1. **Problem 1 (Two marks):** Calculate the mean of the data set: 5, 8, 12, 20, 25. 2. **Solution:** - The mean is calculated by the formula: $$\text{Mean} = \frac{\sum x_i}{n}$$ - Here, $\sum x_i = 5 + 8 + 12 + 20 + 25 = 70$ and $n = 5$. - So, $$\text{Mean} = \frac{70}{5} = 14$$ 3. **Problem 2 (Two marks):** Find the median of the data set: 3, 7, 9, 15, 18. 4. **Solution:** - Arrange data in ascending order (already arranged). - Median is the middle value for odd number of observations. - Here, the middle value is the 3rd value: 9. - So, $$\text{Median} = 9$$ 5. **Problem 3 (Six marks):** A sample of 10 business sales figures (in thousands) is: 12, 15, 14, 10, 18, 20, 22, 17, 19, 16. Calculate the variance and standard deviation. 6. **Solution:** - Step 1: Calculate the mean: $$\text{Mean} = \frac{12 + 15 + 14 + 10 + 18 + 20 + 22 + 17 + 19 + 16}{10} = \frac{163}{10} = 16.3$$ - Step 2: Calculate squared deviations: $$(12-16.3)^2 = 18.49, (15-16.3)^2 = 1.69, (14-16.3)^2 = 5.29, (10-16.3)^2 = 39.69, (18-16.3)^2 = 2.89,$$ $$(20-16.3)^2 = 13.69, (22-16.3)^2 = 32.49, (17-16.3)^2 = 0.49, (19-16.3)^2 = 7.29, (16-16.3)^2 = 0.09$$ - Step 3: Sum of squared deviations: $$18.49 + 1.69 + 5.29 + 39.69 + 2.89 + 13.69 + 32.49 + 0.49 + 7.29 + 0.09 = 122.1$$ - Step 4: Variance (sample): $$s^2 = \frac{122.1}{10 - 1} = \frac{122.1}{9} \approx 13.57$$ - Step 5: Standard deviation: $$s = \sqrt{13.57} \approx 3.68$$ 7. **Problem 4 (Six marks):** Using the Central Limit Theorem, explain why the sampling distribution of the sample mean approaches normality as sample size increases, and calculate the probability that the sample mean of size 36 from a population with mean 50 and standard deviation 12 is greater than 53. 8. **Solution:** - The Central Limit Theorem states that for large sample sizes, the distribution of the sample mean approaches a normal distribution regardless of the population's distribution. - Given $\mu = 50$, $\sigma = 12$, and $n = 36$. - The standard error of the mean is: $$\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{12}{6} = 2$$ - Calculate the z-score for $\bar{x} = 53$: $$z = \frac{53 - 50}{2} = 1.5$$ - Using standard normal tables, $P(Z > 1.5) = 1 - 0.9332 = 0.0668$. - So, the probability that the sample mean is greater than 53 is approximately 6.68%.