1. **Problem statement:** We have 5 executives with years of service: 20, 22, 26, 24, 28. We want to find how many samples of size 2 are possible using combinations. 2. **Formula:** The number of combinations of choosing $r$ items from $n$ is given by $$\binom{n}{r} = \frac{n!}{r!(n-r)!}$$ where $n=5$ and $r=2$. 3. **Calculate:** $$\binom{5}{2} = \frac{5!}{2!3!} = \frac{5 \times 4 \times 3!}{2 \times 1 \times 3!} = \frac{20}{2} = 10$$ 4. **Answer:** There are 10 possible samples of size 2. --- 1. **List all samples of size 2 and compute their means:** - (20, 22): mean = $\frac{20+22}{2} = 21$ - (20, 26): mean = $\frac{20+26}{2} = 23$ - (20, 24): mean = $\frac{20+24}{2} = 22$ - (20, 28): mean = $\frac{20+28}{2} = 24$ - (22, 26): mean = $\frac{22+26}{2} = 24$ - (22, 24): mean = $\frac{22+24}{2} = 23$ - (22, 28): mean = $\frac{22+28}{2} = 25$ - (26, 24): mean = $\frac{26+24}{2} = 25$ - (26, 28): mean = $\frac{26+28}{2} = 27$ - (24, 28): mean = $\frac{24+28}{2} = 26$ --- 1. **Sampling distribution of the means:** The means are: 21, 22, 23, 23, 24, 24, 25, 25, 26, 27. --- 1. **Population mean:** $$\bar{x} = \frac{20 + 22 + 26 + 24 + 28}{5} = \frac{120}{5} = 24$$ 2. **Mean of sample means:** $$\frac{21 + 22 + 23 + 23 + 24 + 24 + 25 + 25 + 26 + 27}{10} = \frac{240}{10} = 24$$ They are equal, illustrating the unbiased nature of sample means. --- 1. **Dispersion comparison:** - Population variance: $$\sigma^2 = \frac{(20-24)^2 + (22-24)^2 + (26-24)^2 + (24-24)^2 + (28-24)^2}{5} = \frac{16 + 4 + 4 + 0 + 16}{5} = \frac{40}{5} = 8$$ - Sample means variance: Calculate variance of sample means: mean is 24, deviations squared sum = $(21-24)^2 + (22-24)^2 + (23-24)^2 + (23-24)^2 + (24-24)^2 + (24-24)^2 + (25-24)^2 + (25-24)^2 + (26-24)^2 + (27-24)^2 = 9 + 4 + 1 + 1 + 0 + 0 + 1 + 1 + 4 + 9 = 30$; variance = $\frac{30}{10} = 3$ Sample means have less dispersion than the population. --- 1. **Normality of population distribution:** The population values (20, 22, 24, 26, 28) are evenly spaced and small in number, so the distribution is not bell-shaped or normal. 2. **Normality of sample means distribution:** The sample means show a more symmetric distribution and start to show a bell-shaped tendency, illustrating the Central Limit Theorem. --- **Final answers:** - a) 10 samples - b) Samples and means listed above - c) Sampling distribution: 21, 22, 23, 23, 24, 24, 25, 25, 26, 27 - d) Population mean = 24, mean of sample means = 24 - e) Population variance = 8, sample means variance = 3 (less dispersion) - f) Population distribution is not normal - g) Sample means distribution tends toward normality