Discrete and Continuous Random Variables I can find the standard deviation of discrete random...

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Discrete and Continuous Random Variables I can find the standard deviation of discrete random variables. I can find the probability of a continuous random variable. 6.1b h.w: pg pg 354: 14, 18, 19, 23, 25 Slide 2 The Variance of a Discrete Random Variable Recall: Variance and standard deviation are measures of spread. Slide 3 If X is a discrete random variable with mean , then the variance of X is The standard deviation is the square root of the variance. Slide 4 Example: Selling of Aircraft Gain Communication sells aircraft communications units to both the military and the civilian markets. Next years sales depend on market conditions that can not be predicted exactly. Slide 5 Gains follows the modern practice of using probability estimates of sales. The military estimates the sales as follows: Units sold:1000 3000 5000 10,000 Probability: 0.1 0.3 0.4 0.2 Take X to be the number of military units sold. Slide 6 Compute x : xx = (1000)(0.1) + (3000)(0.3) + (5000)(0.4) + (10000)(.2) = 100 + 900 + 2000 + 2000 = 5000 units Slide 7 Calculate the variance of X: x 2 = (x i - x ) 2 p i = (1000 5000) 2 (0.10) + finish = 7,800,000 Slide 8 Standard Deviation x = sqrt 7,800,000 = 2792.8 The standard deviation is the measure of how variable the number of units sold is. Slide 9 To find the variance with calculator: In notes for your info. Try it if you want on your own. Slide 10 Recall: If we use a table or a calculator to select digits 0 and 1, the result is a discrete random variable which we can count. What is the probability of 0.3 X 0.7 ? Infinite possible values! Slide 11 Now we will assign values as areas under a density curve. Slide 12 Continuous Random Variables A continuous random variable X takes all values in a given interval of numbers. The probability distribution of a continuous random variable is shown by a density curve. Slide 13 The probability that X is between an interval of numbers is the area under the density curve between the interval endpoints. The probability that a continuous random variable X is exactly equal to a number is zero Slide 14 Example: Uniform Distribution (of random digits between 0 and 1) Note: P(X 0.5 or X 0.8) = 0.7 We can add non-overlapping parts. Slide 15 Normal Distributions as Probability Distributions Recall N(,) is the shorthand notation for the normal distribution having mean and standard deviation . Slide 16 If X has the N(,) then the standardized variable Z = (X ) / is a standard normal random variable having the distribution N(0,1). Slide 17 Example: Drugs In School An opinion poll asks a SRS of 1500 of U.S. adults what they think is the most serious problem facing our schools. Suppose 30% would say drugs. The population parameter p is approximately N(0.3,.0118). Slide 18 What is the probability that the poll differs from the truth about the population by more than 2 percentage points? More than one way to do this. = P(p 0.32) Theshaded region = P(p 0.32) Slide 19 Confirm the z-scores and the area of the shaded region. Standardize the values. P(p < 0.28) = P( z < (0.28 0.30)/0.0118 ) = P(z < -1.69) Slide 20 Use z-score to find the area. Calc: 2nd VARS(DIST):normalcdf(-EE99, -1.69) = 0.0455 Slide 21 P(p > 0.32) = 0.0455 also why? P(p 0.32) = 0.0455 + 0.0455 = 0.0910 Conclusion: The probability that the sample will miss the truth by more than 2 percentage points is 0.0910. Slide 22 Or, use the complement to get middle or unshaded region: 1 P(-1.69 < z < 1.69) = 1 - normcdf(-1.69,1.69) = 1 - 0.9090 =.0901 with complement rule Slide 23 Exercise: Car Ownership Chose an American at random and let the random variable X be the number of cars (including SUVs and light trucks) they own. Here is the probability model if we ignore the few households that own more than 5 cars. Slide 24 Probability model a) Verify that this is a legitimate discrete distribution. Number of cars X 012345 Prob.0.090.360.350.130.050.02 Slide 25 Display the distribution in a probability histogram. (2 min) Slide 26 b) Say in words what the event {X 1} is. The event that the household owns at least one car. Find P(X 1) = P(X = 1) + P(X = 2) + + P(X=5) = 0.91 Or, 1 P(X = 0) = 1 0.09 = 0.91 Slide 27 c) A housing company builds houses with two-car garages. What percent of households have more cars than the garage can hold? P(X > 2) = P(X=3) + P(X=4) + P(X+5) = 0.20 20% of households have more cars than the garage can hold.