The binomial distribution models the number of successes \( k \) in \( n \) independent trials, each with success probability \( p \). Its probability mass function (PMF) is: \[ P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k} \] The exponential family form for a distribution is: \[ p(x | \theta) = h(x) \exp\left(\eta(\theta) \cdot T(x) - A(\theta)\right) \] To express the binomial distribution in this form, rewrite the PMF as: \[ P(X = k) = \binom{n}{k} (1 - p)^n \left(\frac{p}{1 - p}\right)^k = \binom{n}{k} (1 - p)^n \exp\left(k \ln\left(\frac{p}{1 - p}\right)\right) \] Question:
What is the natural parameter \( \eta(p) \) for the binomial distribution when \( p = 0.6 \)? Round to 2 decimal places.

1. Identify \( \eta(p) \) from the exponential form.
2. Compute \( \eta(p) \) for \( p = 0.6 \).