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Mean 320 Normal Probabilities Using x's and z's St'd Dev. 20 USING x P(X< 300) = P(X>335) = P(320<X<350) = P(325<X<355) = P(308<X<347) = P(275<X<285) = USING z When x = 300, z = P(Z < z) = When x = 335, z = P(Z > z) = When xu = 350, zu = When xl = 320, zl = P(zl<Z<zu) = When xu = 355, zu = When xl = 325, zl = P(zl<Z<zu) = When xu = 347, zu = When xl = 308, zl = P(zl<Z<zu) = When xu = 285, zu = When xl = 275, zl = P(zl<Z<zu) = Mean 320 CALCULATING x AND z VALUES St'd Dev. 20 FOR GIVEN PROBALITIES CALCULATING x x such that P(X< x) = .1000 x such that P(X< x) = .1250 x such that P(X> x) = .1250 CALCULATING z then x z such that P(Z< z) = .1000 Corresponding x = z such that P(Z< z) = .1250 Corresponding x = z such that P(Z> z) = .1250 Corresponding x = VALUES Mean 320 Normal Probabilities Using x's and z's St'd Dev. 20 USING x P(X< 300) = 0.158655 <== =NORMDIST(300,320,20,TRUE) P(X>335) = 0.226627 <== =1-NORMDIST(335,320,20,TRUE) P(320<X<350) = 0.433193 <== =NORMDIST(350,320,20,TRUE)-NORMDIST(320,320,20,TRUE) P(325<X<355) = 0.361235 <== =NORMDIST(355,320,20,TRUE)-NORMDIST(325,320,20,TRUE) P(308<X<347) = 0.637239 <== =NORMDIST(347,320,20,TRUE)-NORMDIST(308,320,20,TRUE) P(275<X<285) = 0.027835 <== =NORMDIST(285,320,20,TRUE)-NORMDIST(275,320,20,TRUE) USING z When x = 300, z = -1 <== =(300-B1)/B2 P(Z < z) = 0.158655 <== =NORMSDIST(B20) When x = 335, z = 0.75 <== =(335-B1)/B2 P(Z > z) = 0.226627 <== =1-NORMSDIST(B23) When xu = 350, zu = 1.5 <== =(350-B1)/B2 When xl = 320, zl = 0 <== =(320-B1)/B2 P(zl<Z<zu) = 0.433193 <== =NORMSDIST(B26)-NORMSDIST(B27) When xu = 355, zu = 1.75 <== =(355-B1)/B2 When xl = 325, zl = 0.25 <== =(325-B1)/B2 P(zl<Z<zu) = 0.361235 <== =NORMSDIST(B30)-NORMSDIST(B31) When xu = 347, zu = 1.35 <== =(347-B1)/B2 When xl = 308, zl = -0.6 <== =(308-B1)/B2 P(zl<Z<zu) = 0.637239 <== =NORMSDIST(B34)-NORMSDIST(B35) When xu = 285, zu = -1.75 <== =(285-B1)/B2 When xl = 275, zl = -2.25 <== =(275-B1)/B2 P(zl<Z<zu) = 0.027835 <== =NORMSDIST(B38) - NORMSDIST(B39) Mean 320 CALCULATING x AND z VALUES St'd Dev. 20 FOR GIVEN PROBALITIES CALCULATING x x such that P(X< x) = .1000 294.36897 <== =NORMINV(0.1,B1,B2) x such that P(X< x) = .1250 296.99301 <== =NORMINV(0.125,B1,B2) x such that P(X> x) = .1250 343.00699 <== =NORMINV(1-.125,B1,B2) CALCULATING z then x z such that P(Z< z) = .1000 -1.281552 <== =NORMSINV(0.1) Corresponding x = 294.36897 <== =B1+ B18*B2 z such that P(Z< z) = .1250 -1.150349 <== =NORMSINV(0.125) Corresponding x = 296.99301 <== =B1+ B21*B2 z such that P(Z> z) = .1250 1.1503494 <== =NORMSINV(1-.125) Corresponding x = 343.00699 <== =B1+ B24*B2 VALUES

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posted: | 3/25/2013 |

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