University Physics V · topic UPV-01.02
Probability distributions, moments, and expectation values
Connect the ideas, make your working visible, and test the result. Use the hints when you need a starting point; open the full solution after your attempt.
Further practice
Work through a fresh case before tackling the extension.
Probability distributions, moments and expectation values: predict a finite change
A probability distribution assigns each outcome its long-run share over identical trials — dimensionless Pₙ for discrete outcomes, a density ρ(x) with the inverse unit of x for continuous ones — and moments ⟨xᵏ⟩ = ∫xᵏρ dx compress that weight into location, spread and shape. Use the supplied topic model Y_out = ((1-w)*exp(-2*(s*w/sig)^2)+w*exp(-2*(s*(1-w)/sig)^2))/(sig*sqrt(2*pi)), where Y_out is DENSITY AT MEAN (nm⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; exp(x) means eˣ. s is Hump position ±s, measured in nm. Keep w (Right-hump weight w) = 0.5; sig (Hump width σₕ) = 0.35 nm fixed. For this exercise use only 0.2125 nm ≤ s ≤ 1.7875 nm on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. Calculate Y_out at s = 0.559 nm and 0.7795 nm. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δs across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
DENSITY AT MEAN against s
Predictions from the supplied Probability distributions, moments and expectation values model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): DENSITY AT MEAN / nm⁻¹
Horizontal axis (x): s: Hump position ±s / nm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| s: Hump position ±s / nm | DENSITY AT MEAN / nm⁻¹ |
|---|---|
| 0.2125 | 0.9479745404095744 |
| 0.60625 | 0.2542910383900601 |
| 1 | 0.019240284214943552 |
| 1.3937499999999998 | 0.0004106179753637183 |
| 1.7874999999999999 | 0.0000024717837787474024 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of s separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At s = 0.559 nm, substitute into Y_out = ((1-w)*exp(-2*(s*w/sig)^2)+w*exp(-2*(s*(1-w)/sig)^2))/(sig*sqrt(2*pi)): Y₀ = 0.318367 nm⁻¹.
- At s = 0.7795 nm, the same fixed-parameter model gives Y₁ = 0.0954456 nm⁻¹.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.222922 nm⁻¹; the input change is Δs = 0.2205 nm.
- Average sensitivity is ΔY_out/Δs = -1.01098 nm⁻¹ per nm. Its negative sign means that this response decreases over the specified interval.
- This is a secant slope between two settings. It equals a point derivative only for a locally linear relation or in a justified small-interval limit.
Answer Y₀ = 0.318367 nm⁻¹, Y₁ = 0.0954456 nm⁻¹; ΔY_out = -0.222922 nm⁻¹; average sensitivity = -1.01098 nm⁻¹ per nm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Extra challenging questions
Work through the linked stages and explain what each result means.
Probability distributions, moments and expectation values: infer an input with uncertainty
A probability distribution assigns each outcome its long-run share over identical trials — dimensionless Pₙ for discrete outcomes, a density ρ(x) with the inverse unit of x for continuous ones — and moments ⟨xᵏ⟩ = ∫xᵏρ dx compress that weight into location, spread and shape. Use the supplied topic model Y_out = ((1-w)*exp(-2*(s*w/sig)^2)+w*exp(-2*(s*(1-w)/sig)^2))/(sig*sqrt(2*pi)), where Y_out is DENSITY AT MEAN (nm⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; exp(x) means eˣ. s is Hump position ±s, measured in nm. Keep w (Right-hump weight w) = 0.5; sig (Hump width σₕ) = 0.35 nm fixed. For this exercise use only 0.2125 nm ≤ s ≤ 1.7875 nm on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. A reading gives Y_out = 0.0041379811 nm⁻¹, with a bounded reading uncertainty ±0.00062033 nm⁻¹. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown s by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0001575 nm, using a numerical bracket if required. Find the full input interval consistent with the bounded reading uncertainty and compare it with a symmetric input-error assumption. Do not assume that the supplied calibration points prove monotonicity outside this branch.
DENSITY AT MEAN against s
Predictions from the supplied Probability distributions, moments and expectation values model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): DENSITY AT MEAN / nm⁻¹
Horizontal axis (x): s: Hump position ±s / nm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| s: Hump position ±s / nm | DENSITY AT MEAN / nm⁻¹ |
|---|---|
| 0.2125 | 0.9479745404095744 |
| 0.60625 | 0.2542910383900601 |
| 1 | 0.019240284214943552 |
| 1.3937499999999998 | 0.0004106179753637183 |
| 1.7874999999999999 | 0.0000024717837787474024 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
The model's endpoint predictions bracket the reading. Solve Y_out(s) = Y_measured and then solve again at Y_measured − u and Y_measured + u; a nonlinear inverse can give unequal input deviations.
Reveal the worked solution
- Endpoint predictions are Y_out(0.2125) = 0.947975 nm⁻¹ and Y_out(1.7875) = 0.00000247178 nm⁻¹. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives s_line = 0.2125 + (0.00413798 − 0.947975)(1.7875 − 0.2125)/(0.00000247178 − 0.947975) = 1.78063 nm. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives s = 1.17325 nm; substituting back gives Y_out = 0.0041379811 nm⁻¹.
- The interpolation error in the input is |s_line − s| = 0.607379 nm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.0035176511 nm⁻¹ and 0.0047583111 nm⁻¹. Sort the roots even if the response decreases: 1.15857 nm ≤ s ≤ 1.19009 nm.
- The lower and upper input deviations are 0.0146764 nm and 0.016837 nm. They differ because the inverse response is nonlinear; a symmetric error bar would only be a local approximation. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer s ≈ 1.17325 nm; compatible interval [1.15857, 1.19009] nm. Endpoint interpolation gives 1.78063 nm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.