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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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