the valu
In an experiment the value of thickness of a
wire was found to be 1.54, 1.53, 1.44, 1.54, 1.56
and 1.45 in successive measurements. Then the
percentage error is
(1) 28%
(2).09 %
(3) 9%
(4) 2.8%​

Respuesta :

Answer:

[tex]\% Error = 2.6\%[/tex]

Explanation:

Given

[tex]x: 1.54, 1.53, 1.44, 1.54, 1.56, 1.45[/tex]

Required

Determine the percentage error

First, we calculate the mean

[tex]\bar x = \frac{\sum x}{n}[/tex]

This gives:

[tex]\bar x = \frac{1.54+ 1.53+ 1.44+ 1.54+ 1.56+ 1.45}{6}[/tex]

[tex]\bar x = \frac{9.06}{6}[/tex]

[tex]\bar x = 1.51[/tex]

Next, calculate the mean absolute error (E)

[tex]|E| = \sqrt{\frac{1}{6}\sum(x - \bar x)^2}[/tex]

This gives:

[tex]|E| = \sqrt{\frac{1}{6}*[(1.54 - 1.51)^2 +(1.53- 1.51)^2 +.... +(1.45- 1.51)^2]}[/tex]

[tex]|E| = \sqrt{\frac{1}{6}*0.0132}[/tex]

[tex]|E| = \sqrt{0.0022}[/tex]

[tex]|E| = 0.04[/tex]

Next, calculate the relative error (R)

[tex]R = \frac{|E|}{\bar x}[/tex]

[tex]R = \frac{0.04}{1.51}[/tex]

[tex]R = 0.026[/tex]

Lastly, the percentage error is calculated as:

[tex]\% Error = R * 100\%[/tex]

[tex]\% Error = 0.026 * 100\%[/tex]

[tex]\% Error = 2.6\%[/tex]