Answer:
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Step-by-step explanation:
The images for the complete question is attached to this solution.
The set S for the variables contains {5, 8, 11, 14}
For the (a) part, the set of possible samples of 2 without replacement include
(5,8), (5,11), (5,14), (8,5), (8,11), (8,14), (11,5), (11,8), (11,14), (14,5), (14,8), (14,11)
Note that it is impossible to pick a variable twice because the sampling is done without replacement.
- There are 12 different possible samples in this case.
Mean = x = μₛₐ = (x₁ + x₂)/2
The sample means of the sampling distribution include
(6.5), (8.0), (9.5), (6.5), (9.5), (11.0), (8.0), (9.5), (12.5), (9.5), (11.0), (12.5)
Sample means = x = μₛₐ; frequency = f
The sampling distribution of the means is now expressed in the table below.
The probabilities, p, are obtained from the relative frequency of each sample mean.
p = (f/Σf)
Σf = 12
μₛₐ | f | p
6.5 | 2 | (1/6)
8.0 | 2 | (1/6)
9.5 | 4 | (1/3)
11.0 | 2 | (1/6)
12.5| 2 | (1/6)
b) For random sample of 2 with replacements, the possible samples include
(5,5), (5,8), (5,11), (5,14), (8,5), (8,8), (8,11), (8,14), (11,5), (11,8), (11,11), (11,14), (14,5), (14,8), (14,11), (14,14)
- Note that it is possible to pick a variable twice because the sampling is done with replacement.
- There are truly 16 different possible samples in this case.
Mean = x = μₛₐ = (x₁ + x₂)/2
The sample means of the sampling distribution include
(5.0), (6.5), (8.0), (9.5), (6.5), (8.0), (9.5), (11.0), (8.0), (9.5), (11.0), (12.5), (9.5), (11.0), (12.5), (14.0)
Sample means = x = μₛₐ; frequency = f
The sampling distribution of the means is now expressed in the table below.
The probabilities, p, are obtained from the relative frequency of each sample mean.
p = (f/Σf)
Σf = 16
μₛₐ | f | p
5.0 | 1 | (1/16)
6.5 | 2 | (1/8)
8.0 | 3 | (3/16)
9.5 | 4 | (1/4)
11.0 | 3 | (3/16)
12.5| 2 | (1/8)
14.0| 1 | (1/16)
Hope this Helps!!!
