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.13 3. Continuous Random Variables p.16 4. The Normal/Gaussian Distribution p.18

BSTT523: Pagano & Gavreau, Chapter 7 2

1. 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)

1 0.416 2 0.330 3 0.158 4 0.058 5 0.021 6 0.009 7 0.004

8+ 0.004 Total 1.000

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(𝑥𝑖) ≤ 1 for all 𝑥𝑖 Non-negative ii. ∑ 𝑓(𝑥𝑖){𝑎𝑙𝑙 𝑥𝑖} = 1 Exhaustive iii. 𝑃�𝑋 = 𝑥𝑖 ∪ 𝑋 = 𝑥𝑗� = 𝑓(𝑥𝑖) + 𝑓(𝑥𝑗) Additive Cumulative Distribution Function (CDF): 𝐹(𝑥) = 𝑃(𝑋 ≤ 𝑥) = ∑ 𝑃(𝑋 = 𝑥𝑖)𝑥𝑖≤𝑥 = ∑ 𝑓(𝑥𝑖)𝑥𝑖≤𝑥 Note: i. PDF 𝑓(𝑥𝑖) = 𝑃(𝑋 = 𝑥𝑖) = 𝐹(𝑥𝑖) − 𝐹(𝑥𝑖−1) ii. 𝑃(𝑋 < 𝑥𝑖) = 𝐹(𝑥𝑖−1) iii. 𝑃(𝑎 < 𝑋 ≤ 𝑏) = 𝐹(𝑏) − 𝐹(𝑎)

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)

1 0.416 0.416 2 0.330 0.746 3 0.158 0.904 4 0.058 0.962 5 0.021 0.983 6 0.009 0.992 7 0.004 0.996

8+ 0.004 1.000 Q1. Prob. that a child picked at random was mother’s 1st or 2nd 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?

BSTT523: Pagano & Gavreau, Chapter 7 5

2. Some Discrete Distributions Bernoulli Distribution

Bernoulli Variable: Binary Variable

𝑋 = �1, 𝑠𝑢𝑐𝑐𝑒𝑠𝑠0, 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

𝑓(𝑥) = �𝑝 𝑖𝑓 𝑋 = 1𝑞 𝑖𝑓 𝑋 = 0

= 𝑝𝑥𝑞1−𝑥, 𝑥 = 0,1

= 𝑝𝑥(1 − 𝑝)1−𝑥, 𝑥 = 0,1 The Bernoulli distribution has one parameter = 𝑝

BSTT523: Pagano & Gavreau, Chapter 7 6

If X follows a Bernoulli distribution, then Mean: 𝜇 = 𝐸(𝑋) = 𝑝 Variance: 𝜎2 = 𝑉𝑎𝑟(𝑋) = 𝑝𝑞 = 𝑝(1 − 𝑝) Examples of Bernoulli variables:

Ex. 1: flip a coin 𝑋 = �1,𝐻𝑒𝑎𝑑𝑠0, 𝑇𝑎𝑖𝑙𝑠

Ex. 2: roll a die, interested in 3’s 𝑋 = �1,𝑑𝑖𝑒 𝑓𝑎𝑙𝑙𝑠 𝑜𝑛 30, 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒

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 𝑞 = 1 − 𝑝 𝑋~𝐵𝐼𝑁(𝑛,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 = 𝑝2𝑞3

Q2. Other results with 2 successes out of 5? Number Sequence

1 11000 2 10100 3 10010 4 10001 5 01100 6 01010 7 01001 8 00110 9 00101

10 00011 There are 10 ways to get 2 successes out of 5 The probability of each sequence is 𝑝2𝑞3 P(Sequence 1 or 2 or … or 10) = 10𝑝2𝑞3

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!

2!(5−2)! = 5·4

2·1 = 10

Back to binomial distribution question: 𝑋~𝐵𝐼𝑁(5,𝑝); 𝑓(2)=? 𝑓(5)=? Ans: 𝑓(2) = 5C2p2q3 = 10p2q3 𝑓(5) = 5C5p5q0 = 5!

5!0!p5q0 = 1p51= p5

BSTT523: Pagano & Gavreau, Chapter 7 9

Binomial PDF 𝑓(𝑥) : 𝑋~𝐵𝐼𝑁(𝑛,𝑝) P(x successes in n Bernoulli trials) 𝑓(𝑥) = nCx pxqn-x , x = 0, 1, …, n = 0, otherwise

Number of Successes 𝑥

Probability 𝑓(𝑥)

0 nC0 qn 1 nC1 pqn-1

. . . . . . 𝑥 nCx pxqn-x

. . . . . . n-1 nCn-1 pn-1q

𝑛 nCn pn Total 1

Important Binomial distribution features: Mean: 𝜇 = 𝐸(𝑋) = 𝑛𝑝 Variance: 𝜎2 = 𝑉𝑎𝑟(𝑋) = 𝑛𝑝𝑞

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!4!6!

