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Random Variables Numerical Quantities whose values are determine by the outcome of a random experiment Discrete Random Variables Discrete Random Variable: A random variable usually assuming an integer value. • a discrete random variable assumes values that are isolated points along the real line. That is neighbouring values are not “possible values” for a discrete random variable Note: Usually associated with counting • The number of times a head occurs in 10 tosses of a coin • The number of auto accidents occurring on a weekend • The size of a family Continuous Random Variables Continuous Random Variable: A quantitative random variable that can vary over a continuum • A continuous random variable can assume any value along a line interval, including every possible value between any two points on the line Note: Usually associated with a measurement • Blood Pressure • Weight gain • Height Probability Distributions of a Discrete Random Variable Probability Distribution & Function Probability Distribution: A mathematical description of how probabilities are distributed with each of the possible values of a random variable. Notes: The probability distribution allows one to determine probabilities of events related to the values of a random variable. The probability distribution may be presented in the form of a table, chart, formula. Probability Function: A rule that assigns probabilities to the values of the random variable Comments: Every probability function must satisfy: 1. The probability assigned to each value of the random variable must be between 0 and 1, inclusive: 0 p( x) 1 2. The sum of the probabilities assigned to all the values of the random variable must equal 1: p( x) 1 3. Pa X b x b p( x) x a p(a) p(a 1) p(b) Mean and Variance of a Discrete Probability Distribution • Describe the center and spread of a probability distribution • The mean (denoted by greek letter m (mu)), measures the centre of the distribution. • The variance (s2) and the standard deviation (s) measure the spread of the distribution. s is the greek letter for s. Mean of a Discrete Random Variable • The mean, m, of a discrete random variable x is found by multiplying each possible value of x by its own probability and then adding all the products together: m xpx x x1 px1 x2 px2 xk pxk Notes: The mean is a weighted average of the values of X. The mean is the long-run average value of the random variable. The mean is centre of gravity of the probability distribution of the random variable 0.3 0.2 0.1 1 2 3 4 5 6 7 8 m 9 10 11 Variance and Standard Deviation Variance of a Discrete Random Variable: Variance, s2, of a discrete random variable x is found by multiplying each possible value of the squared deviation from the mean, (x - m)2, by its own probability and then adding all the products together: s 2 x - m 2 px 2 x 2 x px - xpx x x x 2 px - m 2 x Standard Deviation of a Discrete Random Variable: The positive square root of the variance: s s2 Example The number of individuals, X, on base when a home run is hit ranges in value from 0 to 3. x 0 1 2 3 Total p (x ) xp(x) 0.429 0.000 0.286 0.286 0.214 0.429 0.071 0.214 1.000 0.929 p(x) xp(x) x 2 0 1 4 9 2 x p(x) 0.000 0.286 0.857 0.643 1.786 2 x p( x) • Computing the mean: m xpx 0.929 x Note: • 0.929 is the long-run average value of the random variable • 0.929 is the centre of gravity value of the probability distribution of the random variable • Computing the variance: s 2 x - m 2 px 2 x 2 x px - xpx x x 1.786 - .929 0.923 2 • Computing the standard deviation: s s2 0.923 0.961 The Binomial distribution 1. We have an experiment with two outcomes – Success(S) and Failure(F). 