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Lesson #16 Standardizing a Normal Distribution
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X ~ N( , 2 ) X - Z = ~ N( , )
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P(X < c) c = P(X < c) =
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P(X < 60) X = diastolic blood pressure, X ~ N(77, 11.6 2 ) 7788.665.4 60 = P(Z < -1.47)=.0708
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P(X > 90) 7788.665.4 90 = 1 - P(Z < 1.12)= 1 -.8686 =.1314 = 1 - P(X < 90)
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P(60 < X < 90) = P(Z < 1.12) - P(Z < -1.47) =.8686 -.0708 =.7978 = P(X < 90) - P(X < 60)
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7788.665.4 6090.0708.1314 P(60 < X < 90) = 1 – [ P(X 90) ] = 1 – [.0708 +.1314 ] =.7978
![P(60 < X < 90) = 1 – [ P(X 90) ] = 1 – [ ] =.7978](http://images.slideplayer.com/16/5092331/slides/slide_7.jpg "P(60 < X < 90) = 1 – [ P(X 90) ] = 1 – [ ] =.7978")
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CHD (D): X ~ N(244, 51 2 ) No CHD (D’): X ~ N(219, 41 2 ) D’ D 219244
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D’ 219 P(X > 260 | CHD) D 244 = 1 - P(X < 260 | CHD) = 1 - P(Z < 0.31) = 1 -.6217 =.3783 260
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P(X > 260 | CHD) D’ 219 D 244 260 = P( + | D) + = Se
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D’ 219 P(X > 260 | no CHD) D 244 = 1 - P(X < 260 | no CHD) = 1 - P(Z < 1.00) = 1 -.8413 =.1587 260
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P(X > 260 | no CHD) 260 = P( + | D’) D’ 219 D 244 + = 1 - P( | D’) = 1 - Sp Sp =.8413
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P(X < 260 | CHD) = P( | D) = 1 - P( + | D) = 1 - Se = 1 -.3783 =.6217
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D 244 260 D’ 219 + Se Sp
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.95 = Sp = P( | D’) = P( X < c | no CHD).95 P( Z < 1.65)
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.95 = Sp = P( | D’) = P( X < c | no CHD).95 P( Z < 1.65) c 286.65 287
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