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Either formula for V1x2 can be used. Figure 3-6 illustrates that two probability distributions can differ even though they have identical means and variances. EXAMPLE 3-9 In Example 3-4, there is a chance that a bit transmitted through a digital transmission channel is received in error. Let X equal the number of bits in error in the next four bits transmitted. The possible values for X are 50, 1, 2, 3, 46 . Based on a model for the errors that is presented in the following section, probabilities for these values will be determined. Suppose that the probabilities are P1X P1X 02 12 0.6561 0.2916 P1X P1X 22 32 0.0486 0.0036 P1X 42 0.0001
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Figure 3-6 The probability distributions illustrated in Parts (a) and (b) differ even though they have equal means and equal variances.
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CHAPTER 3 DISCRETE RANDOM VARIABLES AND PROBABILITY DISTRIBUTIONS
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Now E1X2 0f 102 1f 112 2f 122 010.65612 110.29162 0.4 3f 132 4f 142 210.04862 310.00362 410.00012
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Although X never assumes the value 0.4, the weighted average of the possible values is 0.4. To calculate V1X 2, a table is convenient.
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The alternative formula for variance could also be used to obtain the same result. EXAMPLE 3-10 Two new product designs are to be compared on the basis of revenue potential. Marketing feels that the revenue from design A can be predicted quite accurately to be $3 million. The revenue potential of design B is more difficult to assess. Marketing concludes that there is a probability of 0.3 that the revenue from design B will be $7 million, but there is a 0.7 probability that the revenue will be only $2 million. Which design do you prefer Let X denote the revenue from design A. Because there is no uncertainty in the revenue from design A, we can model the distribution of the random variable X as $3 million with probability 1. Therefore, E1X 2 $3 million. Let Y denote the revenue from design B. The expected value of Y in millions of dollars is E1Y 2 $710.32 $210.72 $3.5
Because E(Y) exceeds E(X), we might prefer design B. However, the variability of the result from design B is larger. That is,
5.25 millions of dollars squared
3.52 2 10.32
3.52 2 10.72
Because the units of the variables in this example are millions of dollars, and because the variance of a random variable squares the deviations from the mean, the units of 2 are millions of dollars squared. These units make interpretation difficult. Because the units of standard deviation are the same as the units of the random variable, the standard deviation is easier to interpret. In this example, we can summarize our results as the average deviation of Y from its mean is $2.29 million.
3-4 MEAN AND VARIANCE OF A DISCRETE RANDOM VARIABLE
EXAMPLE 3-11
The number of messages sent per hour over a computer network has the following distribution:
x f 1x2 number of messages 10 0.08 11 0.15 12 0.30 13 0.20 14 0.20 15 0.07
Determine the mean and standard deviation of the number of messages sent per hour. V1X 2 E1X 2 10 10.082
2V1X 2
p p 1.36
11 10.152
15 10.072
12.5 12.52 1.85
The variance of a random variable X can be considered to be the expected value of a specific function of X, namely, h1X 2 1X 2 2 . In general, the expected value of any function h1X 2 of a discrete random variable is defined in a similar manner.
Expected Value of a Function of a Discrete Random Variable
If X is a discrete random variable with probability mass function f 1x2, E3h1X 2 4 a xh1x 2 f 1x 2
(3-4)
EXAMPLE 3-12
In Example 3-9, X is the number of bits in error in the next four bits transmitted. What is the expected value of the square of the number of bits in error Now, h1X 2 X 2 . Therefore, E3h1X2 4 02 0.6561 12 32 0.2916 0.0036 22 42 0.0486 0.0001 0.52
In the previous example, the expected value of X 2 does not equal E1X 2 squared. However, in aE1X 2 b. This the special case that h1X 2 aX b for any constants a and b, E 3h1X 2 4 can be shown from the properties of sums in the definition in Equation 3-4.
EXERCISES FOR SECTION 3-4
3-37. If the range of X is the set {0, 1, 2, 3, 4} and P(X x) 0.2 determine the mean and variance of the random variable. 3-38. Determine the mean and variance of the random variable in Exercise 3-13. 3-39. Determine the mean and variance of the random variable in Exercise 3-15. 3-40. Determine the mean and variance of the random variable in Exercise 3-17. 3-41. Determine the mean and variance of the random variable in Exercise 3-19. 3-42. Determine the mean and variance of the random variable in Exercise 3-20. 3-43. Determine the mean and variance of the random variable in Exercise 3-22. 3-44. Determine the mean and variance of the random variable in Exercise 3-23. 3-45. The range of the random variable X is 3 0, 1, 2, 3, x4 , where x is unknown. If each value is equally likely and the mean of X is 6, determine x.