1. **Problem statement:** How many different samples of size $n=3$ can be selected from populations of sizes $N=4, 8, 20, 50, 15$? 2. **Formula used:** The number of different samples of size $n$ from a population of size $N$ without replacement is given by the combination formula: $$\text{Number of samples} = \binom{N}{n} = \frac{N!}{n!(N-n)!}$$ 3. **Calculations:** - For $N=4$, $n=3$: $$\binom{4}{3} = \frac{4!}{3!\cancel{1!}} = \frac{4 \times 3!}{3!} = 4$$ - For $N=8$, $n=3$: $$\binom{8}{3} = \frac{8!}{3!5!} = \frac{8 \times 7 \times 6 \times 5!}{3! \times 5!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$$ - For $N=20$, $n=3$: $$\binom{20}{3} = \frac{20 \times 19 \times 18}{3 \times 2 \times 1} = 1140$$ - For $N=50$, $n=3$: $$\binom{50}{3} = \frac{50 \times 49 \times 48}{3 \times 2 \times 1} = 19600$$ - For $N=15$, $n=3$: $$\binom{15}{3} = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} = 455$$ --- 1. **Problem statement:** List all possible samples of size 2 from the population $\{2,3,6,8,11\}$ and compute their means. 2. **Samples and means:** - Sample: $(2,3)$, Mean: $\frac{2+3}{2} = 2.5$ - Sample: $(2,6)$, Mean: $4$ - Sample: $(2,8)$, Mean: $5$ - Sample: $(2,11)$, Mean: $6.5$ - Sample: $(3,6)$, Mean: $4.5$ - Sample: $(3,8)$, Mean: $5.5$ - Sample: $(3,11)$, Mean: $7$ - Sample: $(6,8)$, Mean: $7$ - Sample: $(6,11)$, Mean: $8.5$ - Sample: $(8,11)$, Mean: $9.5$ 3. **Constructing the sampling distribution of the sample means:** - List unique sample means and count their frequencies: - $2.5$: 1 - $4$: 1 - $4.5$: 1 - $5$: 1 - $5.5$: 1 - $6.5$: 1 - $7$: 2 - $8.5$: 1 - $9.5$: 1 - Total samples: 10 - Probability for each mean = Frequency / Total samples | Sample Mean | Frequency | Probability | |-------------|-----------|-------------| | 2.5 | 1 | 0.1 | | 4 | 1 | 0.1 | | 4.5 | 1 | 0.1 | | 5 | 1 | 0.1 | | 5.5 | 1 | 0.1 | | 6.5 | 1 | 0.1 | | 7 | 2 | 0.2 | | 8.5 | 1 | 0.1 | | 9.5 | 1 | 0.1 |