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Probability distributions Chapter 4

Probability distributions Chapter 4. Prof. Felix Apfaltrer fapfaltrer@bmcc.cuny.edu Office:N518 Phone: 212-220 74 21 Office hours: Tue, Thu 10:30am-12 pm.

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Probability distributions Chapter 4

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  1. Probability distributions Chapter 4 Prof. Felix Apfaltrer fapfaltrer@bmcc.cuny.edu Office:N518 Phone: 212-220 74 21 Office hours: Tue, Thu 10:30am-12 pm

  2. A random variable is a variable (typically represented by x) that has a single numerical value, determined by chance, for each outcome of a procedure. A probability distribution is a graph, table, or a formula that gives the probability for each value of the random variable. Gender of children: A study consists of randomly selecting 14 newborn babies and counting the number of girls in the sample. If we assume that having a boy or a girl is equally likely, and let x = number of girls among the 14 babies then x is a random variable because its value depends on chance. The possible values are x =0,1,2,3,…,11,12,13,14. A probability distribution is shown to the left. Random variables and distributions

  3. A discreterandom variable has either a finite or countable number of values. Countable means it might be infinite, but you can still “count” them (there are gaps between them). A continuousrandom variable has infinitely many values without gaps between them (like interval subsets of the real numbers). Examples: Discrete random variables: Number of eggs a hen lays per day. cannot lay 2.3 eggs one day! R.v.: # not known for sure in advance! Number of people attending the Columbus Day Parade. Discrete r.v.: counting the number of people. Random: we do not know in advance exactly how many are going. (but we might have an estimate) The sum of the faces when we roll two dice. The points in hand of Black Jack. The average number of eggs per hen per day in a farm with 10 hens. Discrete and Continuous Random Variables (r.v.’s) • Continuous random variables: • Amount of milk a cow produces a day. • Continuous r.v.: She might yield 1.345 gallons, or 1.34512 (no gaps in measurement). • The humidity at a given day. • Continuous r.v.: Percentage of humidity can be 75.34%. • The daily closing value of the Dow Jones Industrial Average index. • The daily ocean temperature at a marine laboratory investigating whales.

  4. Probability histogram Probability histogram • Very similar to relative frequency histogram • Instead of percent (relative frequency) probability is shown. • The values 0, 1, 2, …, 13, 14, are at the center of the rectangles -> base = 1 • area = height*base = height

  5. ∑P(x) = 1 where x assumes all possible values. 0 ≤ P(x) ≤ 1 for every individual value of x. Discussion: x takes all possible values, so it represents all options in the sample space For table ‘girls’, sum is 0.999, almost 1 except for rounding errors. All P(x) between 0 and 1 because they are probabilities! Requirements of Probability Distributions Example: Does the table represent a probability distribution? • All values between 0 and 1. Good! • ∑P(x) = 0.2+0.3+0.4+0.5 = 1.4 . Uups! • is not 1. • Therefore, it is not a probability distribution. Does the functionP(x) = x/9 represent a probability distribution? • P(2) =2/9, P(3) =3/9, P(4) =4/9, • ∑P(x) = 2/9 + 3/9 + 4/9 = (2+3+4)/9 = 9/9 = 1 • It is 1. • Therefore, the function does representa probability distribution. P(x) = x/9 for x = 2,3, & 4

  6.  = ∑ x•P(x) mean  2 = ∑ (x –  )2•P(x) variance = ∑ [ x 2 •P(x) ]–  2 variance (alternative formula)  = √ ∑ [ x 2 •P(x) ]–  2 standard deviation Rationale: Mean, Variance and Standard Deviation for Distributions

  7. Round off at 1 more decimal than data! Minimum usual value  – 2 Maximum usual value  + 2 Example: In previous calculation,  = 7,  =1.9. Minimum usual value:  – 2 = 7 – 2(1.9) = 3.2 Maximum usual value:  + 2 = 7 + 2(1.9) = 10.8 For the group of 14 babies, the usual values for the number of girls fall between 3.2 and 10.8. Mean, Variance for Distributions (round-off and unusual values) Rare event rule: If, under a given assumption, the probability of an event is extremely low, we conclude that the assumption is most likely incorrect. With probabilities: • x successes among n trials are unusually high if P(x or more) <0.05 • x successes among n trials are unusually low if P(x or less) <0.05 Example (Gender Selection): Getting 13 or more girls. P(13 or more girls) =P(13)+P(14) = 0.001+0.000 = 0.001 unusually high.

