Probability and Binomial Distributions in Random Sampling - Prof. Edith Seier, Study notes of Statistics

Download Probability and Binomial Distributions in Random Sampling - Prof. Edith Seier and more Study notes Statistics in PDF only on Docsity! ROBE Math 1530 – 005, 009, & 014 Fall 2003 Lecture Notes for The Binomial Distributions (part of ch. 17) Consider a city with thousands of cars in it. Suppose that 30% of the cars in the city are red. Now suppose you pick one car at random from the city. What is the probability that it is a red car? P(red) = P(not red) = . Suppose I pick two cars at random. The cars are independent of one another. Separately: P(first car is red) = P(second car is red) = . For two or more independent events, probabilities multiply. Together: P(both cars are red) = P(first is red and second is red) = P(first is red) × P(second is red) = × = P(both are not red) = × = P(first is red and second is not) = × = P(first is not and second is red) = × = P(one red in the two-car sample) = ?? Now - the idea ‘one red car out of the two chosen’ is an Event with two possible outcomes in it - “first is red and second is not” or “first is not and second is red.” Probabilities must be calculated accordingly. P(one red in the two-car sample) = P(“ first is red and second is not “or “ first is not and second is red “) Because these two are disjoint, we can just add .. = P(first is red and second is not) + P(first is not and second is red) = + = Observe: “pick two cars at random” consists of 3 possibilities: ‘both red’ ‘both not’ and ‘one red, one not’ their probabilities add up + + . just like they should = . Now, suppose I pick three cars at random. Third verse, same as the first, a little bit longer and a whole lot worse: P(all three are red) = P(first is red) × P(second is red) × P(third is red) = × × = . P(none are red) = × × = . The event “one is red and other two are not” consists of R-N-N, N-R-N, and N-N-R P(R-N-N) = 0.3 x 0.7 x 0.7 = 0.147 P(N-R-N) = 0.7 x 0.3 x 0.7 = . P(N-N-R) = 0.3 x 0.7 x 0.7 = . These are all three possible arrangements of one red car and two not-red cars. Each one’s individual probability is 0.3 × 0.7 × 0.7, so.. P(one is red and two are not) = P(RNN or NRN or NNR) = 0.147 + + = . = probability of the event “one car is red and two are not” Enough. Picking cars at random from a large city, looking only at red/not red are Bernoulli trials. Bernoulli trials with a fixed number n of observations makes the Binomial setting: 1. There is a fixed number n of independent trials. For example ‘suppose you pick 3 cars...’ 2. There are only two outcomes for each observation, ‘success’ or ‘failure.’ For example, ‘red’ car or ‘not red’ car 3. The probability p of success is the same for each observation. In a large population of cars, pulling out one (red) car does not really change the probability that the next car is red (or not). ± p. 320 ‘Binomial distribution’ (Binomial probability model) Let X be the count of successes out of a fixed number n of Bernoulli trials with probability p of success on any one trial. Then X can be 0, 1, 2, 3, ... , n (X is a variable, it can only take on these certain values!) We say X has the binomial probability model with parameters n and p. Suppose x is one of 0, 1, 2, ... n. P(X = x) can be computed from n, p, and x. Ex Say your random phenomenon is “pick a car at random out of a large city,” you are going to pick 3 cars, and p = 0.3 for a red car. p. 320 again Binomial mean and standard deviation: If X has the binomial distribution with parameters n and p, then X is a variable, one that can take on several different values, some of them more likely than others. The mean, variance, and standard deviation of X’s distribution are µ = n × p σ 2 = n × p × (1 - p) so σ =  ppn  1 This ONLY works for the binomial distribution Ex Suppose X is binomial with n = 10 and p = 0.3 (you pick 10 cars and see how many are red) µ = 10 (0.3) = 3 On average, you will find 3 red cars. σ  7.03.010  = 2.1 = 1.449 with standard deviation 1.449 p. 321 The normal approximation to binomial distributions: Sometimes even the binomial probability formula is not really practical. Suppose you have a 50-question, multiple-choice test with 4 choices for each question. You are going to guess on every single question. ‘Guessing every single time’ means that your answers are independent of each other. The number of guesses (or trials!! ) is fixed at n = 50. On any one question, there are only two outcomes - either you get it right or wrong. ‘4 choices’ means the probability that you guess right is p = 1/4 = 0.25 (for ea. question) Aha! Test-guessing is a binomial setting!! P(you get 30 questions right) = 50 C 30 (0.25) 30 (0.75) 20 = 0.000000129 not terribly likely What is the probability that you pass? Say ‘pass’ means you get 30 or more right. Then P(you pass) = P(X ≥ 30) = P(X = 30 or X = 31 or X = 32 or ... or X = 50) = P(X = 30) + P(X = 31) + P(X = 32) + ... + P(X = 50) = 0.000000129 + 0.000000027 + 0.000000005 + .... eew = twenty calculations to do, with calculator or software because few tables go to n = 50, and a messy source of likely arithmetic error Note p. 322, the success/failure condition: When n × p and n × (1 - p) are both ≥ 10, the binomial distribution is approximately Normal!! (mean and standard deviation from n and p ) Does our 50-guesses test fit this? n × p = 50 (0.25) = 12.5 10, OK and n × (1 - p) = 50(0.75) = 37.5 10, also OK So, the distribution of X = ‘the number of correct answers on a 50-guessed-questions test’ is approximately Normal with... mean µ = n p = 50(0.25) = 12.5 and standard deviation σ =  ppn  1 =  75.025.050  = √9.375 = 3.0618 now - P(X ≥ 30) = do normal calculations! z =     0618.3 5.1230  x = P( X > 30 ) = P( z > 5.72 ) = … Let’s try a different one. You have been doing well in a course, and you figure out that if you get 21 or more out of 50 on the Final Exam, you will pass, and if you get 20 or fewer, you will fail. What is the probability of failing the course if you guess every question?