chap05PRN econ 325

Pdf of z z0 z0 fz z area fz0 f z0 upper tail area 1

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PDF of Z z0 0 -z0 f(z) z Area = F(z0) - F(-z0) Upper Tail Area = 1 - F(z0) Lower Tail Area = F(-z0) By symmetry of the normal distribution the area in the “lower tail” is identical to the area in the “upper tail.” Econ 325 – Chapter 5 18 Now suppose the random variable to work with is: ) , ( N ~ X 2 σ μ For two numerical values a and b , with a < b , a probability of interest is: ) b X a ( P < < This probability statement can be transformed to a probability statement about the standard normal random variable Z . This is done as follows: σ μ - - σ μ - = σ μ - < < σ μ - = σ μ - < σ μ - < σ μ - = < < a F b F b Z a P b X a P ) b X a ( P The Appendix Table can now be used to look-up the cumulative probabilities for the standard normal distribution.
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Econ 325 – Chapter 5 19 Example Let the continuous random variable X be the amount of money spent on clothing in a year by a university student. It is known that: ) , ( N ~ X 2 σ μ with $380 = μ and 50 $ = σ Questions and Answers s Find ) X ( P 400 < . This gives the probability that a randomly chosen student will spend less than $400 on clothing in a year. First state the probability as a probability about the standard normal variable Z : ) ( F Z ( P Z P X P ) X ( P 0.4 0.4) 50 380 400 400 400 = < = - < = σ μ - < σ μ - = < Now look-up the answer in the Appendix Table. The table gives: 0.6554 0.4 = ) ( F Therefore, 0.6554 400 = < ) X ( P Econ 325 – Chapter 5 20 A graph gives a helpful illustration of the use of the statistical tables for this problem. PDF of Z 0.4 0 f(z) z Area = F(0.4) = 0.6554 Now check the answer with Microsoft Excel by selecting Insert Function: NORMDIST(x, X X , σ μ , cumulative) Enter the values: NORMDIST(400, 380, 50 ,1) This returns the probability: 0.6554
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Econ 325 – Chapter 5 21 s Find ) X ( P 360 > . This gives the probability that a randomly chosen student will spend more than $360 on clothing in a year. Express the problem in the form of a probability statement about the standard normal variable Z : ) ( F Z ( P Z ( P Z P X P ) X ( P 0.4 symmetry by 0.4) 0.4) 50 380 360 360 360 = < = - > = - > = σ μ - > σ μ - = > This is identical to the probability calculated for ) X ( P 400 < . That is, 0.6554 400 360 = < = > ) X ( P ) X ( P This result holds since the normal distribution is symmetric about the mean $380 = μ . Econ 325 – Chapter 5 22 The graph below demonstrates that because of symmetry about the mean: ) X ( P ) X ( P 400 360 < = > Also, ) X ( P ) X ( P 400 360 > = < PDF of ) , ( N ~ X 2 50 380 400 380 360 f(x) x P(X < 360) = 1 - 0.6554 P(X > 400)= 1 - 0.6554
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Econ 325 – Chapter 5 23 s Find ) X ( P 400 300 < < .
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PDF of Z z0 z0 fz z Area Fz0 F z0 Upper Tail Area 1 Fz0...

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