A company maintains three offices in a certain region each s

A company maintains three offices in a certain region, each staffed by two employees. Information concerning yearly salaries (1000s of dollars) is as follows:

(a) Suppose two of these employees are randomly selected from among the six (without replacement). Determine the sampling distribution of the sample mean salary

X.

(Enter your answers for p(x) as fractions.)

(b) Suppose one of the three offices is randomly selected. Let X₁ and X₂ denote the salaries of the two employees. Determine the sampling distribution of

X.

(Enter your answers as fractions.)

(c) How does

E(X)

from parts (a) and (b) compare to the population mean salary ??

E(X)

from part (a) is  ---Select--- greater than less than equal to ?, and

E(X)

from part (b) is  ---Select--- greater than less than equal to ?.

Solution

Suppose two of these employees are randomly selected from among the six (without replacement). Determine the sampling distribution of the sample mean salary .
S={(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,3),(2,4),(2,5),(2,6),(3,1),(3,2),(3,4),(3,5),(3,6), (4,1),(4,2),(4,3),(4,5),(4,6),(5,1),(5,2),(5,3),(5,4),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5)}. There are 30 outcomes.

21.65   when (1,2),(2,1),(1,4),(4,1),(2,6),(6,2),(4,6),(6,4) with 8 outcomes
19.95   when (1,3),(3,1),(3,6),(6,3) with 4 outcomes
17.75      when (1,5),(5,1),(5,6),(6,5) with 4 outcomes
19.70     when (1,6),(6,1),(4,5),(5,4),(2,5),(5,2) with 6 outcomes
21.90   when (2,3),(3,2),(3,4),(4,3) with 4 outcomes
23.60   when (2,4),(4,2) with 2 outcomes
18.0   when (3,5),(5,3) with 2 outcomes
:   17.75   18   19.7   19.95   21.65   21.90   23.60   otherwise
:   4/30   2/30   6/30   4/30   8/30   4/30   2/30   0
E( )= =613/30=20.43

b) Suppose one of the three offices is randomly selected. Determine the sampling distribution of the sample mean salary .
S={(1,2),(2,1),(3,4),(4,3),(5,6),(6,5)}. There are 6 outcomes

21.65 when (1,2),(2,1) with probability 1/3
21.90 when (3,4),(4,3) with probability 1/3
17.75 when (5,6),(6,5) with probability 1/3
E( )=61.3/3=20.43
Population mean = (19.7+23.6+20.2+23.6+15.8+19.7)/6=20.43