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chap05PRN econ 325

# A graph of the pdf illustrates the problem 025 4 3 2

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A graph of the PDF illustrates the problem: 0.25 0 4 3 2 1 0 f(x) x Area = P(X < 1) = F(1) The area of a box is calculated as: (height)∙(width). The answer is: 0.25 0) (0.25)(1 = - = = < ) 1 ( F ) 1 X ( P

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Econ 325 – Chapter 5 7 s 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 illustrates the problem: 0.25 0 4 3.5 3 2 1 0.5 0 f(x) x Area = P(0.5 < X < 3.5) The range probability is: 0.75 0.5) (0.25)(3.5 = - = < < ) 5 . 3 X 5 . 0 ( P Also note: ) 5 . 0 ( F ) 5 . 3 ( F ) 5 . 3 X 5 . 0 ( P - = < < Therefore, the probability that an emergency is outside the 1.5 mile limit is: 0.25 0.75 1 = - = < < - ) 5 . 3 X 5 . 0 ( P 1 Econ 325 – Chapter 5 8 Another way of getting the answer is to calculate: ) 5 . 3 X ( P ) 5 . 0 X ( P > + < The graph of the PDF shows: ) 5 . 3 X ( P ) 5 . 0 X ( P > = < Therefore: 0.25 ) (0.25)(0.5 2 = = < = > + < ) 5 . 0 X ( P 2 ) 5 . 3 X ( P ) 5 . 0 X ( P
Econ 325 – Chapter 5 9 Chapter 5.2 Expectations Summary information about a probability distribution is provided by the mean and variance. ) X ( E 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: ) X ( E X = μ A measure of dispersion is the variance : 2 X 2 2 X 2 X ) X ( E ] ) X ( [ E ) X ( Var μ - = μ - = = σ The standard deviation of a random variable X is defined as: 0 ) X ( Var 2 X X > = σ = σ Econ 325 – Chapter 5 10 Recall the rules introduced for discrete random variables. That is, for constant fixed numbers a and b : X b a ) X ( E b a ) X b a ( E μ + = + = + and ) X ( Var b ) X b a ( Var 2 = + As a special case, the standardized random variable is defined as: X X X Z σ μ - = The properties of Z are: 0 ) X ( E 1 X E ) Z ( E X X X X = μ - σ = σ μ - = and 1 ) X ( Var 1 X Var ) Z ( Var 2 X X X = σ = σ μ - = That is, the standardized random variable Z has mean 0 and variance 1 .

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Econ 325 – Chapter 5 11 Chapter 5.3 The Normal Distribution The continuous random variable that follows the normal distribution has popularity in applied work. The probability density function (PDF) for a normally distributed random variable X with mean X μ and variance 2 X σ is: μ - σ - σ π = 2 X 2 X 2 X ) x ( 2 1 exp 2 1 ) x ( f for < < - x Graph of the PDF for a Normal Distribution E(X) f(x) x The shape of the PDF is a symmetric, bell-shaped curve centered on the mean X μ .
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A graph of the PDF illustrates the problem 025 4 3 2 1 fx x...

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