Update Automated MNLP evaluation report (2026-06-01)

EVAL_REPORT.md

@@ -2,7 +2,7 @@
 
 - **Model repo:** [`cs-552-2026-flab/general_knowledge_model`](https://huggingface.co/cs-552-2026-flab/general_knowledge_model)
 - **Owner(s):** group **flab**
-- **Generated at:** 2026-05-29T06:54:15+00:00 (UTC)
+- **Generated at:** 2026-06-01T06:13:57+00:00 (UTC)
 - **Pipeline:** [mnlp-project-ci](https://github.com/eric11eca/mnlp-project-ci)
 
 _This PR is opened automatically by the course CI. It is **non-blocking** — you do not need to merge it. The next nightly run will refresh this file._
@@ -12,7 +12,7 @@ _This PR is opened automatically by the course CI. It is **non-blocking** — yo
 | Benchmark | Accuracy | Status |
 |---|---:|---|
 | Math | — | not run |
-| Knowledge | 0.2200 | ok |
+| Knowledge | 0.4400 | ok |
 | Multilingual | — | not run |
 | Safety | — | not run |
 
@@ -24,30 +24,124 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
 
 **Correct** (1 shown)
 
-- **reference**: `B`
+- **reference**: `D`
 - **overall** (1/1 completions correct)
-- **extracted** (✓): `B`
+- **extracted** (✓): `D`
 - **completion**:
 
     ```text
     <think>
+    Okay, let's try to figure out this problem. So, we have a solution that's 200.00 cm³, which is 0.200 liters. It contains 1.00 g of KH2PO4 and 1.00 g of Na2HPO4·2H2O. We need to find the concentration of orthophosphate ions, which I think is the phosphate ion, H2PO4^-? Wait, no. Wait, orthophosphate ions would be the phosphate ion, which is PO4^3-? Wait, no. Let me think. The problem says "
+    
+    ... [truncated 821 chars] ...
+    
+    The pH is determined using the Henderson-Hasselbalch equation for the buffer system:
+    
+    $$
+    \text{pH} = \text{p}K_{a2} + \log\left(\frac{[\text{HPO}_4^{2-}]}{[\text{H}_2\text{PO}_4^-]}\right)
+    $$
+    
+    - **pKa₂** (from the given Ka₂ = 6.2 × 10⁻⁸):
+      $$
+      \text{p}K_{a2} = -\log(6.2 \times 10^{-8}) \approx 7.21
+      $$
+    
+    - **Ratio:**
+      $$
+      \frac{[\text{HPO}_4^{2-}]}{[\text{H}_2\text{PO}_4^-]} = \frac{0.0282}{0.036735} \approx 0.768
+      $$
+    
+    - **pH:**
+      $$
+      \text{pH} = 7.21 + \log(0.768) \approx 7.21 - 0.113 = 7.097
+      $$
+    
+    ---
+    
+    ### Step 4: Use the Ka₃ Expression to Find [PO₄³⁻]
+    
+    The dissociation reaction is:
+    
+    $$
+    \text{HPO}_4^{2-} \rightleftharpoons \text{H}^+ + \text{PO}_4^{3-}
+    $$
+    
+    The equilibrium expression is:
+    
+    $$
+    K_{a3} = \frac{[\text{H}^+][\text{PO}_4^{3-}]}{[\text{HPO}_4^{2-}]}
+    $$
+    
+    Solving for [PO₄³⁻]:
     
-    </think>
+    $$
+    [\text{PO}_4^{3-}] = \frac{K_{a3} \cdot [\text{HPO}_4^{2-}]}{[\text{H}^+]}
+    $$
     
-    The correct option is \boxed{B}.
+    - **Kₐ₃ = 1.8 × 10⁻¹²**
+    - **[HPO₄²⁻] = 0.0282 M**
+    - **[H⁺] = 10⁻⁷·⁰⁹⁷ ≈ 8.13 × 10⁻⁸ M**
+    
+    $$
+    [\text{PO}_4^{3-}] = \frac{(1.8 \times 10^{-12}) \cdot 0.0282}{8.13 \times 10^{-8}} \approx \frac{5.076 \times 10^{-14}}{8.13 \times 10^{-8}} \approx 6.24 \times 10^{-7}\ \text{M}
+    $$
+    
+    ---
+    
+    ### Final Answer
+    
+    $$
+    \boxed{D}
+    $$
     ```
 
 **Incorrect** (1 shown)
 
-- **reference**: `D`
+- **reference**: `A`
 - **overall** (0/1 completions correct)
-- **extracted** (✗): `B`
+- **extracted** (✗): `H`
 - **completion**:
 
     ```text
     <think>
+    Okay, let's try to figure out the power spectral density (PSD) of the function f(t) given by the convolution of h(σ) and φ(t - σ). The question asks to express G_f(ω) in terms of G_φ(ω) and H(ω), where H(ω) is the Fourier transform of h(t). 
+    
+    First, I remember that the power spectral density is related to the autocorrelation function, but here we have a convolution. Let me recall the prope
+    
+    ... [truncated 822 chars] ...
+    
+    # Key Concepts
+    
+    1. **Convolution in Time Domain and Fourier Transform in Frequency Domain**:
+       - The Fourier transform of the convolution of two functions is the product of their individual Fourier transforms:
+         $$
+         \mathcal{F}\{f(t)\} = \mathcal{F}\{h(t)\} \cdot \mathcal{F}\{\phi(t)\} = H(\omega) \cdot \Phi(\omega).
+         $$
+    
+    2. **Power Spectral Density (PSD)**:
+       - The **PSD** of a signal is the **Fourier transform of its autocorrelation function**.
+       - For a linear system with impulse response $ h(t) $, the **PSD of the output** $ f(t) $ is given by:
+         $$
+         G_f(\omega) = |H(\omega)|^2 \cdot G_\phi(\omega),
+         $$
+         where:
+         - $ G_\phi(\omega) $ is the power spectral density of $ \phi(t) $,
+         - $ H(\omega) $ is the Fourier transform of $ h(t) $,
+         - $ |H(\omega)|^2 $ is the power spectral density of $ h(t) $.
+    
+    ---
+    
+    ### Final Answer
+    
+    From the above reasoning, the correct expression for the power spectral density of $ f(t) $ is:
+    
+    $$
+    G_f(\omega) = |H(\omega)|^2 \cdot G_\phi(\omega).
+    $$
     
-    </think>
+    However, none of the provided choices exactly match this expression. Among the given options, the **closest** and **most logically consistent** choice is:
     
-    The correct option is \boxed{B}.
+    $$
+    \boxed{H}
+    $$
     ```