# First state the probability as a probability about

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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. Now check the answer with Microsoft Excel by selecting Insert Function: NORM.DIST(x, X X , σ μ , cumulative) Enter the values: NORM.DIST(400, 380, 50 ,1) This returns the probability: 0.6554
Econ 325 – Chapter 5 21 ! 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 > = <
Econ 325 – Chapter 5 23 ! Find ) X ( P 400 300 < < . This gives the probability that a randomly chosen student will spend between \$300 and \$400 on clothing in a year. The range probability is calculated as: ) X ( P ) X ( P ) X ( P 300 400 400 300 < < = < < A graph gives a helpful picture of the calculations. Econ 325 – Chapter 5 24 From the previous calculations: 0.6554 400 = < ) X ( P Now find: ) ( F 1 Z ( P 1 Z ( P Z P X P ) X ( P 1.6 symmetry by 1.6) 1.6) 50 380 300 300 300 = < = < = < = σ μ < σ μ = < A look-up in the Appendix Table gives: 0.9452 1.6 = ) ( F The answer is: 0.60 0.9452 1 0.6554 400 300 = = < < ) ( ) X ( P
Econ 325 – Chapter 5 25 " Finding Cutoff Points or Critical Values A problem that has been presented is: What is the probability that values will occur in some range ? Another problem is: What numerical value corresponds to a probability of 10% ? That is, find the value b such that: 10 . 0 ) b X ( P = > A graph of the problem is below. Note: the upper tail probability can be set to any level of interest. The value of 10% is chosen here. Econ 325 – Chapter 5 26 A probability result is: ) b X ( P 1 ) b X ( P < = > Therefore, as shown in the above graph, the problem is to find the value b such that: 90 . 0 ) b X ( P = < A result is: σ μ = σ μ < = < b F b Z P ) b X ( P The Appendix Table gives 90 . 0 ) ( F = 1.28 (some approximation was used). Therefore, 1.28 = σ μ b Rearranging gives: σ + μ = 1.28 b
Econ 325 – Chapter 5 27 The cutoff point (or critical value) b can be computed with Microsoft Excel with the function: NORM.INV(probability, X X , σ μ ) Cutoff points from the standard normal distribution are computed with the function: NORM.S.INV(probability) For example, to find the value 0 z such that 90 . 0 ) z ( F 0 = with Microsoft Excel select Insert Function: NORM.S.INV( 0.9 ) or NORM.INV( 0.9 , 0 , 1 ) Both these functions return the answer 0 z = 1.2816 Econ 325 – Chapter 5 28 Example: student clothing expenditure exercise Continued ! Find a range of dollar clothing expenditure that includes 80% of all students.

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• Spring '10
• WHISTLER
• Normal Distribution, Probability theory, probability density function, Cumulative distribution function

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