Mass — Question 9

Question 9

Hard measurement bounds produce an interval of possible masses

A calibration supplies independent hard bounds $9.8\le F\le10.2$ N and $1.9\le a\le2.1$ m/s$^2$. The vectors are collinear in the same direction, and $m=F/a$.

1. Find the tight lower and upper mass bounds.

2. Compare the central-value estimate with the interval.

3. Why should the interval not automatically be called a one-standard-deviation uncertainty?

Question 9 - Solution

Hard measurement bounds produce an interval of possible masses

(a) Use monotonicity of the ratio.

$$\boxed{m_{\min}=\frac{F_{\min}}{a_{\max}}=\frac{9.8}{2.1}=4.667\,\mathrm{kg}},\quad m_{\max}=\frac{10.2}{1.9}=5.368\,\mathrm{kg}$$ The minimum is $9.8/2.1=4.667$ kg and maximum is $10.2/1.9=5.368$ kg. Both extreme combinations are allowed by independent hard bounds.

(b) Compute the central estimate separately.

$$m_0=\frac{10}{2}=5\,\mathrm{kg},\quad m-m_0\in[-0.333,0.368]\,\mathrm{kg}$$ The nominal ratio is $10/2=5$ kg, inside the interval. The deviations from 5 kg are about $-0.333$ and $+0.368$ kg, not exactly symmetric.

(c) Preserve the meaning of the input.

$$m\in\left[\frac{F_{\min}}{a_{\max}},\frac{F_{\max}}{a_{\min}}\right]\quad(F,a>0)$$ Hard bounds assert allowed ranges, not a probability distribution. No standard deviation or confidence level is supplied, so statistical language would add an unsupported assumption.