Question

solve to check work please

4. Compute the mean and the Standard deviation of the probability distribution table of: х Y 0 1 2 3 0.2 0.5 0.2 0.1 I 4.B. Compute the variance square and Standard Deviation for the POISSON Distribution of: LAMBDA -2.5 , t 3 hr. and then find P(3) ?

Answer 1

4) bolur. Probability distribution is given by : Х 0 2 3 Y = P(x=de) 0.2 0.2 0.1 :. mean = EXCP (X=2)) = 0*0:2+ 1*0.5 +2*0.2 +3*0:1 oto 5 +0.4+0.3 .. Mean = X = E(X)= 1.2 2 : E(X2) = E x2* P(X=) 0²40.2 + 1 24.5+ 2 + 0.2 + 3 + 0.1 = 0+0.5 +0.8 +o.g E (X²) = 2.2 Vor(x) = E(X2)-( E(X) = 2.2-(1-22 = 2.2-1.44 .: Vai (x) = 616 : Standard deviation Varcx) 50.76 = 0.87 178 Mean= 1.2 and standard deviation standard deviation = 0.87178

4.B) Xa poisson (LAMBOA = 2.5) « p (=2,5) Vaviance = 0 = 2:5 variance square= d = (2-5)2= 6:25 Standard deviations d 12.5 = 15811 Now we know pdf of poisson is given by P ( T=) - e-do Ł: 0,1,2, t P(39 = P(T-3) = € 25x12.579 31, e-2.5 * 15.625 1 * 2 * 3 = 0.213763 P(3) = 0.213763