Select the MOST appropriate statement for the population parameters of the Normal distribution shown below

Question

BlackTom AI · Accepted Answer

The normal PDF peak equals 1/(σ√(2π)), so the graph’s maximum value directly gives the standard deviation when the mean is at the center (x = 12.5).

1) Relevant concepts
- For N(μ, σ^2) the density at the mean μ is f(μ) = 1/(σ√(2π)).
- Variance = σ^2. Options give either σ or σ^2; we must match the plotted peak height and horizontal spread.

2) Step-by-step reasoning
- From the plot the center is at x = 12.5, so μ = 12.5.
- Read the peak height from the y-axis: the top grid line is 0.9 and the curve peak is about 0.8 (visually ≈0.8). Use f(μ) ≈ 0.8.
- Solve 1/(σ√(2π)) = 0.8 for σ.

3) Calculations (intermediate steps)
- √(2π) ≈ 2.5066.
- σ = 1 / (0.8 × 2.5066) = 1 / 2.0053 ≈ 0.4987 ≈ 0.5.
- Check consistency at x = 12 (one half unit left): z = (12 − 12.5)/0.5 = −1, so f(12) = (1/(0.5√(2π))) e^(−1/2) ≈ (1/1.2533)×0.6065 ≈ 0.4839, which matches the plotted value near 0.48–0.50. This confirms σ ≈ 0.5.

4) Why the correct answer is correct
- The peak height and the rate the curve falls off horizontally both match σ ≈ 0.5 and μ = 12.5. The computed σ from the peak equals 0.5, so the correct parameter statement is mean 12.5 and standard deviation 0.5.

5) Why other options are incorrect
- Mean 12.5 and variance 1.5 ⇒ σ = √1.5 ≈ 1.225: this would give a peak ≈ 1/(1.225·2.5066) ≈ 0.325, much lower than the plotted ~0.8.
- Mean 12.5 and σ = 1.5 ⇒ peak ≈ 1/(1.5·2.5066) ≈ 0.266, too low and the curve would be much wider than plotted.
- Mean 12.5 and variance 0.5 ⇒ σ = √0.5 ≈ 0.707, which yields peak ≈ 1/(0.707·2.5066) ≈ 0.565, noticeably lower than the observed ~0.8.

Conclusion: The most appropriate statement is Mean of 12.5 and Standard Deviation of 0.5, which matches the provided answer.

Select the MOST appropriate statement for the population parameters of the Normal distribution shown below单项选择题

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