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Unformatted text preview: eight)∙(width). The answer is: 6 P(X < 1) = F(1) = (0.25)(1 − 0) = 0.25 Chapter 6 (d) The rescue team’s base is at the mid‐point of this stretch of river. Find the probability that a given emergency arises more than 1.5 miles from this base. First, calculate the probability that an emergency arises within 1.5 miles from the base. A graph of the PDF is shown: f(x) Area = P(0.5 < X < 3.5) 0.25 0
0 0.5 1 2 3 3.5 4 x The answer is: P(0.5 < X < 3.5) = (0.25)(3.5 − 0.5) = 0.75 Also note: P(0.5 < X < 3.5) = F(3.5) − F(0.5) Therefore, the probability that an emergency is outside the 1.5 mile limit is: 1 − P(0.5 < X < 3.5) = 1 − 0.75 = 0.25 7 Chapter 6 Chapter 6.2 Expectations Summary information about a probability distribution is provided by the mean and variance. E(X ) is the expected value of a random variable X. The expected value can be viewed as the average of the observed values from a “large” number of trials of a random experiment. The mean of a random variable X is denoted by: μ X = E( X ) A measure of dispersion is the variance: σ 2 = Var(X ) = E[ (X − μ X )2 ]
X = E(X 2 ) − μ 2
The standard deviation of a random variable X is defined as: 8 σ X = σ 2 = Var ( X ) > 0 X Chapter 6 Recall the rules introduced for discrete random variables. That is, for constant fixed numbers a and b: E(a + b X ) = a + b E(X ) = a + b μ X and Var (a + b X ) = b2 Var (X ) As a special case, the standardized random variable is defined as: Z= X − μX σX The properties of Z are: ⎡ X − μX ⎤
E(Z) = E ⎢
⎥ = σ E(X −...
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This note was uploaded on 02/06/2014 for the course ECON ECON 325 taught by Professor Whistler during the Spring '10 term at The University of British Columbia.
- Spring '10