2. Let p denote the probability of S (Success). 3. In this case q=1-p denotes the probability of Failure(F). 4. This experiment is repeated n times independently. 5. X denote the number of successes occuring in the n repititions. The possible values of X are 0, 1, 2, 3, 4, … , (n – 2), (n – 1), n and p(x) for any of the above values of x is given by: n x n x n- x n- x px p 1 - p p q x x X is said to have the Binomial distribution with parameters n and p. Summary: X is said to have the Binomial distribution with parameters n and p. 1. X is the number of successes occurring in the n repetitions of a Success-Failure Experiment. 2. The probability of success is p. 3. The probability function n x n- x px p 1 - p x Example: 1. A coin is tossed n = 5 times. X is the number of heads occurring in the 5 tosses of the coin. In this case p = ½ and 5 1 x 1 5- x 5 1 5 5 1 px 2 2 2 32 x x x x 0 1 2 3 4 5 p(x) 1 32 5 32 10 32 10 32 5 32 1 32 Note: 5 5! x x ! 5 - x ! 5 5! 1 0 0! 5 - 0 ! 5 5! 5! 5 1 1! 5 - 1! 4! 5 5! 5 4 10 2 2!3! 2 1 5 5! 5 4 10 3 3!2! 2 1 5 5! 5 4 4!1! 5 5! 1 5 0!5! 0.4 p (x ) 0.3 0.2 0.1 0.0 1 2 3 4 number of heads 5 6 Computing the summary parameters for the distribution – m, s2, s x 0 1 2 3 4 5 Total p (x ) 0.03125 0.15625 0.31250 0.31250 0.15625 0.03125 1.000 p(x) xp(x) 0.000 0.156 0.625 0.938 0.625 0.156 2.500 xp(x) x 2 0 1 4 9 16 25 2 x p(x) 0.000 0.156 1.250 2.813 2.500 0.781 7.500 2 x p( x) • Computing the mean: m xpx 2.5 x • Computing the variance: s 2 x - m 2 px 2 x 2 x px - xpx x x 7.5 - 2.5 1.25 2 • Computing the standard deviation: s s2 1.25 1.118 Example: • A surgeon performs a difficult operation n = 10 times. • X is the number of times that the operation is a success. • The success rate for the operation is 80%. In this case p = 0.80 and • X has a Binomial distribution with n = 10 and p = 0.80. 10 x 10- x px 0.80 0.20 x Computing p(x) for x = 1, 2, 3, … , 10 x p (x ) x p (x ) 0 0.0000 6 0.0881 1 0.0000 7 0.2013 2 0.0001 8 0.3020 3 0.0008 9 0.2684 4 0.0055 10 0.1074 5 0.0264 The Graph 0.4 p (x ) 0.3 0.2 0.1 0 1 2 3 4 5 6 7 Number of successes, x 8 9 10 Computing the summary parameters for the distribution – m, s2, s x 0 1 2 3 4 5 6 7 8 9 10 Total p (x ) 0.0000 0.0000 0.0001 0.0008 0.0055 0.0264 0.0881 0.2013 0.3020 0.2684 0.1074 1.000 xp(x) 0.000 0.000 0.000 0.002 0.022 0.132 0.528 1.409 2.416 2.416 1.074 8.000 xp(x) x2 x 2 p(x) 0 1 4 9 16 25 36 49 64 81 100 0.000 0.000 0.000 0.007 0.088 0.661 3.171 9.865 19.327 21.743 10.737 65.600 2 x p( x) • Computing the mean: m xpx 8.0 x • Computing the variance: s 2 x - m 2 px 2 x 2 x px - xpx x x 65.6 - 8.0 1.60 2 • Computing the standard deviation: s s2 1.25 1.118 Notes The value of many binomial probabilities are found in Tables posted on the Stats 244 site. The value that is tabulated for n = 1, 2, 3, …,20; 25 and various values of p is: c n x 10- x PX c p 1 - p px x 0 x x 0 c p0 p1 p2 pc Hence pc Tabled value for c - Tabled value for c - 1 The other table, tabulates p(x). Thus when using this table you will have to sum up the values Example Suppose n = 8 and p = 0.70 and we want to compute P[X = 5] = p(5) n =5 c 0.05 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 0.95 0 0.663 0.430 0.168 0.058 0.017 0.004 0.001 0.000 0.000 0.000 0.000 1 0.943 0.813 0.503 0.255 0.106 0.035 0.009 0.001 0.000 0.000 0.000 2 0.994 0.962 0.797 0.552 0.315 0.145 0.050 0.011 0.001 0.000 0.000 3 1.000 0.995 0.944 0.806 0.594 0.363 0.174 0.058 0.010 0.000 0.000 4 1.000 1.000 0.990 0.942 0.826 0.637 0.406 0.194 0.056 0.005 0.000 5 1.000 1.000 0.999 0.989 0.950 0.855 0.685 0.448 0.203 0.038 0.006 6 1.000 1.000 