1. **Problem Statement:** We have a sample of complaints per day over 10 days: 3, 5, 2, 4, 6, 3, 4, 7, 5, 4. We want to find the Maximum Likelihood Estimate (MLE) of the Poisson parameter $\lambda$, which represents the average rate of complaints per day. 2. **Formula for MLE of $\lambda$ in Poisson Distribution:** The MLE for $\lambda$ is the sample mean: $$\hat{\lambda} = \frac{1}{n} \sum_{i=1}^n X_i$$ where $X_i$ are the observed counts and $n$ is the number of observations. 3. **Calculate the sample mean:** Sum of complaints: $$3 + 5 + 2 + 4 + 6 + 3 + 4 + 7 + 5 + 4 = 43$$ Number of days $n = 10$ 4. **Compute MLE:** $$\hat{\lambda} = \frac{43}{10} = 4.3$$ 5. **Interpretation:** The MLE estimate of $\lambda$ is 4.3 complaints per day, meaning the average daily complaints rate is estimated to be 4.3 based on the observed data. **Final answer:** $$\hat{\lambda} = 4.3$$