In statistical inference, a point estimate is a single numerical value calculated from sample observations that serves as the best single guess or estimate of an unknown true population parameter (e.g. sample mean $\bar{x}$ as a point estimate of population mean $\mu$). In contrast, an estimator is the rule or mathematical formula used, while an interval estimate provides a range of values.

Option analysis:

• Statistic: Any numerical characteristic of a sample, broader than an estimate.
• Point estimate: Correct, the specific single numerical value assigned to an unknown parameter.
• Variable: A characteristic that varies across units.
• Estimator: The rule or mathematical function (random variable) used to calculate the estimate.

Karl Pearson's empirical relationship between mean, median, and mode for a moderately skewed unimodal distribution is:

Rearranging in terms of differences from the mean:

Given that $\text{Mean} - \text{Mode} = 36$:

Option analysis:

• $6$: Arithmetic miscalculation.
• $12$: Correct, calculated as $36 / 3 = 12$.
• $24$: Arithmetic miscalculation.
• $36$: Value of Mean - Mode, not Mean - Median.

Covariance is the fundamental statistical measure of the joint variability between two random variables, defined as $\text{Cov}(X,Y) = E[(X - E[X])(Y - E[Y])]$. It evaluates the extent to which two variables vary together. While correlation is the dimensionless, standardized measure of linear association (scaled between $-1$ and $+1$), covariance is the raw mathematical measure of joint co-movement.

Option analysis:

• Covariance: Correct, standard statistical definition for joint variability of two random variables.
• Correlation: Standardized covariance, measure of strength of linear relationship.
• Regression: Predicts the dependent variable from independent predictor(s).
• Coefficient of determination: $R^2$, proportion of variance explained by regression model.

The two draws from separate ponds are statistically independent events:

• Probability of drawing a marked fish from the first pond:
  $$P(A) = \frac{40}{200} = \frac{1}{5}$$
• Probability of drawing a marked fish from the second pond:
  $$P(B) = \frac{50}{300} = \frac{1}{6}$$
• Probability that both fishes are marked:
  $$P(A \cap B) = P(A) \times P(B) = \frac{1}{5} \times \frac{1}{6} = \frac{1}{30}$$
  
  Option analysis:
• $1/90$: Incorrect calculation.
• $1/60$: Incorrect calculation.
• $1/30$: Correct, $(1/5) \times (1/6) = 1/30$.
• $1/11$: Incorrect addition/ratio error.