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Lesson #16 Standardizing a Normal Distribution

Lesson #16 Standardizing a Normal Distribution. m - s. m. m + s. 0. -1. 1. X ~ N( m , s 2 ). X. - m. Z =. ~ N( 0 , 1 ). s. =. c. P(X < c ). P(X < c ) =. 60. X = diastolic blood pressure, X ~ N(77 , 11.6 2 ). P(X < 60). = P(Z < -1.47). = .0708. 65.4. 77. 88.6.

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Lesson #16 Standardizing a Normal Distribution

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  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

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