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Unformatted text preview: : ⎛ X − μ 300 − μ ⎞
P(X < 300) = P⎜
300 − 380 ⎞
= P⎜ Z <
⎠ = P(Z < − 1.6)
= 1 − P(Z < 1.6) by symmetry
= 1 − F(1.6)
A look‐up in Appendix Table 1 gives: F(1.6) = 0.9452 The answer is: P(300 < X < 400) = 0.6554 − (1 − 0.9452) = 0.60 23 Chapter 6 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: P(X > b ) = 0.10 where the probability of 10% can be varied to any level of interest. A graph of the problem is below. PDF of X ~ N(μ , σ ) 2 f(x) P(X < b) = 0.9 Upper Tail Area = 0.10 E(X)
x 24 b Chapter 6 A probability result is: P( X > b ) = 1 − P( X < b ) Therefore, as shown in the above graph, the problem is to find the value b such that: P(X < b ) = 0.90 A result is: b− μ⎞
P(X < b ) = P⎜ Z <
⎝ b− μ⎞
Appendix Table 1 gives F(1.28 ) = 0.90 (some approximation was used). Therefore, Rearranging gives: 25 b−μ
σ b = μ + 1.28 σ Chapter 6 The cutoff point (or critical value) b can be computed with Microsoft Excel with the function: NORMINV(probability, μ X , σ X ) Cutoff points from the standard normal distribution are computed with the function: NORMSINV(probability) For example, to find the value z0 such that F( z0 ) = 0.90 with Microsoft Excel select...
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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