  8. The expected value of a discrete random variable is denoted by E, and it represents the average value of the outcomes. E = E[X] = ∑ {x•P(x) } Example (NJ pick 3 game): Bet $ 0.50 and select a 3 digit number between 000 and 999. If you get the number, you collect $275. Your net gain is then $274.50. Suppose that you bet $0.50 on the number 007. What is your expected value of gain or loss? Expected Value A: Each outcome is equally likely. P(win) = 1/1000 = 0.001 P(loss) = 999/1000 = 0.999 E[X] = ∑ x•P(x) = ∑ x•P(x) =274.50 • 0.001 + (-0.50) • 0.999 win loss = 0.2745 - 0.4995 = - 0.225 On average you will be loosing 22.5 cents every time you play.

  9. The Bernoulli probability distribution results from a procedure such that: there is one trial, like one flip of a coin there are only two outcomes (heads/tails, 0/1, red/white, success/failure) Examples: Tossing one coin (or bean) 1 trial outcomes: heads or tails Birth of one child: 1 trial Outcomes: boy or girl Tossing one die, win if it’s 6, loose 1-5 1 trial outcomes: win or loose Suppose you pay $1 to play and get $3 back if ‘6’comes out. Weather tomorrow 1 trial (day) Outcomes: rain or shine Bernoulli Distribution Probabilities: • P(X=heads)=0.5 • P(X=tails)=0.5 • P(girl)=0.513 = p success probability • P(boy)=0.487 = q = (1– p) failure prob • X=“number of girls” in one birth: 0 or1 • = 0P(0)+1P(1) = 0 q + 1p = p  2= 0 2P(0)+12P(1) – p 2 =p – p 2 =p(1 – p) = pq • P( win) =1/6 = p , P(loose)=5/6 = q • X=“number of wins” in one toss: 0 or 1  = 0P(0)+1P(1) = p = 1/6,  2=pq= 5/36 • Expectation: E[X] =3•1/6 + (-1)•6/6 = – 3/6 On average you will be loosing 50 cents per play

  10. A procedure has a binomial probability distribution if: each trial must have all outcomes in 2 categories the procedure has a fixed number of trials the trials are independent the probabilities must remain constant for each trial Notation for binomial probability distributions: 2 categories: S success (p prob. of success) F failure (q prob. of failure) Probabilities: P(S) = p P(F) = q =1– p n fixed number of trials x :: X = x X denotes the random variable, x denotes number of successes in n trials p probability of success (success is arbitrary, can be good or not) q probability of failure P(x) = P( X = x )probability of getting exactly x successes among n trials P( X ≤ x ) probability of getting x or less successes among n trials B( n , p ) binomial distribution with n trials and probability of success p Binomial Distributions Note:B(n,p) = sum of n independent Bernoulli distributions with probability of success p X = Y1 + Y2 + …+ Yn  X =  Y1 +  Y2 +…+  Y n= p + p +…+ p = np  2X =  2Y1+ 2Y2 +…+ 2Y n= pq + pq +…+ pq = npq

  11. Remember: Poll and test samples usually done without replacement -> dependent If sample small enough (< 5% of population), then it is safe to assume independence (even though there is no independence) Multiple choice answers: (answered at random, options: a,b,c,d,e, 4 questions) P(3 answers correct) Binomially distributed? Number of trials fixed n = 4. Trials independent. (answers do not depend on previous ones). 2 outcomes: right, or wrong. One answer correct, p=1/5=0.2; q = 0.8. YES! Binomial Distributions: Examples Use binomial formula

  12. Use table A-1: Hence, P(3) = 0.0256 Question: What is the probability that at least 3 answers are correct? ‘at least 3 answers correct’ = {X≥3} = {X=3 or X= 4} P(X ≥ 3 ) = P(X =3 ) + P(X = 4 ) = 0.0256 + 0.0016 = 0.0272 Binomial Distributions: Examples Continued HW: READ BOX, p. 186 p.186 s. 4-3 #3,4,7,8,9, 21-24,33,37*,40* Mean, variance and expectation:  X = np = 4 ( 0.2) = 0.8  2X = npq = 4 ( 0.2) (0.8) = 0.64 ->  X = 0.8 Suppose that someone you pay $1000 if the person that answers at random won’t answer 3 or more answers correctly, and that you receive $100 otherwise. What is your expected loss/gain? E[ X] = -1000 (0.0272) + 100 ( 1- 0.0272) = - 27.2 + 77.28 = 49.92

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