For a negative binomial distribution, the mean and variance are related bya.μ<σ2𝜇<𝜎2b.μ=σ2𝜇=𝜎2c.μ>σ2𝜇>𝜎2d.None of all

To solve this problem, we need to understand the properties of the negative binomial distribution.

1. Break Down the Problem

The mean (μ) and variance (σ²) of the negative binomial distribution must be compared to determine their relationship.

2. Relevant Concepts

For a negative binomial distribution with parameters $r$ (number of successes) and $p$ (probability of success):

• Mean (μ) is given by: $\mu = \frac{r(1-p)}{p}$
• Variance (σ²) is given by: $\sigma^2 = \frac{r(1-p)}{p^2}$

3. Analysis and Detail

From the formulas:

• The variance can be rewritten in terms of the mean: $\sigma^2 = \frac{1}{p} \mu$ This implies that the variance (σ²) will always be greater than the mean (μ) since $p$ is a probability (0 < p ≤ 1), causing $\frac{1}{p} > 1$.

4. Verify and Summarize

Thus, it can be confirmed that: $\mu < \sigma^2$

Final Answer

Therefore, the correct option is a. $\mu < \sigma^2$.