )? If X~N(μ, σ 2 ) and Z = (X μ) σ, then Z~N(0, 1). Example 7.7 Systolic Blood Pressure (SBP) (p.181 P&G) X = SBP for year old males; X~N(μ, σ 2 ) with μ=129 mm Hg and σ=19.8 mm Hg. Find x which is the cutoff for the upper 2.5% of the SBP distribution; i.e. find x such that P(X > x) =.025. Solution: By Table A.3 we know that P(Z 1.96) =.025. (x μ) σ = 1.96 (x 129) 19.8 = 1.96 x = (1.96)(19.8) = What proportion of men in this population have SBP>150 mmhg? Solution: P(X > 150) = P (x μ) σ > ( ) 19.8 = P(Z > 1.06) = %

21 BSTT523: Pagano & Gavreau, Chapter 7 21 Example 7.8 Breath study (Diskin et al.) X = Ammonia concentration in parts per billion (ppb) μ=491 ppb, σ = 119 ppb; i.e. X~N(491, ) P(292 X 649) =? Solution 1: P(292 X 649) = P X μ σ = P( 1.67 Z 1.33) = 1 P(Z 1.67) P(Z 1.33) = =.861 Solution 2: SAS: ProbNorm(x) N(0,1) CDF PDF( NORMAL, x) N(0,1) PDF PDF( NORMAL, x, μ, σ) N(μ, σ) PDF CDF( NORMAL, x) N(0,1) CDF CDF( NORMAL, x, μ, σ) N(μ, σ) CDF

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