Prepare for your exams

Study with the several resources on Docsity

Earn points to download

Earn points by helping other students or get them with a premium plan

Guidelines and tips

Prepare for your exams

Study with the several resources on Docsity

Earn points to download

Earn points by helping other students or get them with a premium plan

Community

Ask the community

Ask the community for help and clear up your study doubts

University Rankings

Discover the best universities in your country according to Docsity users

Free resources

Our save-the-student-ebooks!

Download our free guides on studying techniques, anxiety management strategies, and thesis advice from Docsity tutors

The concept of probability, focusing on probability models, rules, discrete and continuous sample spaces, and random variables. It explains how to calculate probabilities for events and intervals in both discrete and continuous sample spaces, and discusses the importance of probability distributions, particularly the normal distribution.

Typology: Study notes

2009/2010

1 / 25

Download Understanding Probability Theory: Sample Spaces, Discrete & Continuous Probabilities, Rand and more Study notes Statistics in PDF only on Docsity! Introducing probability BPS chapter 10 © 2006 W. H. Freeman and Company Objectives (BPS chapter 10) Introducing probability The idea of probability Probability models Probability rules Discrete sample space Continuous sample space Random variables Personal probability The trials are independent only when you put the coin back each time. It is called sampling with replacement. Two events are independent if the probability that one event occurs on any given trial of an experiment is not affected or changed by the occurrence of the other event. When are trials not independent? Imagine that these coins were spread out so that half were heads up and half were tails up. Close your eyes and pick one. The probability of it being heads is 0.5. However, if you don’t put it back in the pile, the probability of picking up another coin and having it be heads is now less than 0.5. Probability models mathematically describe the outcome of random processes. They consist of two parts: 1) S = Sample Space: This is a set, or list, of all possible outcomes of a random process. An event is a subset of the sample space. 2) A probability for each possible event in the sample space S. Probability models Example: Probability Model for a Coin Toss S = {Head, Tail} Probability of heads = 0.5 Probability of tails = 0.5 Important: It’s the question that determines the sample space. Sample space A. A basketball player shoots three free throws. What are the possible sequences of hits (H) and misses (M)? H H H - HHH M … M M - HHM H - HMH M - HMM … S = {HHH, HHM, HMH, HMM, MHH, MHM, MMH, MMM } Note: 8 elements, 23 B. A basketball player shoots three free throws. What is the number of baskets made? S = {0, 1, 2, 3} C. A nutrition researcher feeds a new diet to a young male white rat. What are the possible outcomes of weight gain (in grams)? S = [0, ∞] = (all numbers ≥ 0) Note: Discrete data contrast with continuous data that can take on any one of an infinite number of possible values over an interval. Dice are good examples of finite sample spaces. Finite means that there is a limited number of outcomes. Throwing 1 die: S = {1, 2, 3, 4, 5, 6}, and the probability of each event = 1/6. Discrete sample space Discrete sample spaces deal with data that can take on only certain values. These values are often integers or whole numbers. In some situations, we define an event as a combination of outcomes. In that case, the probabilities need to be calculated from our knowledge of the probabilities of the simpler events. Example: You toss two dice. What is the probability of the outcomes summing to five? There are 36 possible outcomes in S, all equally likely (given fair dice). Thus, the probability of any one of them is 1/36. P(the roll of two dice sums to 5) = P(1,4) + P(2,3) + P(3,2) + P(4,1) = 4 * 1/36 = 1/9 = 0.111 This is S: {(1,1), (1,2), (1,3), ……etc.} The gambling industry relies on probability distributions to calculate the odds of winning. The rewards are then fixed precisely so that, on average, players lose and the house wins. The industry is very tough on so-called “cheaters” because their probability to win exceeds that of the house. Remember that it is a business, and therefore it has to be profitable. Probability distribution for a continuous random variable Because the probability of drawing one individual at random depends on the frequency of this type of individual in the population, the probability is also the shaded area under the curve. The shaded area under the density curve shows the proportion, or percent, of individuals in the population with values of X between x1 and x2. % individuals with X such that x1 < X < x2 P(y < 0.5 or y > 0.8) = P(y < 0.5) + P(y > 0.8) = 1 − P(0.5 < y < 0.8) = 0.7 The probability of a single event is zero: P(y = 1) = (1 − 1)*1 = 0 Intervals The probability of a single event is meaningless for a continuous sample space. Only intervals can have a non-zero probability, represented by the area under the density curve for that interval. Height = 1 y The probability of an interval is the same whether boundary values are included or excluded: P(0 ≤ y ≤ 0.5) = (0.5 − 0)*1 = 0.5 P(0 < y < 0.5) = (0.5 − 0)*1 = 0.5 P(0 ≤ y < 0.5) = (0.5 − 0)*1 = 0.5 We generate two random numbers between 0 and 1 and take Y to be their sum. Y can take any value between 0 and 2. The density curve for Y is: 0 1 2 Height = 1. We know this because the base = 2, and the area under the curve has to equal 1 by definition. The area of a this triangle is ½ (base*height). What is the probability that Y is < 1? What is the probability that Y < 0.5? 0 1 20.5 0.25 0.5 0.125 0.125 1.5 Y Previously, we wanted to calculate the proportion of individuals in the population with a given characteristic. N (µ, ) = N (64.5, 2.5) Distribution of women’s heights ≈ Example: What's the proportion of women with a height between 57" and 72"? That’s within ± 3 standard deviations of the mean , thus that proportion is roughly 99.7%. Since about 99.7% of all women have heights between 57" and 72", the chance of picking one woman at random with a height in that range is also about 99.7%. What is the probability, if we pick one woman at random, that her height will be some value X? For instance, between 68” and 70” P(68 < X < 70)? Because the woman is selected at random, X is a random variable. As before, we calculate the z- scores for 68 and 70. For x = 68", For x = 70", z (x ) 4.1 5.2 )5.6468( z z (70 64.5) 2.5 2.2 The area under the curve for the interval [68" to 70"] is 0.9861 − 0.9192 = 0.0669. Thus, the probability that a randomly chosen woman falls into this range is 6.69%. P(68 < X < 70) = 6.69% 0.9861 0.9192 N(µ, ) = N(64.5, 2.5) Inverse problem: Your favorite chocolate bar is dark chocolate with whole hazelnuts. The weight on the wrapping indicates 8 oz. Whole hazelnuts vary in weight, so how can they guarantee you 8 oz. of your favorite treat? You are a bit skeptical... To avoid customer complaints and lawsuits, the manufacturer makes sure that 98% of all chocolate bars weight 8 oz. or more. The manufacturing process is roughly normal and has a known variability = 0.2 oz. How should they calibrate the machines to produce bars with a mean such that P(x < 8 oz.) = 2%? = ?x = 8 oz. Lowest 2% = 0.2 oz.