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) = 012131415.71828. 2eand.14159. 3wherexe21) x ( f2x) 2/ 1 (16The norm

4、al distribution is bell shaped, and symmetrical around .Why symmetrical? Let = 100. Suppose x = 110. 2210) 2/ 1 (100110) 2/ 1 (e21e21)110( fNow suppose x = 902210) 2/ 1 (10090) 2/ 1 (e21e21)90( f1109017How does the standard deviation affect the shape of f(x)?= 2 =3 =4 = 10 = 11 = 12How does the expe

5、cted value affect the location of f(x)?182221?2xbap aXbedx192,NX dzzfbZapba2021xxXZE(Z) = = 0V(Z) = 2 = 1Every normal variablewith some and , canbe transformed into this Z.Therefore, once probabilities for Zare calculated, probabilities of any normal variable can be found.22P(45X60) = P( )45X60 - 50

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

7、z0 = 1z0 =-.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

8、.02910%0%20-2(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-.5x

9、0 1 2 3 4 5Exponential distribution for l = .5, 1, 200.511.522.5abP(axb) = e-la - e-lb 3839404142Compute Exponential probabilities43442/ ) 1(2t1)!2()!1() t ( f4500.050.10.150.2-6-4-2024600.050.10.150.2-6-5-4-3-2-10123456 = 3 = 104647Degrees of Freedom13.0786.31412.70631.82163.65721.8862.924.3036.965

10、9.925.101.3721.8122.2282.7643.169.2001.2861.6531.9722.3452.6011.2821.6451.962.3262.576tAt.100t.05t.025t.01t.005A = .05A = .05-tAThe t distribution issymmetrical around 0=1.812=-1.812480)(2!1)2/(1)(221)2/(22/22ef4900.00020.00040.00060.00080.0010.00120.00140.00160.001805101520253035 = 5 = 105005101520253035A 2A51Degrees offreedom10.00003930.0001571.3.841466.63497.87944.102.155852.55821.18.30723.209325.1882.05101520253035=.05A =.992.995 2.990 2.05 2.010 2.005A2ATo find 2 for which P(22)=.01, lookup the column labeled21-.01 or 2.99 2 2.05.05520FF1