Automated MNLP evaluation report (2026-05-18)
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by zechen-nlp - opened
- EVAL_REPORT.md +40 -23
EVAL_REPORT.md
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- **Model repo:** [`cs-552-2026-emainelpe/general_knowledge_model`](https://huggingface.co/cs-552-2026-emainelpe/general_knowledge_model)
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- **Owner(s):** group **emainelpe**
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- **Generated at:** 2026-05-
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- **Pipeline:** [mnlp-project-ci](https://github.com/eric11eca/mnlp-project-ci)
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_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._
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| Benchmark | Accuracy | Status |
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| Math | — | not run |
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| Knowledge | 0.
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| Multilingual | — | not run |
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| Safety | — | not run |
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**Correct** (1 shown)
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- **reference**: `
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- **overall** (1/1 completions correct)
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- **extracted** (✓): `
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- **completion**:
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```text
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<think>
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Okay, so I
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First, I
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... [truncated
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</think>
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\boxed{
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```
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**Incorrect** (1 shown)
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- **reference**: `
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- **overall** (0/1 completions correct)
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- **extracted** (✗): `
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- **completion**:
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```text
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<think>
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Okay, so I
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First, the
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... [truncated 823 chars] ...
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```
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- **Model repo:** [`cs-552-2026-emainelpe/general_knowledge_model`](https://huggingface.co/cs-552-2026-emainelpe/general_knowledge_model)
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- **Owner(s):** group **emainelpe**
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- **Generated at:** 2026-05-18T02:30:25+00:00 (UTC)
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- **Pipeline:** [mnlp-project-ci](https://github.com/eric11eca/mnlp-project-ci)
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_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._
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| Benchmark | Accuracy | Status |
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|---|---:|---|
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| Math | — | not run |
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| Knowledge | 0.3900 | ok |
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| Multilingual | — | not run |
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| Safety | — | not run |
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**Correct** (1 shown)
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- **reference**: `A`
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- **overall** (1/1 completions correct)
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- **extracted** (✓): `A`
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- **completion**:
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```text
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<think>
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Okay, so I need to figure out the power spectral density of the function f(t), which is defined as the convolution of h(σ) and φ(t - σ) over all σ from negative infinity to positive infinity. The question gives me the power spectral density of φ(t) as Gφ(ω) and asks me to express Gf(ω) in terms of Gφ(ω) and the Fourier transform of h(t), which I'll denote as H(ω).
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First, I remember that t
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... [truncated 823 chars] ...
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H(ω) * Gφ(ω)|^2, which is |H(ω)|^2 * |Gφ(ω)|^2. So, that would be |H(ω)|^2 * |Gφ(ω)|^2. But wait, the question says "in terms of the power spectral density of φ(t) and the Fourier transform of h(t)." So, it's |H(ω)|^2 times Gφ(ω)^2, which would be option A or something else?
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Looking at the options, option A is Gf(ω) = |H(ω)|^2 * Gφ(ω)^2. But wait, in the question, it's written as Gf(ω) = ... so maybe it's not squared? Wait, no, the power spectral density is the square of the Fourier transform, so if the Fourier transform is H(ω) * Gφ(ω), then the PSD is |H(ω) * Gφ(ω)|^2, which is |H(ω)|^2 * |Gφ(ω)|^2. So that would be Gf(ω) = |H(ω)|^2 * Gφ(ω)^2.
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But looking at the options, option A is Gf(ω) = |H(ω)|^2 * Gφ(ω)^2, which seems to fit. Wait, but the question says "the power spectral density of f(t) in terms of the power spectral density of φ(t) and the Fourier transform of h(t)." So, it's |H(ω)|^2 times Gφ(ω)^2, which is exactly what option A is.
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Wait, but I thought the power spectral density of the convolution is the product of the power spectral densities. So, if f(t) = h(t) * φ(t), then Gf(ω) = |H(ω)|^2 * |Gφ(ω)|^2. So, that's correct. So, option A is correct.
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</think>
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\boxed{A}
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```
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**Incorrect** (1 shown)
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- **reference**: `C`
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- **overall** (0/1 completions correct)
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- **extracted** (✗): `<no answer>`
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- **completion**:
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```text
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<think>
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Okay, so I have this problem about probability and radioactive decay. Let me try to break it down step by step.
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First, the problem says that we can produce a single atom, which is an atomic nucleus, through beta decay. Beta decay is when a nucleus emits an electron, right? So, the nucleus starts off, and then it decays over time. The probability that this atom decays within 100 minutes is
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... [truncated 823 chars] ...
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*100) = 0.68 => λ = -ln(0.68)/100 ≈ 0.00385 per minute.
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Now, we need the probability of decay between 50 and 150 minutes, which is from t = 50 to t = 150 minutes.
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The probability of decay between t1 and t2 is P(t2) - P(t1).
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So, compute P(150) - P(50).
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P(150) = 1 - e^(-λ*150) = 1 - e^(-0.00385*150) = 1 - e^(-0.5775).
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Compute e^(-0.5775): e^(-0.5) = 0.6065, e^(-0.6) ≈ 0.5488, so 0.5775 is between 0.5 and 0.6.
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Let me approximate e^(-0.5775). Let's use the fact that ln(0.55) ≈ -0.5978, ln(0.56) ≈ -0.5798, ln(0.57) ≈ -0.5625.
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So, e^(-0.5775) is between 0.56 and 0.57.
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Let me compute it as follows: let x = -0.5775, so e^x = e^(-0.5775).
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We can use linear approximation between x = -0.5798 (which is ln(0.56)) and x = -0.5625 (ln(0.57)).
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Wait, no, actually, the value of e^(-0.5775) is between e^(-0.57) and e^(-0.58).
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Wait, perhaps it's easier to use a calculator-like approach. Since I don't have a calculator here, I'll approximate it.
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Let me note that e^(-0.5775) ≈ 1 / e^(0.5775).
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We know that e^0.5 ≈ 1.6487, e^0.6 ≈ 1.8221, e^0.5775 ≈ ?
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Let me use the Taylor series around a known point. Let's take a = 0.57, which is ln(0.57) ≈ -0.5625. Wait, this might not be the best approach
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```
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