Chapter 16 Random Variables Ranya Kaluarachchi George Weng

Chapter 16 Random Variables Ranya Kaluarachchi George Weng Period 3

Random Variables A random variable assumes any of several different values as a result of some random event. Two Types: Discrete random variables can take one of a FINITE number of DISTINCT OUTCOMES. Ex. Number of credit hours (2 hours, 3 hours, 4 hours) Continuous random variables can take ANY NUMERIC VALUE within a RANGE of values. Ex. Cost of books this term (80. 00$ - 120. 00$)

What Can Go Wrong? Probability models are still models. They may not show what will actually happen in reality. Don’t assume everything follows a Normal model. Independent variables You can add expected values for any two random variables. You can only add variances of independent random variables. Variances of independent random variables add; standard deviations don’t. The variance of independent random variable’s sum or difference is always the sum of their variances.

Expected Value E(X) = μ that’s the symbol they use, mu Example from textbook: life insurance An insurance company will pay you 10, 000$ if you die that year, 5000$ if you’re disabled, or 0$ if neither occurs. One out of every 1000 people die, and two out of every 1000 are disabled. What are the outcomes? Payout x 10000 5000 0 What are the chances? Probability P(X = x) 1/1000 2/1000 997/1000 Expected Value = E(X) = 10, 000(1/1000) + 5000(2/1000) + 0(997/1000) E(X) = 20$ in other words, each policy must cost at least 20$ for the company to break even “just multiply each possible value by the probability that it occurs, and find the sum”

Standard Deviation So, continuing the example… to find standard deviation, first find deviation. μ (expected value) = $20 Payout “x” 10000 5000 0 Probability “P(X=x)” 1/1000 2/1000 997/1000 Deviation “(x - μ)” 10000 - 20 = 9980 5000 - 20 = 4980 0 - 20 = -20 Next, we square each deviation. The variance is the expected value of those squared deviations. Var(X) = 9980²(1/1000) + 4980²(2/1000) + (-20)²(997/1000) = 149, 600 Then take the square root of the variance to get standard deviation. SD(X) = sqrt(149, 600) ~~ 386. 78$

Rules for adding and subtracting E(X ± c) = E(X) ± c Var(X ± c) = Var(X) E(a*X) = a*E(X) Var(a*X) = a²Var(X)

Homework Problem #13 Kids. A couple plans to have children until they get a girl, but they agree that they will not have more than three children even if all are boys. (Assume boys and girls are equally likely. ) 13. Find the standard deviation of the number of children the couple may have. # of Children they’ll have 1 2 3 P(Children) 0. 5 0. 25 Deviation “(x - μ)” 1 - 1. 75 = -0. 75 2 - 1. 75 = 0. 25 3 - 1. 75 = 1. 25 E(X) = 1(0. 5) + 2(0. 25) + 3(0. 25) = 1. 75 Deviation = (1 -1. 75) Var(X) = (-0. 75)²(0. 5) + (0. 25)²(0. 25) + (1. 25)²(0. 25) = 0. 6875 SD(X) = √ 0. 6875 = 0. 829

Homework Problem #17 Defects. A consumer organization inspecting new cars found that many had appearance defects(dents, scratches, paint chips, etc. ). While none had more than three of these defects, 7% had three, 11% two, and 21% one defect. Find the expected number of appearance defects in a new car and the standard deviation. # of defects 3 2 1 0 P(defects) 7/100 11/100 21/100 61/100 Deviation 3 - 0. 64 = 2. 36 2 - 0. 64 = 1. 36 1 - 0. 64 = 0. 36 0 - 0. 64 = -0. 64 E(X) = 3(7/100) + 2(11/100) + 1(21/100) + 0(61/100) = Var(X) = 2. 36²(7/100) + 1. 36²(11/100) + 0. 36²(21/100) + -0. 64²(61/100) = 0. 8704 SD(X) = √ 0. 8704 = 0. 933