Question #188184

A researcher is conducting a study about the effectiveness of a new

product to the households in a certain Barangay. His population

consists of the numbers 8, 4, 2, 1, 13, and 10. Perform the following

tasks:

1. Determine the number of sets of all possible random samples using the

combination formula

2. List all the possible samples and compute the mean of each sample

3. Construct the sampling distribution.

4. Construct the histogram


Expert's answer

1.

Number=N!n!(Nn)!n=1Number=6!1!(61)!=61=6n=2Number=6!2!(62)!=5×62=15n=3Number=6!3!(63)!=4×5×62×3=20n=4Number=6!4!(64)!=3×4×5×62×3×4=15n=5Number=6!5!(65)!=2×3×4×5×62×3×4×5=6n=6Number=6!6!(66)!=11=1Number = \frac{N!}{n!(N-n)!} \\ n = 1 \\ Number = \frac{6!}{1!(6-1)!} = \frac{6}{1} = 6 \\ n = 2 \\ Number = \frac{6!}{2!(6-2)!} = \frac{5 \times 6}{2} = 15 \\ n = 3 \\ Number = \frac{6!}{3!(6-3)!} = \frac{4 \times 5 \times 6}{2 \times 3} = 20 \\ n = 4 \\ Number = \frac{6!}{4!(6-4)!} = \frac{3 \times 4 \times 5 \times 6}{2 \times 3 \times 4} = 15 \\ n = 5 \\ Number = \frac{6!}{5!(6-5)!} = \frac{2 \times 3 \times 4 \times 5 \times 6}{2 \times 3 \times 4 \times 5} = 6 \\ n = 6 \\ Number = \frac{6!}{6!(6-6)!} = \frac{1}{1} = 1

2.






3. The sampling distribution



4.


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