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ENGR 323 BHW #4 Tien Tai Problem 3-15 Many manufacturers have quality control programs that include inspection of incoming materials for defects. Suppose a computer manufacturer receives computer boards in lots of five. Two boards are selected from each lot for inspection. We can represent possible outcomes of the selection process by pairs. For example, the pair (1,2) represents the selection of boards 1 and 2 for inspection. a. List 10 different possible outcomes In part a, we are asked to find out all outcomes. So I list all the possible combinations below. (1,2) (1,3) (1,4) (1,5) (2,3) (2,4) (2,5) (3,4) (3,5) (4,5) Figure 1. Total Outcomes b. Suppose that board 1 and 2 are only defective boards in a lot of five. Two boards are to be chosen at random. Define X to be the number of defective boards observed among those inspected. Find the probability distribution for X. In part b, we are asked to find the pmf of the function, and by definition from probability textbook, we know that the probability mass function (pmf) of a discrete random variable is defined for every number y by p(y) = P(X = x) = P(all s ∈ζ: Y(s) = y). * For every possible value x of X, P(X=x) specifies the probability of observing that value when the experiment is performed. So, Let X = No. of defective boards The pmf is constructed below p ( x ) = 0 .3 x = 0 0 .6 x = 1 0 .1 x = 2 0 Otherwise 1 The result from the pmf is given from the result of part a. Example. Probability of none of the boards are defective p(0)=0.3 The sum of blue highlighted = No. of defective boards picked for examination Probability of Getting Defected Boards After we calculate pmf, then we graph it. The graph is referred to figure 2. 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0 0.5 1 1.5 2 2.5 No. of Defected Boards Figure 2 A pmf Graph of Problem 3.15 c. Let F(x) denote the cdf of X. First determine F(0) = P(X≤0), F(1), F(2) then F(x) for all other x. We can now calculate cdf from the pmf we calculated from part b. By definition on p.98 in the probability textbook, we know that the cumulative distribution function (cdf) F(y) of a discrete rv variable Y with pmf p(x) is defined for every number y by F ( x ) = P ( X ≤ x) = ∑ p( y ) \ x: x≤ y For any number y, F(y) is the probability that the observed value of Y will be at most y 2 Therefore, we can construct the cumulative distribution function from the pmf we just calculated, and the cdf is listed on following page. x < 0 1 < x ≤ 0 2 < x ≤ 1 x ≥ 2 0 0 .3 F ( x ) = 0 .9 1 After we know the value of cdf, we can construct a graph of cdf (see figure 3) Probability of Getting Defected Boards 1.5 1 0.5 0 -1 0 1 2 -0.5 -1 No. of Defected Boards Figure 3 Graph of cdf Problem 3.15 Compute E(X), V(X) Beth also asked me to find the expected value E(X) and sample variance V(X) of the function d) To calculate E(X), we need to use the equation that’s listed on p.104 3 3 E (X ) = µ x = ∑ x ⋅ p(x) = (0 ⋅ 0.3 ) + (1 ⋅ 0.6 ) + (2 ⋅ 0.1 ) = 0.6 + 0.2 = 0.8 x ∈D The expected value of defective boards = 0.8 boards e) To calculate V(X), we need to use the equation that’s listed on p. 110 V(X) = s 2 = E(X 2 ) − [E(X)]2 = [(02 ⋅ 0.3) + (1 2 ⋅ 0.6) + (2 2 ⋅ 0.1)] − [0.8]2 = 0 + 0.6 + 0.4 = 1 − 0.64 = 0.36 Sample Variance of defective board =0.36 board 2 4