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In five successive measurements, the mass of a ball is measured to be $2.61 \mathrm{~g}, 2.58 \mathrm{~g}$, $2.40 \mathrm{~g}, 2.73 \mathrm{~g}$ and $2.80 \mathrm{~g}$. The absolute error in the measurement is
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The correct answer is:
$0.11 \mathrm{~g}$
Mean mass of the ball is
$\bar{M}=\frac{m_1+m_2+m_3+m_4+m_5}{5}$
$=\frac{2.61+2.58+2.40+2.73+2.80}{5}$
$\bar{M}=2.624 \mathrm{~g} \simeq 2.62 \mathrm{~g}$
$\therefore$ Mean absolute error,
$\Delta \bar{M}=\frac{|2.62-2.61|+|2.62-2.58|+|2.62-2.40|+|2.62-2.73|+|2.62-2.80|}{5}$
$=\frac{0.01+0.04+0.22+0.11+0.18}{5}$
$=\frac{0.56}{5}=0.112 \simeq 0.11 \mathrm{~g}$
$\bar{M}=\frac{m_1+m_2+m_3+m_4+m_5}{5}$
$=\frac{2.61+2.58+2.40+2.73+2.80}{5}$
$\bar{M}=2.624 \mathrm{~g} \simeq 2.62 \mathrm{~g}$
$\therefore$ Mean absolute error,
$\Delta \bar{M}=\frac{|2.62-2.61|+|2.62-2.58|+|2.62-2.40|+|2.62-2.73|+|2.62-2.80|}{5}$
$=\frac{0.01+0.04+0.22+0.11+0.18}{5}$
$=\frac{0.56}{5}=0.112 \simeq 0.11 \mathrm{~g}$
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