There are two basic kinds of random variables continuous and discrete.
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1 Summary of Lectures 5 and 6 Random Variables The random variable is usually represented by an upper case letter, say X. A measured value of the random variable is denoted by the corresponding lower case letter; in this case x. There are two basic kinds of random variables continuous and discrete. Discrete Random Variables Probability Mass Function The probability mass function of a random variable X is a description of the probabilities associated with the possible values of X. P(X = x) = p(x) For any probability mass function, the sum of all probabilities over its entire sample space must equal one. Therefore, p(x) = The cumulative distribution function is the sum of all probabilities up to a given value x, F(x) = P(X x) = p(x) all X x So the probability that X lies between a and b is given by, Also, it must be noted p(x) 0 P(a < X b) = F(b) F(a) The mean value of any random variable X is also called the expected value X and is denoted by E(X) or μ. It can be seen that the sample is given by, x = xn x N The same can be transformed to compute population mean by,
2 E(X) = xp(x) Similarly, the variance of a random variable X is denoted by V(X) and σ 2. The square root of the variance is the standard deviation (σ). The standard deviation computed from a random sample is denoted by s. Transforming this as a population variance, Which can be shown to be the same as, s 2 = n x(x x ) 2 N σ 2 = p(x)(x μ) 2 V(X) = E(X 2 ) E(X) 2 Continuous random variables The probability distribution of a continuous random variable is described in terms of a probability density function, f(x). It must be noted f(x) does not give us probability. It gives us probability per unit value of x. That is why we call it probability density function. So the probability that X lies between a and b is given by, Also, ) f(x) 0 2) f(x)dx = b P(a < X < b) = f(x)dx a Cumulative distribution function Another way to define the probability distribution of a random variable is in terms of probability that X is less than or equal to x. With this definition, F(x) = P(X x) = x f(u)du
3 P(a < X < b) = F(b) F(a) Mean and Variance The mean value of any random variable X is also called the expected value X and is denoted by E(X) or μ. E(X) = μ = xf(x)dx Similarly, the variance of a random variable X is denoted by V(X) and σ 2. The square root of the variance is the standard deviation (σ). The standard deviation computed from a random sample is denoted by s. V(X) = E(X 2 ) E(X) 2 The p th percentile of a random variable X is the value xp such that p of the values of X are below it. That is, x p f(x)dx = p
4 E 243 Recitation 4 Spring 205 February 20, 205 Examples ) A lot contains ten good and three defective articles. Four articles are selected at random from the lot. Find the probability mass function for X, the number of defective articles among those selected. 2) A chemical supply company ships a certain solvent in 0-gallon drums. Let X represent the number of drums ordered by a randomly chosen customer. Assume X has the following probability mass function. x p(x) k a) Find the value of k b) Find the mean number of drums ordered c) Find the variance of the number of drums ordered d) Find the probability that a customer order less than 3 drums 3) Consider the following probability density function for the continuous random variable, X: f(x) = 6x 5 for 0 < x < f(x) = 0 otherwise. a) Find the probability that 0.5 < X <. b) Find the cumulative probability distribution function, F(x). c) Find the mean and standard deviation of X. d) Find the 30 th percentile value of X Practice Problems 4) A certain type of component is packaged in lots of four. Let X represent the number of properly functioning components in a randomly chosen lot. The probability mass function of X is given by, p(x) = cx for x =, 2, 3, or 4 = 0 for all other values of x a) Find the value of c b) Find the cumulative probability for X=2 c) Find the mean and variance of X 5) A box of articles contains four good articles and one defective article. To locate the defective article, one at a time is drawn from the box and tested (and not replaced). Let X denote the number of test in which the defective article is located. Determine the probability mass function of X.
5 3. No. of ways of selecting 4 articles from 3 = 4 C 3 No. of ways of selecting defective articles from 3 = x C No. of ways of selecting 4-x good articles from 0 = Using multiplicative rule, P(X = x) = 3 x C 0 4 x C 4 x 0 C 3 4 C for x = 0,, 2, 3 2. a) p(x) = Here, k = k = 0.8 = 0.2 b) E(X) = x p(x) = = = 2.3 c) V(X) = E(X 2 ) E(X) 2 E(X 2 ) = = = 7. V(X) = =.8 d) P(X < 3) = P(X = ) + P(X = 2) = = f(x) = 6x 5 0 < x < = 0 otherwise 0.5 a) P(0.5 < X < ) = 6 x 5 dx = [ 6 x6 ] = = 63 64
6 = x b) F(X) = 6 x 5 dx = [ 6 x6 ] x = x 6 c) E(X) = x 6 x 5 dx = [ 6 x7 0 7 ] 0 V(X) = x 2 6 x 5 dx ( ) 0 = 6 = [ 6 x8 ] 6 8 ( 0 7 )2 = 3 4 (6 7 )2 = 0.05 σ x = x d) 0.3 6x 5 dx = 0.3 = [ 6 x6 ] x = [x ] [0] = x = 0.3 x = a) p(x) = c + 2c + 3c + 4c = c = 0 = 0. b) P(X 2) = P(X = ) + P(X = 2) = = 0.3 ( 0.) + (2 0.) c) E(X) = x p(x) = = = 3 V(X) = E(X 2 ) E(X) 2
7 E(X 2 ) = = = 0 V(X) = = 5. P(X = ) = 5 P(X = 2) = = 5 P(X = 3) = = 5 P(X = x) = 5 x =,2,3,4,5 = 0 otherwise
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