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Example · Example 5

Q.In an experiment, the acceleration due to gravity gg is measured five times, in m s−2^{-2}: 9.79, 9.82, 9.81, 9.78, 9.809.79,\ 9.82,\ 9.81,\ 9.78,\ 9.80. Find the mean value, the mean absolute error, and the percentage error.

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Step 1 — mean value.

gˉ=9.79+9.82+9.81+9.78+9.805=49.005=9.80 m s−2\bar g = \frac{9.79+9.82+9.81+9.78+9.80}{5} = \frac{49.00}{5} = 9.80\ \text{m s}^{-2}

Step 2 — absolute deviation of each reading from the mean.

∣9.79−9.80∣=0.01, ∣9.82−9.80∣=0.02, ∣9.81−9.80∣=0.01, ∣9.78−9.80∣=0.02, ∣9.80−9.80∣=0.00|9.79-9.80|=0.01,\ |9.82-9.80|=0.02,\ |9.81-9.80|=0.01,\ |9.78-9.80|=0.02,\ |9.80-9.80|=0.00

Step 3 — mean absolute error.

Δgmean=0.01+0.02+0.01+0.02+0.005=0.065=0.012 m s−2≈0.01 m s−2\Delta g_{\text{mean}} = \frac{0.01+0.02+0.01+0.02+0.00}{5} = \frac{0.06}{5} = 0.012\ \text{m s}^{-2} \approx 0.01\ \text{m s}^{-2}

(rounded to match the two-decimal precision of the original readings) …

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