Engineering Statistics - IE 261

1. Engineering Statistics - IE 261 Chapter 3 Discrete Random Variables and Probability Distributions URL: http://home.npru.ac.th/piya/ClassesTU.html http://home.npru.ac.th/piya/webscilab

2. 3-1 Discrete Random Variables

3. 3-1 Discrete Random Variables Example 3-1

4. 3-2 Probability Distributions and Probability Mass Functions Figure 3-1Probability distribution for bits in error.

5. 3-2 Probability Distributions and Probability Mass Functions Definition

6. Example 3-5

7. Example 3-5 (continued)

8. 3-3 Cumulative Distribution Functions Definition

9. Example 3-8

10. Example 3-8 Figure 3-4Cumulative distribution function for Example 3-8.

11. 3-4 Mean and Variance of a Discrete Random Variable Definition

12. 3-4 Mean and Variance of a Discrete Random Variable Figure 3-5A probability distribution can be viewed as a loading with the mean equal to the balance point. Parts (a) and (b) illustrate equal means, but Part (a) illustrates a larger variance.

13. Proof of Variance:

14. 3-4 Mean and Variance of a Discrete Random Variable Figure 3-6The probability distribution illustrated in Parts (a) and (b) differ even though they have equal means and equal variances.

15. Example 3-9 There is a chance that a bit transmitted through a digital transmissionchannel is received in error. Let X equal the number of bits in error inthe next four bits transmitted. The possible values for X are {0, 1, 2, 3, 4}Suppose: P(X = 0) = 0.6561 P(X = 1) = 0.2916 P(X = 2) = 0.0486 P(X = 3) = 0.0036 P(X = 4) = 0.0001 Find the mean and the variance of X

16. Example 3-9 (Solution) SCILAB -->x = [0 1 2 3 4]; -->fx = [0.6561 0.2916 0.0486 0.0036 0.0001]; -->MeanX = sum(x.*fx) MeanX = 0.4 -->VarX = sum((x.^2).*fx) - MeanX^2 VarX = 0.36

17. Example 3-9 (Solution)

18. Example 3-11 -->x = [10:15]; fx = [0.08 0.15 0.3 0.2 0.2 0.07]; -->MeanX = sum(x.*fx) MeanX = 12.5 -->VarX = sum((x.^2).*fx) - MeanX^2 VarX = 1.85