1.000 0.999 0.991 0.965 0.894 0.745 0.497 0.187 0.057 7 1.000 1.000 1.000 1.000 0.999 0.996 0.983 0.942 0.832 0.570 0.337 8 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 Table value for n = 8, p = 0.70 and c =5 is 0.448 = P[X 5] P[X = 5] = p(5) = P[X 5] - P[X 4] = 0.448 – 0.194 = .254 We can also compute Binomial probabilities using Excel The function =BINOMDIST(x, n, p, FALSE) will compute p(x). The function =BINOMDIST(c, n, p, TRUE) c n x 10- x will compute PX c p 1 - p px x 0 x x 0 p0 p1 p2 pc c Mean,Variance & Standard Deviation • The mean, variance and standard deviation of the binomial distribution can be found by using the following three formulas: 1. m np 2. s npq np1 - p 2 3. s npq np1 - p Example Example: Find the mean and standard deviation of the binomial distribution when n = 20 and p = 0.75 Solutions: 1) n = 20, p = 0.75, q = 1 - 0.75 = 0.25 m np (20)(0.75) 15 s npq (20)(0.75)(0.25) 3.75 1936 . 2) These values can also be calculated using the probability function: 20 p ( x ) (0.75) x (0.25)20 x for x 0, 1, 2, ... , 20 x Table of probabilities x 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Total p (x ) 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0002 0.0008 0.0030 0.0099 0.0271 0.0609 0.1124 0.1686 0.2023 0.1897 0.1339 0.0669 0.0211 0.0032 1.000 xp(x) 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.001 0.006 0.027 0.099 0.298 0.731 1.461 2.361 3.035 3.035 2.276 1.205 0.402 0.063 15.000 x2 x 2 p(x) 0 1 4 9 16 25 36 49 64 81 100 121 144 169 196 225 256 289 324 361 400 0.000 0.000 0.000 0.000 0.000 0.000 0.001 0.008 0.048 0.244 0.992 3.274 8.768 18.997 33.047 45.525 48.559 38.696 21.691 7.632 1.268 228.750 • Computing the mean: m xpx 15.0 x • Computing the variance: s 2 x - m 2 px 2 x 2 x px - xpx x x 228.75 - 15.0 3.75 2 • Computing the standard deviation: s s2 3.75 1.936 Histogram m 0.3 s p(x) 0.2 0.1 0 2 4 6 8 10 12 -0.1 no. of successes 14 16 18 20 Probability Distributions of Continuous Random Variables Probability Density Function The probability distribution of a continuous random variable is describe by probability density curve f(x). Notes: The Total Area under the probability density curve is 1. The Area under the probability density curve is from a to b is P[a < X < b]. Normal Probability Distributions P(a x b) a m b x Normal Probability Distributions • The normal probability distribution is the most important distribution in all of statistics • Many continuous random variables have normal or approximately normal distributions The Normal Probability Distribution Points of Inflection s m - 3s m - 2s m - s m m s m 2s m 3s Main characteristics of the Normal Distribution • Bell Shaped, symmetric • Points of inflection on the bell shaped curve are at m – s and m + s. That is one standard deviation from the mean • Area under the bell shaped curve between m – s and m + s is approximately 2/3. • Area under the bell shaped curve between m – 2s and m + 2s is approximately 95%. There are many Normal distributions depending on by m and s Normal m = 100, s =20 0.03 Normal m = 100, s = 40 Normal m = 140, s =20 f(x) 0.02 0.01 0 0 50 100 x 150 200 The Standard Normal Distribution m = 0, s = 1 0.4 0.3 0.2 0.1 0 -3 -2 -1 0 1 2 3 • There are infinitely many normal probability distributions (differing in m and s) • Area under the Normal distribution with mean m and standard deviation s can be converted to area under the standard normal distribution • If X has a Normal distribution with mean m and standard deviation s than has a standard normal distribution z X -m s has a standard normal distribution. • z is called the standard score (z-score) of X. Example: Suppose