1. Continuous Random Variables
Lecture 25
Section 7.5.3
Mon, Oct 23, 2006

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.
Remember,
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 0.25 and 0.75?
f(x) 1 x
0.3
0.9 1

11. Example
What is the probability that the random number is between 0.25 and 0.75?
f(x) 1 x
0.25
0.75 1

12. Example
What is the probability that the random number is between 0.3 and 0.9?
f(x) 1 x
0.25
0.75 1

13. Example
What is the probability that the random number is between 0.3 and 0.9?
Probability = 60%.
f(x) 1
Area = 0.5 x
0.25
0.75 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 0.25 and 0.75.
Use a TI-83 histogram and Trace to do this.
Or use Excel to generate values of X.

15. Hypothesis Testing (n = 1)
An experiment is designed to determine whether a random variable X has the distribution U(0, 1) or U(0.5, 1.5).
H0: X is U(0, 1).
H1: X is U(0.5, 1.5).
One value of X is sampled (n = 1).

16. Hypothesis Testing (n = 1)
An experiment is designed to determine whether a random variable X has the distribution U(0, 1) or U(0.5, 1.5).
H0: X is U(0, 1).
H1: X is U(0.5, 1.5).
One value of X is sampled (n = 1).
If X is more than 0.75, then H0 will be rejected.

17. Hypothesis Testing (n = 1)
Distribution of X under H0:
Distribution of X under H1:
0.5 1
1.5
0.5 1
1.5

18. Hypothesis Testing (n = 1)
What are  and ? 1
0.5 1
1.5 1
0.5 1
1.5

19. Hypothesis Testing (n = 1)
What are  and ? 1
0.5
0.75 1
1.5 1
0.5
0.75 1
1.5

20. Hypothesis Testing (n = 1)
What are  and ? 1
0.5
0.75 1
1.5
Acceptance Region
Rejection Region 1
0.5
0.75 1
1.5

21. Hypothesis Testing (n = 1)
What are  and ? 1
0.5
0.75 1
1.5 1
0.5
0.75 1
1.5

22. Hypothesis Testing (n = 1)
What are  and ?
 = ¼ = 0.25 1
0.5
0.75 1
1.5 1
0.5
0.75 1
1.5

23. Hypothesis Testing (n = 1)
What are  and ?
 = ¼ = 0.25 1
0.5
0.75 1
1.5
 = ¼ = 0.25 1
0.5
0.75 1
1.5

24. Example
Now suppose we use the TI-83 to get two random numbers from 0 to 1, and then add them together.
Let X2 = the sum of the two random numbers.
What is the pdf of X2?

25. Example
The graph of the pdf of X2.
f(y) 1 y 1 2

26. Example
The graph of the pdf of X2.
f(y) 1
Area = 1 y 1 2

27. Example What is the probability that X2 is between 0.5 and 1.5? f(y) 1
0.5 1
1.5 2

28. Example What is the probability that X2 is between 0.5 and 1.5? f(y) 1
0.5 1
1.5 2

29. 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

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

31. 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

32. 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.
Or use Excel to generate pairs of values of X.

33. Hypothesis Testing (n = 2)
An experiment is designed to determine whether a random variable X has the distribution U(0, 1) or U(0.5, 1.5).
H0: X is U(0, 1).
H1: X is U(0.5, 1.5).
Two values of X are sampled (n = 2). Let X2 be the sum. If X2 is more than 1.5, then H0 will be rejected.

34. Hypothesis Testing (n = 2)
Distribution of X2 under H0:
Distribution of X2 under H1: 1 2 3 1 2 3

35. Hypothesis Testing (n = 2)
What are  and ? 1 2 3 1 2 3

36. Hypothesis Testing (n = 2)
What are  and ? 1 2 3
1.5 1 2 3
1.5

37. Hypothesis Testing (n = 2)
What are  and ? 1 2 3
1.5 1 2 3
1.5

38. Hypothesis Testing (n = 2)
What are  and ? 1 2 3
 = 1/8 = 0.125
1.5 1 2 3
1.5

39. Hypothesis Testing (n = 2)
What are  and ? 1 2 3
 = 1/8 = 0.125
1.5 1 2 3
 = 1/8 = 0.125
1.5

40. Conclusion
By increasing the sample size, we can lower both  and  simultaneously.

41. Example
Now suppose we use the TI-83 to get three random numbers from 0 to 1, and then add them together.
Let X3 = the sum of the three random numbers.
What is the pdf of X3?

42. Example
The graph of the pdf of X3.
f(y) 1 y 1 2 3

43. Example
The graph of the pdf of X3.
f(y) 1
Area = 1 y 1 2 3

44. Example What is the probability that X3 is between 1 and 2? f(y) 1 y 11 2 3

45. Example What is the probability that Y is between 1 and 2? f(y) 1 y 11 2 3

46. Example The probability equals the area under the graph from 1 to 2. 11 2 3

47. Verification Use Excel to generate 10000 sums of three values.
See if about 2/3 of the values lie between 1 and 2.

48. Hypothesis Testing (n = 2)
An experiment is designed to determine whether a random variable X has the distribution U(0, 1) or U(0.5, 1.5).
H0: X is U(0, 1).
H1: X is U(0.5, 1.5).
Three values of X3 are sampled (n = 3). Let X3 be the sum. If X3 is more than 1.5, then H0 will be rejected.

49. Hypothesis Testing (n = 2)
Distribution of X3 under H0:
Distribution of X3 under H1: 1 2 3 4 1 1 2 3 4

50. Hypothesis Testing (n = 2)
Distribution of Y under H0:
Distribution of Y under H1:
 = 0.07 1 2 3 4 1
 = 0.07 1 2 3 4

51. Example
Suppose we get 12 random numbers, uniformly distributed between 0 and 1, from the TI-83 and add them all up.
Let X12 = sum of 12 random numbers from 0 to 1.
What is the pdf of X12?

52. Example
It turns out that the pdf of X12 is nearly exactly normal with a mean of 6 and a standard of 1.
N(6, 1) x 3 4 5 6 7 8 9

53. Example What is the probability that the sum will be between 5 and 7?
P(5 < X12 < 7) = P(–1 < Z < 1)
= – =

54. Example What is the probability that the sum will be between 4 and 8?
P(4 < X12 < 8) = P(–2 < Z < 2)
= – =

55. Experiment
Use the Excel spreadsheet Sum12.xls to generate values of X, where X is the sum of 12 random numbers from U(0, 1).
We should see a value between 5 and 7 about 68.27% of the time.
We should see a value between 4 and 8 about 95.45% of the time.
We should see a value between 3 and 9 about 99.73% of the time.