1、1234011/21/32/31/2 + 1/2 = 11/3 + 1/3 + 1/3 = 11/4 + 1/4 + 1/4 + 1/4 = 1The probability of each value5011/21/32/31/2 + 1/2 = 11/3 + 1/3 + 1/3 = 11/4 + 1/4 + 1/4 + 1/4 = 1As the number of values increases the probability of each value decreases. This is so because the sum of all the probabilities rem

2、ains 1. When the number of values approaches infinity (because X is continuous) the probability of each value approaches 0.The probability of each value6x1x2Area = 1P(x1=X=x2)712) ab ()X(V2baE(X). bxaab1) x ( f289200050001/3000f(x) = 1/(5000-2000) = 1/3000 for x: 2000,5000 x2500 3000P(2500X3000) = (

3、3000-2500)(1/3000) = .166710200050001/3000f(x) = 1/(5000-2000) = 1/3000 for x: 2000,5000 x4000P(X4000) = (5000-4000)(1/3000) = .33311200050001/3000f(x) = 1/(5000-2000) = 1/3000 for x: 2000,5000 x2500P(X=2500) = (2500-2500)(1/3000) = 01213141516The normal distribution is bell shaped, and symmetrical

4、around m.mWhy symmetrical? Let m = 100. Suppose x = 110. 2210) 2/ 1 (100110) 2/ 1 (e21e21)110( fNow suppose x = 902210) 2/ 1 (10090) 2/ 1 (e21e21)90( f1109017The effects of m m and How does the standard deviation affect the shape of f(x)?= 2 =3 =4m = 10m = 11 m = 12How does the expected value affect

5、 the location of f(x)?182221?2xbap aXbedxm192,mNmX dzzfbZapba2021E(Z) = m = 0V(Z) = 2 = 1Every normal variablewith some m and , canbe transformed into this Z.Therefore, once probabilities for Zare calculated, probabilities of any normal variable can be found.22P(45X60) = P( )45X60 m- 50 - 501010= P(

6、-0.5 Z 0Z = 0Z = z0P(0Zz0)24P(45X60) = P( )45X60 m- 50 - 501010= P(-.5 Z 1)z0 = 1z0 = -.5We need to find the shaded area25P(-.5Z0)+ P(0Z1)P(45X60) = P( )45X60 m- 50 - 501010z00.1.0.050.060.00.00000.00400.01990.02390.10.03980.04380.05960.636.1.00.34130.34380.35310.3554.P(0Z1= P(-.5Z1) =z=0z0 = 1z0 =-

7、.5.341326-z0+z00P(-z0Z0) = P(0Zz0)27z00.1.0.050.060.00.00000.00400.01990.02390.10.03980.04380.05960.636.0.50.1915.3413.5-.5.191528z00.1.0.050.060.00.00000.00400.01990.02390.10.03980.04380.05960.636.0.50.1915.1915.1915.1915.1915.3413.5-.5P(-.5Z1) = P(-.5Z0)+ P(0Z1) = .1915 + .3413 = .53281.02910%0%20

8、-2(i) P(X 0 ) = P(Z ) = P(Z2) =ZX.47720.5 - P(0Z2) = 0.5 - .4772 = .02283010%0%-1(ii) P(X 0 ) = P(Z ) 0 - 1010= P(Z1) =ZX.34130.5 - P(0Z1) = 0.5 - .3413 = .1587Find Normal Probabilities131zAA320.05Z0.0500.451.6450.05-Z0.05332222npq3435363700.511.522.5f(x) = 2e-2xf(x) = 1e-1xf(x) = .5e-.5x0 1 2 3 4 5

9、Exponential distribution for l = .5, 1, 200.511.522.5abP(axb) = e-la - e-lb 3839404142Compute Exponential probabilities43444500.050.10.150.2-6-4-2024600.050.10.150.2-6-5-4-3-2-10123456n n = 3n n = 104647Degrees of Freedom13.0786.31412.70631.82163.65721.8862.924.3036.9659.925.101.3721.8122.2282.7643.

10、169.2001.2861.6531.9722.3452.6011.2821.6451.962.3262.576tAt.100t.05t.025t.01t.005A = .05A = .05-tAThe t distribution issymmetrical around 0=1.812=-1.812484900.00020.00040.00060.00080.0010.00120.00140.00160.001805101520253035n n = 5n n = 105005101520253035Ac c2A5105101520253035=.05A =.99c2.995 c2.990 c2.05 c2.010 c2.005Ac2ATo find c2 for which P(c2nc2)=.01, lookup the column labeledc21-.01 or c2.99c c2 2.05.05