Step 1: Rather than listing formulas from memory, think about what mean equal to variance represents statistically. It means the spread of the data grows in exact step with the average count, which happens naturally in a process of rare, independent events counted over a fixed interval.
Step 2: This description matches a Poisson process exactly, since a Poisson variable counts rare independent events (defects, arrivals, germination failures) in a fixed interval, and both its average count and its spread are governed by the single rate parameter $\lambda$, forcing mean and variance to be numerically identical.
Step 3: Contrast this with the Binomial distribution, where there is an upper cap n on the number of successes, and the variance $np(1-p)$ is always squeezed smaller than the mean $np$ because of the $(1-p)$ shrinkage factor.
Step 4: The Normal distribution has mean and variance as two independently chosen parameters, so there is no built-in reason they should match. The Chi-square distribution, being a sum of squared standard normal variables with k degrees of freedom, always has variance $2k$, double its mean $k$. The F-distribution mixes two degrees of freedom in a ratio and its variance never reduces to its mean.
Step 5: Of all five, only the Poisson distribution has this intrinsic mean-variance equality, confirming option 3.\[\boxed{\text{Poisson: mean} = \text{variance} = \lambda}\]