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Probability Distributions

Probability Distributions. D E F I N I T I O N S Variable: A variable is an entity that can assume multiple values. Random Variable: A random variable is a variable that takes on different values based on chance. In other words, there are probabilities associated with the possible

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Probability Distributions

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  1. Probability Distributions

  2. D E F I N I T I O N S Variable: A variable is an entity that can assume multiple values. Random Variable: A random variable is a variable that takes on different values based on chance. In other words, there are probabilities associated with the possible values of the variable (i.e. a random variable has an associated probability distribution).

  3. D E F I N I T I O N S continued Discrete Random Variable: A discrete random variable is a random variable that can only assume certain discrete values. There may be a finite number of possible values, or the number of possible values may be countably infinite. These are characterized by a probability mass function. Continuous Random Variable: A continuous random variable is a random variable that can assume any of the values in an interval. The probabilities are associated with subintervals inside the interval over which the random variable is defined. For a continuous random variable, the probability of a point is zero. These are characterized by probability density functions.

  4. D E F I N I T I O N S continued Cumulative Distribution Function: A cumulative distribution function is a formula (or a table) that may be used to obtain the probability of obtaining a value of the associated random variable that is less than or equal to a specific value of the random variable. The probability values for such a function range from 0 to 1.

  5. D I S C R E T E Let X be a discrete random variable. Then, 1) P(X)  0 for all values which are defined for X. 2)  P(X) = 1 x C O N T I N U O U S Let X be a continuous random variable. Then, 1) P(a  X  b)  0 for any subinterval, [a,b], on which X is defined and a < b. P(X=a) = 0. 2)  P(X) dx = 1 x P R O P E R T I E S O F P R O B A B I L I T Y D I S T R I B U T I O N S

  6. MATHEMATICAL EXPECTATION The “Concept” Let X be a discrete random variable. The expected value of the random variable is the "average" of the values of X that would occur over many trials.

  7. Point: A discrete random variable, x, has an associated probability function, P(x). The expected value of the random variable is also called the mean of the distribution and is denoted by . Definition: Let x be some discrete random variable with the probability function, P(x), then the expected value of x, denoted E(x), is defined: E(x) = xP(x) x

  8. An Example Let x = the sum of the top faces when two dice are rolled. Let Di = the value on the top face of die i. The sample space is given by: D1: 1 1 1 1 1 1 2 2 2 2 2 2 3 3 3 3 3 3 4 4 4 4 4 4 5 5 5 5 5 5 6 6 6 6 6 6 D2: 1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6 If you compute all 36 sums, you will find there are: 1 sum of 2; 2 sums of 3; 3 sums of 4; 4 sums of 5; 5 sums of 6;1 sum of 12; 2 sums of 11; 3 sums of 10; 4 sums of 9; 5 sums of 8; and 6 sums of 7.

  9. Probability Distribution Therefore, the probability distribution of x is: x P(x) 2 1/36 3 2/36 4 3/36 5 4/36 6 5/36 7 6/36 8 5/36 9 4/36 10 3/36 11 2/36 12 1/36

  10. Expected Value of x, the mean E(x) = xP(x) x = 2(1/36) + 3(2/36) + 4(3/36) + 5(4/36) + 6(5/36) + 7(6/36) + 8(5/36) + 9(4/36) + 10(3/36) + 11(2/36) + 12(1/36) = 2 + 6 + 12 + 20 + 30 + 42 + 40 + 36 + 30 + 22 + 12 36 = 252/36 = 7. Therefore, = 7.

  11. Variance of the distribution The variance of a distribution (2 ) is defined as the expected value of the squares of the deviations from the mean. Like before, the standard deviation () is the square root of the variance. Definition: Let x be some discrete random variable with the probability function, P(x), then the variance of x, denoted 2 = E[(x-)2], is defined: E[(x-)2] =  (x-)2P(x) x

  12. For our example 2 = (2-7)2(1/36) + (3-7)2(2/36) + (4-7)2(3/36) + (5-7)2(4/36) + (6-7)2(5/36) + (7-7)2(6/36) + (8-7)2(5/36) + (9-7)2(4/36) + (10-7)2(3/36) + (11-7)2(2/36) + (12-7)2(1/36) = 25 + 32 + 27 + 16 + 5 + 0 + 5 + 16 + 27 + 32 + 25 36 = 210/36 = 5.83. Therefore,  2.415229457

  13. Q: Why would anyone want to go to this amount of trouble to calculate the mean and the standard deviation of a probability distribution? A: To approximate the solution of probability problems. True, it is a “pain” to calculate the mean and the variance of a distribution strictly from the definition, but usually it is not this much trouble, because there are “special cases” that are fairly simple calculations.

  14. Special Discrete Probability Distributions • Hypergeometric • Binomial • Poisson

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