1. Lesson #16 Standardizing a Normal Distribution

2.  X ~ N( ,  2 )   X -   Z = ~ N( ,  )

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

4. P(X < 60) X = diastolic blood pressure, X ~ N(77, 11.6 2 ) 7788.665.4 60 = P(Z < -1.47)=.0708

5. P(X > 90) 7788.665.4 90 = 1 - P(Z < 1.12)= 1 -.8686 =.1314 = 1 - P(X < 90)

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

7. 7788.665.4 6090.0708.1314 P(60 < X < 90) = 1 – [ P(X 90) ] = 1 – [.0708 +.1314 ] =.7978

8. CHD (D): X ~ N(244, 51 2 ) No CHD (D’): X ~ N(219, 41 2 ) D’ D 219244

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

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

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

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

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

14. D 244 260 D’ 219 +  Se Sp

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