The table gives 06554 04 f therefore 06554 400 x p

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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 square4 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 square4 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. PDF of ) , ( N ~ X 2 50 380 400 380 300 f(x) x P(300 < X < 400) P(X < 300) 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
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Econ 325 – Chapter 5 25 xrhombus 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. PDF of ) , ( N ~ X 2 σ μ b E(X) f(x) x Upper Tail Area = 0.10 P(X < b) = 0.9 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
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Econ 325 – Chapter 5 27 The cutoff point (or critical value) b can be computed with Microsoft Excel with the function: NORMINV(probability, X X , σ μ ) Cutoff points from the standard normal distribution are computed with the function: NORMSINV(probability) For example, to find the value 0 z such that 90 . 0 ) z ( F 0 = with Microsoft Excel select Insert Function: NORMSINV( 0.9 ) or NORMINV( 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 square4 Find a range of dollar clothing expenditure that includes 80% of all students.
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