1. **Problem statement:** Tim wants to sample 160 fish from a lake containing 450 mirror carp, 300 leather carp, and 850 common carp to investigate their health. 2. **(a) Why stratified random sampling cannot be used:** Stratified random sampling requires the population to be divided into distinct strata (groups) and then samples are taken from each stratum proportionally or equally. However, if the fish cannot be clearly identified or separated into types before sampling, stratified sampling is not possible. 3. **(b) Sampling method to fairly represent each carp type:** To ensure fair representation, Tim can use **quota sampling** or **proportional sampling** where the sample size from each carp type is proportional to their population size. The sample sizes would be: - Mirror carp: $\frac{450}{1600} \times 160 = 45$ - Leather carp: $\frac{300}{1600} \times 160 = 30$ - Common carp: $\frac{850}{1600} \times 160 = 85$ This method is called **proportional stratified sampling** if the strata are known, or **quota sampling** if the strata are identified during sampling. 4. **(c) Estimate standard deviation of weight:** Given: - $\sum fm = 692$ - $\sum fm^2 = 3053$ - Total frequency $n = 160$ Mean weight $\bar{x} = \frac{\sum fm}{n} = \frac{692}{160} = 4.325$ Variance estimate: $$s^2 = \frac{\sum fm^2}{n} - \bar{x}^2 = \frac{3053}{160} - (4.325)^2 = 19.08125 - 18.7026 = 0.37865$$ Standard deviation: $$s = \sqrt{0.37865} \approx 0.615$$ 5. **(d) Effect of correcting transposed weights:** (i) Changing 2.3 to 3.2 increases the weight value, which increases the mean and variance slightly, so the estimated standard deviation will **increase**. (ii) Changing 4.6 to 6.4 increases the weight value significantly, increasing the spread of data, so the estimated standard deviation will **increase more noticeably**. **Reason:** Increasing values, especially outliers, increase the spread (variance) of the data, thus increasing the standard deviation.