a man aged 40-45 is selected at random from a population. • X is the Blood Pressure of the man. • X is random variable. • Assume that X has a Normal distribution with mean m =180 and a standard deviation s = 15. The probability density of X is plotted in the graph below. • Suppose that we are interested in the probability that X between 170 and 210. X -m X - 180 z s 15 170 - m 170 - 180 a -0.667 s 15 210 - m 210 - 180 b 2.000 s 15 Let Hence P170 X 210 P- .667 z 2.000 P170 X 210 P- .667 z 2.000 P170 X 210 P- .667 z 2.000 Standard Normal Distribution Properties: • The total area under the normal curve is equal to 1 • The distribution is bell-shaped and symmetric; it extends indefinitely in both directions, approaching but never touching the horizontal axis • The distribution has a mean of 0 and a standard deviation of 1 • The mean divides the area in half, 0.50 on each side • Nearly all the area is between z = -3.00 and z = 3.00 Notes: Normal Table, Posted on Stats 244 web site, lists the probabilities below a specific value of z Probabilities of other intervals are found using the table entries, addition, subtraction, and the properties above Table, Posted on stats 244 web site z 0 • The table contains the area under the standard normal curve between - and a specific value of z Example Find the area under the standard normal curve between z = - and z = 1.45 0.9265 0 • A portion of Table 3: z 0.00 0.01 0.02 0.03 0.04 1.45 0.05 .. . 1.4 .. . P( z 1.45) 0.9265 0.9265 z 0.06 Example Find the area to the left of -0.98; P(z < -0.98) Area asked for -0.98 0 P ( z < - 0.98) 0.1635 Example Find the area under the normal curve to the right of z = 1.45; P(z > 1.45) Area asked for 0.9265 0 145 . P( z 1.45) 1.0000 - 0.9265 0.0735 z Example Find the area to the between z = 0 and of z = 1.45; P(0 < z < 1.45) 0 145 . P( z < 1.45) 0.9265 - 0.5000 0.4265 • Area between two points = differences in two tabled areas z Notes Use the fact that the area above zero and the area below zero is 0.5000 the area above zero is 0.5000 When finding normal distribution probabilities, a sketch is always helpful Example Example: Find the area between the mean (z = 0) and z = -1.26 Area asked for -126 . 0 P( -1.26 < z < 0) 0.5000 - 0.1038 0.3962 z Example: Find the area between z = -2.30 and z = 1.80 Required Area .-2.30 0 . 1.80 P( -1.26 < z < 1.80) 0.9641 - 0.0107 0.9534 Example: Find the area between z = -1.40 and z = -0.50 Area asked for -1.40 - 0.500 P( -1.40 < z < -0.50) 0.3085 - 0.0808 0.2277 Computing Areas under the general Normal Distributions (mean m, standard deviation s) Approach: 1. Convert the random variable, X, to its z-score. z X -m s 2. Convert the limits on random variable, X, to their z-scores. 3. Convert area under the distribution of X to area under the standard normal distribution. b-m a - m Pa X b P z s s Example Example: A bottling machine is adjusted to fill bottles with a mean of 32.0 oz of soda and standard deviation of 0.02. Assume the amount of fill is normally distributed and a bottle is selected at random: 1) Find the probability the bottle contains between 32.00 oz and 32.025 oz 2) Find the probability the bottle contains more than 31.97 oz Solutions part 1) When x 32.00 ; When x 32.025; 32.00 - m 32.00 - 32.0 0.00 z s 0.02 z 32.025 - m 32.025 - 32.0 1.25 s 0.02 Graphical Illustration: Area asked for 32.0 0 32.025 1.25 x z 32.0 - 32.0 X - 32.0 32.025 - 32.0 < < P ( 32.0 < X < 32.025) P 0.02 0.02 0.02 P ( 0 < z < 1.25) 0. 3944 Example, Part 2) 31.97 - 150 . 32.0 0 x z x - 32.0 3197 . - 32.0 P( z -150) P( x 3197 . ) P . 0.02 0.02 1.0000 - 0.0668 0.9332

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