(.00707)(.1281) = .1903

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

BSTT523: Pagano & Gavreau, Chapter 7 11

Q3. Among the 10 individuals chosen, what is the expected number of smokers? 𝐸(𝑋) = 𝑛𝑝 = 10 ∙ 29 = 2.9 Variance and SD: 𝑉𝑎𝑟(𝑋) = 𝑛𝑝𝑞 = 10 ∙ (. 29) ∙ (. 71) = 2.059 𝑆𝐷 = �𝑛𝑝𝑞 = √2.059 = 1.43 Note: Using Table A.1, what if 𝑝>0.5? 𝑓(𝑥,𝑛,𝑝) = nCx px(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 𝑋𝐶 as “success”. 𝑃(𝑋 ≤ 𝑥), 𝑋~𝐵𝐼𝑁(𝑛,𝑝) = 𝑃(𝑋𝐶 ≥ 𝑛 − 𝑥), 𝑋𝐶~𝐵𝐼𝑁(𝑛, 1 − 𝑝)

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) = 1 − .7393 = .2607 Q3. Among the sample of 12, what is the expected number of people who think childhood obesity is ‘serious’?

𝐸(𝑋) = 𝑛𝑝 = 12 ∙ .55 = 6.6 Q4. What is the variance of the number who think childhood obesity is ‘serious’?

𝑉𝑎𝑟(𝑋) = 𝑛𝑝𝑞 = 12 ∙ (. 55) ∙ (. 45) = 2.97

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−𝜆𝜆𝑥

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.

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−12123

3! = .00177

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−12120

0!+ 𝑒−12121

1!+ 𝑒−12122

2!�

= 1 − .00052225 = .9994775

BSTT523: Pagano & Gavreau, Chapter 7 15

Solution 2: Table A.2 (P.A-6): POISSON PDF 𝑃(𝑋 ≥ 3) = 1 − 𝐹(2) = 1 − (. 0000 + .0001 + .0004) = .9995 Solution 3: SAS: POISSON(λ, x) – CDF PDF(‘POISSON’, x, λ) – PDF CDF(‘POISSON’, x, λ) – CDF

BSTT523: Pagano & Gavreau, Chapter 7 16

3. 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, or ∫ 𝑓(𝑡)𝑑𝑡𝑏

𝑎 .

BSTT523: Pagano & Gavreau, Chapter 7 17

Cumulative density function (CDF) of X: 𝐹(𝑥) = ∫ 𝑓(𝑡)𝑑𝑡𝑥

−∞ Note: Total area under 𝑓(𝑥) = 1, i.e., ∫ 𝑓(𝑡)𝑑𝑡+∞

−∞ = 1 and 𝑓(𝑥) = 𝑑

𝑑𝑥𝐹(𝑥) = 𝐹′(𝑥)

BSTT523: Pagano & Gavreau, Chapter 7 18

4. A special continuous distribution: the Normal or Gaussian Normal PDF:

𝑓(𝑥) = 1√2𝜋𝜎

𝑒−(𝑥−𝜇)2

2𝜎2 , −∞ < 𝑥 < +∞

𝑋~𝑁(𝜇,𝜎2) Characteristics:

Distribution is symmetric around 𝜇

Mean = Median = Mode = 𝜇

Total area under the curve = 1, i.e., ∫ 1√2𝜋𝜎

𝑒−(𝑥−𝜇)2

2𝜎2+∞−∞ = 1

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

𝐸(𝑋) = 𝜇 location parameter

𝑉𝑎𝑟(𝑋) = 𝜎2 scale parameter Standard Normal Distribution:

𝑍~𝑁(0,1) has PDF 𝜙(𝑧) = 1√2𝜋

𝑒−𝑧2

2 , −∞ < 𝑧 < +∞

BSTT523: Pagano & Gavreau, Chapter 7 19

Table A.3: Standard Normal Upper Tail Cumulative Probabilities

𝑃(𝑍 ≥ 𝑧0) = 1 −Φ(𝑧0) , 𝑧0 ≥ 0

where Φ(𝑧) = ∫ 𝜙(𝑡)𝑑𝑡𝑧−∞ is the CDF for 𝑍

for 𝑧0 < 0, Φ(𝑧0) = 𝑃(𝑍 ≤ 𝑧0) = 𝑃(𝑍 ≥ (−𝑧0)) , 𝑧0 ≤ 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) = 0.159 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

BSTT523: Pagano & Gavreau, Chapter 7 20

How is the 𝑁(0,1) distribution related to 𝑁(𝜇,𝜎2) ?

If 𝑿~𝑵(𝝁,𝝈𝟐) and 𝒁 = (𝑿−𝝁)𝝈

, then 𝒁~𝑵(𝟎,𝟏) . Example 7.7 Systolic Blood Pressure (SBP) (p.181 P&G) 𝑋 = SBP for 18-74 year old males; 𝑋~𝑁(𝜇,𝜎2) 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) + 129 = 167.8 What proportion of men in this population have SBP>150 mmHg?

Solution: 𝑃(𝑋 > 150) = 𝑃 �(𝑥−𝜇)𝜎

> (150−129)19.8

�

= 𝑃(𝑍 > 1.06) = 0.145 ⇒ 14.5%

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, 1192)

𝑃(292 ≤ 𝑋 ≤ 649) =? Solution 1: 𝑃(292 ≤ 𝑋 ≤ 649) = 𝑃 �292−491

119≤ 𝑋−𝜇

𝜎≤ 649−491

119�

= 𝑃(−1.67 ≤ 𝑍 ≤ 1.33) = 1 − 𝑃(𝑍 ≤ −1.67) − 𝑃(𝑍 ≥ 1.33) = 1 − .047 − .092 = .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