1.) Why or how are the three approaches equivalent (Arithmetic Mean, Weighted mean, Probabilistic Approach), and in what way?

2.) Discuss briefly in what way the expected value is the same as the mean of means = weighted mean of means; and in what way are they different.

3.) Reflections on the question.

Please do not answer if you are unsure. Please also refer to the attached photo for reference.

Let X's = A=1; B=2; C=5; D=5; E=8; and F=9

\fCOMBINATION (N =6 and n = 2) Sample Mean Sample Mean - E(X) [Sample Mean - E(x)]2 6C2 = 15 (Computation) (Computation) (Computation) AB (1+2)/2 = 1.5 1.5 - 5= -3.500 12.250 2 AC ( 1+5)/2 = 3.0 3.0 - 5= -2.000 4.000 AD ( 1+5)/2 =3.0 3.0-5= -2.000 4.000 4 AE (1+8)/2 =4.5 4.5 -5= -0.500 0.250 AF ( 1+9)/2 =5.0 5.0 - 5= 0.000 0.000 6 BC (2+5)/2 =3.5 3.5 -5= -1.500 2.250 BD ( 1+5)/2 =3.5 3.5 -5= -1.500 2.250 8 BE (2+8)/2 =5.0 5.0 - 5= 0.000 0.000 BF (2+9)/2 =5.5 5.5 - 5= 0.500 0.250 10 CD ( 5+5)/2 =5.0 5.0 - 5= 0.000 0.000 11 CE (5+8)/2 =6.5 6.5 - 5= 1.500 2.250 12 CF (5+9)/2 =7.0 7.0 -5= 2.000 4.00 13 DE ( 5+8)/2 =6.5 6.5 -5= 1.500 2.250 14 DF (5+9)/2 =7.0 7.0 - 5= 2.000 4.000 15 EF (8+9)/2 = 8.5 8.5 - 5= 3.500 12.250 75 0.000 50.000 SUM 6C2 15 15 Mean of Means: E(X) = 5 Variance of the Sample Mean: 02 = 3.333 Standard Deviation of the Sample Mean: mean = 1.826