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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.

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Presentation on theme: "Discrete and Continuous Random Variables I can find the standard deviation of discrete random variables. I can find the probability of a continuous random."— Presentation transcript:

1 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

2 The Variance of a Discrete Random Variable Recall: Variance and standard deviation are measures of spread.

3 If X is a discrete random variable with mean μ, then the variance of X is The standard deviation is the square root of the variance.

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.

5 Gains follows the modern practice of using probability estimates of sales. The military estimates the sales as follows: Units sold: ,000 Probability: Take X to be the number of military units sold.

6 Compute μ x : μxμx = (1000)(0.1) + (3000)(0.3) + (5000)(0.4) + (10000)(.2) = = 5000 units

7 Calculate the variance of X: σ x 2 = ∑(x i - μ x ) 2 p i = (1000 – 5000) 2 (0.10) + … finish = 7,800,000

8 Standard Deviation σ x = sqrt 7,800,000 = The standard deviation is the measure of how variable the number of units sold is.

9 To find the variance with calculator: In notes for your info. Try it if you want on your own.

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!

11 Now we will assign values as areas under a density curve.

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.

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

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.

15 Normal Distributions as Probability Distributions Recall N(μ,σ) is the shorthand notation for the normal distribution having mean μ and standard deviation σ.

16 If X has the N(μ,σ) then the standardized variable Z = (X – μ) / σ is a standard normal random variable having the distribution N(0,1).

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).

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) The”shaded region” = P(p 0.32)

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)/ ) = P(z < -1.69)

20 Use z-score to find the area. Calc: 2nd VARS(DIST):normalcdf(-EE99, -1.69) =

21 P(p > 0.32) = also why? P(p 0.32) = = Conclusion: The probability that the sample will miss the truth by more than 2 percentage points is

22 Or, use the complement to get “middle” or “unshaded” region: 1 – P(-1.69 < z < 1.69) = 1 - normcdf(-1.69,1.69) = =.0901 with complement rule

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.

24 Probability model a) Verify that this is a legitimate discrete distribution. Number of cars X Prob

25 Display the distribution in a probability histogram. (2 min)

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

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) = % of households have more cars than the garage can hold.


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