1. **Problem statement:** We have a database with 100 records, 10 corrupted and 90 uncorrupted. A data analyst randomly selects 8 records. We want the probability distribution for the number of corrupted records selected, and then find the mean and variance of this distribution. 2. **Distribution type:** This is a hypergeometric distribution because we are sampling without replacement from two groups (corrupted and uncorrupted). 3. **Formula for hypergeometric probability:** $$P(X = k) = \frac{\binom{10}{k} \binom{90}{8-k}}{\binom{100}{8}}$$ where $k$ is the number of corrupted records selected, and $k$ can be from 0 to 8 (but limited by the number of corrupted records available). 4. **Calculate probabilities:** For $k=0,1,2,...,8$ (only up to 8 corrupted records or 10 corrupted records, whichever is smaller), calculate: $$P(X=k) = \frac{\binom{10}{k} \binom{90}{8-k}}{\binom{100}{8}}$$ 5. **Mean and variance formulas for hypergeometric distribution:** - Mean: $$\mu = n \frac{K}{N} = 8 \times \frac{10}{100} = 0.8$$ - Variance: $$\sigma^2 = n \frac{K}{N} \left(1 - \frac{K}{N}\right) \frac{N-n}{N-1} = 8 \times \frac{10}{100} \times \left(1 - \frac{10}{100}\right) \times \frac{100-8}{99}$$ 6. **Calculate variance:** $$\sigma^2 = 8 \times 0.1 \times 0.9 \times \frac{92}{99} = 8 \times 0.1 \times 0.9 \times 0.9292929 = 0.6687$$ 7. **Summary:** - Probability distribution: $P(X=k) = \frac{\binom{10}{k} \binom{90}{8-k}}{\binom{100}{8}}$ for $k=0,1,...,8$ - Mean: $0.8$ - Variance: approximately $0.669$ This gives the full probability distribution and its mean and variance for the number of corrupted records selected.