4. (20 points) The length of a component is to be estimated through repeated mea

Question

4. (20 points) The length of a component is to be estimated through repeated measure- ment. (a) Twelve (12) independent measurements are made with an instrument whose un- denote the average of these measurements. Find the certainty is 0.1 mim. Let uncertainty in X. (b) A new measuring device, whose uncertainty is only 0.05 mm, becomes available. Eight (8) independent measurements are made with this device. Let Y denote the average of these measurements. Find the uncertainty in Y. (c) In order to decrease the uncertainty still further, it is decided to combine the and Y. One engineer suggests estimating the length with L1 = . A second engineer argues that since X is based on 12 measurements, estimates while is based on only 8, a better estimate is L2-8, uncertainty in each of these estimates. Which is smaller Find the d) Find the value c such that the weighted average L cX+ (1-c)Y has minimum uncertainty. Find the uncertainty in L3.

Explanation / Answer

Q.4 (a) uncertainty in x = 0.1 mm / sqrt(12) = 0.0288 mm

(b) uncertainty in   = 0.05 mm / sqrt(8) = 0.0177 mm

(c) L1 = 1/2 x + 1/2 = 1/2 * (0.1 + 0.05)= 0.075 mm

Uncertainty in L1 = 0.075/ sqrt(20) = 0.0168 mm

L2 = 12/20 x + 8/20 = 1/20 * (12 * 0.1 + 8 * 0.05)= 0.08 mm

Uncertainty in L2 = 0.08/ sqrt(20) = 0.0179 mm

Here L1 is smaller

(d) Let say for any c,

L3 = cx + (1-c)

uncertaintly = 0.1/c + 0.05/ (1-c)

diff. with c

du/dc = -0.5 c-3/2 + 0.025 (1-c)-3/2 = 0

(1-c)-3/2 = 2c-3/2

c3 = 8(1-c)3

c/(1-c) = 2

c = 2 - 2c

3c = 2

c = 1/3

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