Question 1 0/1 pt 9 3 99 0 Details An unfair die looks like an ordinary six-sided die but the outcomes are not equally likely. The probability distribution of the face value, X, is as follows: 1 2 3 4 5 16 Total P(X = xi) 0. 13 0. 18 0.37 0.15 0.13 0.04 To find the expected value the following table was set up: P(X = xi) ci . P(X = xi) 1 0.13 1 . 0.13 = 0.13 2 2 . 0.18 = 0.36 3 0.37 3 . = 1.11 4 0.15 4 . 0.15 = 5 5 - 0.13 = 0.65 6 0.04 Total: Fill out the blanks in the table above and compute the E[X]: E[X] = (Round the answer to 2 decimals) Question Help: Video B Written Example Message instructor D Post to forum . Question 2 Co/1 pt 9 3 99 0 Details A grocery store has 6 self-checkout stations. The probability distribution of the number of utilized stations, X, is as follows: 2 3 4 5 6 Total P(X = Di) 0. 13 0.17 0.37 0.16 0. 15 0.02 Compute the expected value: E[X] = (Round the answer to 2 decimals) Question Help: Video G Written Example Message instructor D Post to forum . Question 3 0/1 pt 9 3 99 0 Details The household size in a certain neighborhood varies from 1 to 6 The probability distribution of the size of a random household, X, is as follows: Li 6 Total P ( X = =1) . 0.19 0.44 0.17 0.1 The expected value is computed, E X] = 2.98. To find the variance and standard deviation the following table was set up: P(X = Ii) ( ; - E(X]) 2 (Ii - E[X]) P( X = Ii) 0.1 (1 - 2.98)2 = (-1.98) = 3.9204 3.9204 . 0.1 = 0.392 2 (2 - 2.98)2 = (-0.98)2 = 0.9604 0.9604 . 0.19 = 0.1825 3 0.44 (3 - 2.98)2 = (0.02)2 = 0.0004 . 0.44 = 0.0002 4 0.17 (4 - 2.98)2 = (1.02)2 = 1.0404 1.0404 . 0.17 = 5 (5 - 2.98)2 = (2.02) = 4.0804 4.0804 . 0.1 = 0.408 6 9.1204 . 0 = 0 E[X] = 2.98 Total: Fill out the blanks (round to 4 decimals) in the table above and compute the VAR[X] and SD[X]: VAR[X] = (Round the answer to 4 decimals) SD[X] = (Round the answer to 2 decimals) Question Help: Video B Written Example Message instructor D Post to forum . Question 4