1. **Problem Statement:** We have a sample of 10 workers chosen at random from a university where 94% of workers belong to the union. We want to find: (a) The expected number of union members in the sample. (b) The standard deviation of the number of union members in the sample. 2. **Relevant Distribution:** This is a binomial distribution problem where: - $n = 10$ (number of trials or workers sampled) - $p = 0.94$ (probability a worker is a union member) 3. **Formulas:** - Mean (expected value) of binomial distribution: $$E(X) = np$$ - Standard deviation of binomial distribution: $$\sigma = \sqrt{np(1-p)}$$ 4. **Calculations:** (a) Calculate the mean: $$E(X) = 10 \times 0.94 = 9.4$$ (b) Calculate the standard deviation: $$\sigma = \sqrt{10 \times 0.94 \times (1 - 0.94)} = \sqrt{10 \times 0.94 \times 0.06} = \sqrt{0.564} \approx 0.751$$ 5. **Interpretation:** - On average, we expect 9.4 union members in the sample of 10 workers. - The standard deviation of about 0.751 quantifies the typical variation around this mean number. This means most samples of 10 workers will have around 9 or 10 union members, consistent with the high union membership rate.