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## Chapter 7

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**Chapter 7**Random Variables Warm Up page 402 #10,11 HW: Read and Take Notes Do #’s 14,16,17,18,20,25,{22,26}**Random Variable**• A random variable is a variable whose value is a numerical outcome of a random phenomenon. • We denote random variables by capital letters near the end of the alphabet, such as X or Y.**Introduction**• The outcome of an experiment need not be a number, for example, the outcome when a coin is tossed can be 'heads' or 'tails‘, however, we often want to represent outcomes as numbers. • A random variable is a function that associates a unique numerical value with every outcome of an experiment. • The value of the random variable will vary from trial to trial as the experiment is repeated.**7.1 Discrete and Continuous Random Variables**• There are two types of random variables - discrete and continuous. • A random variable has either an associated probability distribution (discrete random variable) or probability density function (continuous random variable).**Examples**• Ex 1: A coin is tossed ten times. The random variable X is the number of tails that are noted. X can only take the values 0, 1, ..., 10, so X is a discrete random variable. • Ex 2:A light bulb is burned until it burns out. The random variable Y is its lifetime in hours. Y can take any positive real value, so Y is a continuous random variable.**Discrete**• A discrete random variable is one which may take on only a countable number of distinct values such as 0, 1, 2, 3, 4, ... • Discrete random variables are usually (but not necessarily) counts. • If a random variable can take only a finite number of distinct values, then it must be discrete. • Examples of discrete random variables include the number of children in a family, the Friday night attendance at a cinema, the number of patients in a doctor's office, the number of defective light bulbs in a box of ten.**Probability Distribution**• A table of values of X (or Y) and the probabilities that represent them is a probability distribution. • They must satisfy 2 requirements: • 1. Each is a number between 0 and 1. • 2. The sum of the probabilities adds to 1.**Probability histograms can be used to display probability**distributions. • The top picture is the probability histogram for random digits 1-9. • The bottom pictures is the probability histogram for Benford’s Law.**Problem 7.1 p395Roll of the die**a. P(rolling a # <3) = 1/6 +1/6 = 2/6 or 1/3 b.Averages in order: .33, .36 ,.39 ,.24 ,.33 ; average of the 5 = .33**Problem 7.2 p396 Three Children**• GGG, GGB, GBG, GBB, BBB, BBG, BGB, BGG; P(of any one) = 1/8 • P(X=2) = 3/8, see table below • See below**Continuous Random Variables**• A continuous random variable is one which takes an infinite number of possible values. • Continuous random variables are usually measurements. • Examples include height, weight, the amount of sugar in an orange, the time required to run a mile.**A continuous random variable is not defined at specific**values. Instead, it is defined over an interval of values, and is represented by the area under a curve (in statistics known as a density curve, in advanced mathematics, this is known as an integral). • The probability of observing any single value is equal to 0, since the number of values which may be assumed by the random variable is infinite.**Suppose a random variable X takes all values over an**interval of real numbers. • Then the probability that X is in the set of outcomes A, P(A), is defined to be the area above A and under a curve.**The curve, which represents a function p(x), must satisfy**the following: • 1: The curve has no negative values (p(x) > 0 for all x) • 2: The total area under the curve is equal to 1. • A curve meeting these requirements is known as a……..wait for it……… • density curve!**The Uniform Distribution**• A random number generator acting over an interval of numbers (a,b) has a continuous distribution. • Since any interval of numbers of equal width has an equal probability of being observed, the curve describing the distribution is a rectangle, with constant height across the interval and 0 height elsewhere. • Since the area under the curve must be equal to 1, the length of the interval determines the height of the curve.**The previous graphs plotted the density curves for random**number generated over the intervals (4 to 5) (top left), (2 to 6) (top right), (5 to 5.5) (lower left), and (3 to 5) (lower right). • The distributions corresponding to these curves are known as uniform distributions.**Normal Density Curves**• Remember that our notation for Normal distributions is N(μ,σ),where μ is the mean and σ is the standard deviation. • Then Z = X- μ/ σ , and Z is a normal random variable with distribution N(0,1). • From this we can standardize any normal distribution plot. Then, using the z-score we can find probabilities of events from the original distribution.**A class of 500 students have a class averages with a normal**distribution. • The next four histograms represent the same grades, but their intervals differ. You could make the width of the intervals narrower and narrower, until finally the smooth curve can be drawn on the graph. This curve is the density function.**A) P(0 ≤ X ≤ 0.4)**= 0.4 B) P(0.4 ≤ X ≤ 1) = 0.6 C) P(0.3 ≤ X ≤ 0.5) = 0.2 D) P(0.3 < X < 0.5) = 0.2 E) P(0.226 ≤ X ≤ 0.713) = .487 F) p. 399 – All continuous probabilities assign 0 to every individual outcome. ≤ vs. < has no change in area under curve. #6 p.401**A) P(X ≤ 0.49)**= 0.49 B) P( X ≥ 0.27) = 1- 0.27 = .73 C) P(0.27 ≤ X ≤ 1.27) = 1.27 - .27 = .73 D) P(0.1 ≤ X ≤ 0.2) OR P(0.8 ≤ X ≤ 0.9) 0.2 – 0.1 = .1 OR 0.9 – 0.8 = .1 So add to get 0.2 E) P( not 0.3 ≤ X ≤ 0.8) Therefore X ≤ 0.3 and ≥ 0.8 = 0.3 + 0.2 = 0.5 F) P( X = 0.5) = 0 #7 p. 402**#8 p. 402**• Did they just say the mean was .40 and the standard deviation was 0.023? • Sounds like I am to use Z = X – μ / σ • At least means no less than OR more than 45%, SO, Z = .45 - .40 / 0.023 = 2.17. • GO to TABLE A….. Z = 2.17 then p = .9850 • SINCE they want more than 45%, we do 1 - .9850 = .0150 is the probability of the sample that believes violence is school’s most serious problem.**b) Less than 35% c) between**• B) Less than 35% means Z = .35 - .40 / 0.023 = - 2.17, so Table A give a p of 0.015. • C) SINCE they want between 35% and 45%, we do 1-.015 -.015 = .97, which is the probability of the sample who think violence is school’ most serious problem.