Here are the solutions to the questions: a) Calculate a 96% confidence interval for the mean weight of all cabbages. Step 1: Calculate the sample mean and sample standard deviation. Given: Sample size, $n = 100$ Sum of weights, $\sum x = 16380$ Sum of squared weights, $\sum x^2 = 2,686,310$ Sample mean: $$ \bar{x} = \frac{\sum x}{n} = \frac{16380}{100} = 163.8 $$ Sample variance: $$ s^2 = \frac{1}{n-1} \left( \sum x^2 - \frac{(\sum x)^2}{n} \right) $$ $$ s^2 = \frac{1}{100-1} \left( 2,686,310 - \frac{(16380)^2}{100} \right) $$ $$ s^2 = \frac{1}{99} \left( 2,686,310 - \frac{268304400}{100} \right) $$ $$ s^2 = \frac{1}{99} \left( 2,686,310 - 2,683,044 \right) $$ $$ s^2 = \frac{3266}{99} \approx 33.00 $$ Sample standard deviation: $$ s = \sqrt{33.00} \approx 5.7446 $$ Step 2: Determine the critical z-value. For a 96% confidence level, $\alpha = 1 - 0.96 = 0.04$. The critical value is $z_{\alpha/2} = z_{0.02}$. Looking up the z-table for an area of $1 - 0.02 = 0.98$, we find $z_{0.02} \approx 2.05$. Step 3: Calculate the margin of error. The margin of error ($E$) is given by: $$ E = z_{\alpha/2} \frac{s}{\sqrt{n}} $$ $$ E = 2.05 \times \frac{5.7446}{\sqrt{100}} $$ $$ E = 2.05 \times \frac{5.7446}{10} $$ $$ E = 2.05 \times 0.57446 \approx 1.1776 $$ Step 4: Construct the confidence interval. The confidence interval is $\bar{x} \pm E$: $$ 163.8 \pm 1.1776 $$ Lower bound: $163.8 - 1.1776 = 162.6224$ Upper bound: