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Continuous Random Variables

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Presentation on theme: "Continuous Random Variables"— Presentation transcript:

1 Continuous Random Variables
Lecture 25 Section 7.5.3 Fri, Oct 29, 2004

2 Continuous Probability Distribution Functions
Continuous Probability Distribution Function (pdf) – For a random variable X, it is a function with the property that the area between the graph of the function and an interval a ≤ x ≤ b equals the probability that a ≤ X ≤ b. In other words, Area = Probability

3 Example The TI-83 will return a random number between 0 and 1 if we enter rand and press ENTER. These numbers have a uniform distribution from 0 to 1. Let X be the random number returned by the TI-83.

4 Example The graph of the pdf of X. f(x) 1 x 1

5 Example What is the probability that the random number is at least 0.3?

6 Example What is the probability that the random number is at least 0.3? f(x) 1 x 0.3 1

7 Example What is the probability that the random number is at least 0.3? f(x) 1 x 0.3 1

8 Example What is the probability that the random number is at least 0.3? f(x) 1 x 0.3 1

9 Example What is the probability that the random number is at least 0.3? Probability = 70%. f(x) 1 Area = 0.7 x 0.3 1

10 Example What is the probability that the random number is between 0.3 and 0.8? f(x) 1 x 0.3 0.8 1

11 Example What is the probability that the random number is between 0.3 and 0.8? f(x) 1 x 0.3 0.8 1

12 Example What is the probability that the random number is between 0.3 and 0.8? f(x) 1 x 0.3 0.8 1

13 Example What is the probability that the random number is between 0.3 and 0.8? Probability = 50%. f(x) 1 Area = 0.5 x 0.3 0.8 1

14 Experiment Use the TI-83 to generate 500 values of X.
Use rand(500) to do this. Check to see what proportion of them are between 0.3 and 0.8. Use a TI-83 histogram and Trace to do this.

15 Example Now suppose we use the TI-83 to get two random numbers from 0 to 1, and then add them together. Let Y = the sum of the two random numbers. What is the pdf of Y?

16 Example The graph of the pdf of Y. f(y) 1 y 1 2

17 Example The graph of the pdf of Y. f(y) 1 Area = 1 y 1 2

18 Example What is the probability that Y is between 0.5 and 1.5? f(y) 1
0.5 1 1.5 2

19 Example What is the probability that Y is between 0.5 and 1.5? f(y) 1
0.5 1 1.5 2

20 Example The probability equals the area under the graph from 0.5 to 1.5. f(y) 1 y 0.5 1 1.5 2

21 Example Cut it into two simple shapes, with areas 0.25 and 0.5. f(y) 1
0.5 1 1.5 2

22 Example The total area is 0.75. The probability is 75%. Area = 0.75
f(y) 1 Area = 0.75 y 0.5 1 1.5 2

23 Verification Use the TI-83 to generate 500 values of Y.
Use rand(500) + rand(500). Use a histogram to find out how many are between 0.5 and 1.5.

24 Example Suppose we get 12 random numbers, uniformly distributed between 0 and 1, from the TI-83 and add them all up. Let X = sum of 12 random numbers from 0 to 1. What is the pdf of X?

25 Example It turns out that the pdf of X is approximately normal with a mean of 6 and a standard of 1. 6 7 8 9 5 4 3 N(6, 1) x

26 Example What is the probability that the sum will be between 5 and 7?
P(5 < X < 7) = P(–1 < Z < 1) = – =

27 Example What is the probability that the sum will be between 4 and 8?
P(4 < X < 8) = P(–2 < Z < 2) = – =

28 Experiment Use the TI-83 to see if we really do get a total between 5 and 7 about 68% of the time. Enter the expression sum(rand(12)) and press ENTER. Do it 500 times and see how often the total is between 5 and 7.

29 Experiment We should see a value between 5 and 7 about 68% of the time. We should see a value between 4 and 8 about 95% of the time. We should see a value between 3 and 9 nearly always (99.7%).


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