1 / 16

Understanding Standardizing the Normal Distribution in Statistical Analysis

This lesson delves into the process of standardizing a normal distribution using the formula Z = (X - μ) / σ. It covers key concepts such as calculating probabilities and interpreting Z-scores. Through examples involving diastolic blood pressure and distinct groups (with and without coronary heart disease), students will learn to compute probabilities related to specific values. The lesson emphasizes the application of standard normal distribution and provides insights into sensitivity and specificity in diagnostic tests.

yepa
Télécharger la présentation

Understanding Standardizing the Normal Distribution in Statistical Analysis

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Lesson #16 Standardizing a Normal Distribution

  2. m - s m m + s 0 -1 1 X ~ N(m , s2) X - m Z = ~ N(0 , 1) s

  3. = c P(X < c) P(X < c) =

  4. 60 X = diastolic blood pressure, X ~ N(77 , 11.62) P(X < 60) = P(Z < -1.47) = .0708 65.4 77 88.6

  5. 65.4 77 88.6 90 P(X > 90) = 1 - P(X < 90) = 1 - P(Z < 1.12) = 1 - .8686 = .1314

  6. P(60 < X < 90) = P(X < 90) - P(X < 60) = P(Z < 1.12) - P(Z < -1.47) = .8686 - .0708 = .7978

  7. .1314 .0708 60 90 = 1 – [ P(X < 60) + P(X > 90) ] P(60 < X < 90) = 1 – [ .0708 + .1314 ] = .7978 65.4 77 88.6

  8. CHD (D): X ~ N(244 , 512) No CHD (D’): X ~ N(219 , 412) D’ D 219 244

  9. D’ D 219 244 P(X > 260 | CHD) = 1 - P(X < 260 | CHD) = 1 - P(Z < 0.31) = 1 - .6217 = .3783 260

  10. D’ D 260 - + 219 244 P(X > 260 | CHD) = P( + | D) = Se

  11. D’ D 219 244 P(X > 260 | no CHD) = 1 - P(X < 260 | no CHD) = 1 - P(Z < 1.00) = 1 - .8413 = .1587 260

  12. D’ D - + 219 244 P(X > 260 | no CHD) = P( + | D’)  Sp = .8413 = 1 - P( - | D’) = 1 - Sp 260

  13. P(X < 260 | CHD) = P( - | D) = 1 - P( + | D) = 1 - .3783 = .6217 = 1 - Se

  14. Sp Se D’ D - + 219 244 260

  15. .95 = Sp = P( - | D’) = P( X < c | no CHD) .95  P( Z < 1.65)

  16. .95 = Sp = P( - | D’) = P( X < c | no CHD) .95  P( Z < 1.65)  c  286.65  287

More Related