Respuesta :
Answer:
Reading is close to (b) 13.44 which is the best estimate of the period
Associated error, [tex]\Delta E =0.178 s[/tex]
Given:
[tex]t_{a} = 13.54 s[/tex]
[tex]t_{b} = 13.44 s[/tex]
[tex]t_{c} = 13.89 s[/tex]
[tex]t_{d} = 13.41 s[/tex]
[tex]t_{e} = 13.17 s[/tex]
[tex]t_{f} = 13.22 s[/tex]
Solution:
1.The best estimate of the period can be calculated by the mean of the measurements and the one closest to the mean is the best estimate of the measurement:
[tex]Mean, \bar {x} = \fra{sum of all observations}{No. of observation}[/tex]
[tex]Mean, \bar {x} = \frac{t_{a} + t_{b} + t_{c} +t_{d} + t_{e} + t_{f}}{6}[/tex]
[tex]Mean, \bar {x} = \frac{13.54 + 13.44 + 13.89 + 13.41 + 13.17 + 13.22}{6}[/tex]
[tex]Mean, \bar {x} = 13.445 s[/tex]
It is close to 13.44 s
2. Associated error is given by:
[tex]\Delta E_{n} = |measured value - actual value|[/tex]
[tex]\Delta E_{n} = |t_{n} - \bar {x}|[/tex]
where
n = a, b,......, e
Now,
[tex]\Delta E_{a} = |t_{a} - \bar {x}| = |13.54 - 13.44| = 0.01[/tex]
[tex]\Delta E_{b} = |t_{b} - \bar {x}| = |13.44 - 13.44| = 0.00[/tex]
[tex]\Delta E_{c} = |t_{c} - \bar {x}| = |13.89 - 13.44| = 0.45[/tex]
[tex]\Delta E_{d} = |t_{d} - \bar {x}| = |13.41 - 13.44| = 0.03[/tex]
[tex]\Delta E_{e} = |t_{e} - \bar {x}| = |13.17 - 13.44| = 0.027[/tex]
[tex]\Delta E_{f} = |t_{f} - \bar {x}| = |13.54 - 13.44| = 0.10[/tex]
Mean Absolute Error, [tex]\Delta E = \frac{\Sigma E_{n}}{6}[/tex]
[tex]\Delta E = \frac{0.01 + 0.00 + 0.45 + 0.03 +0.027 + 1.10}{6}[/tex]
[tex]\Delta E =0.178 s[/tex]
3. The assumption behind the estimation is population is considered to distributed normally.
