Question

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%​

Answer 1

`\% Error = 2.6\%`

Step-by-step explanation:

Given

`x: 1.54, 1.53, 1.44, 1.54, 1.56, 1.45`

Required

Determine the percentage error

First, we calculate the mean

`\bar x = (\sum x)/(n)`

This gives:

`\bar x = (1.54+ 1.53+ 1.44+ 1.54+ 1.56+ 1.45)/(6)`

`\bar x = (9.06)/(6)`

`\bar x = 1.51`

Next, calculate the mean absolute error (E)

`|E| = \sqrt{(1)/(6)\sum(x - \bar x)^2}`

This gives:

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

`|E| = \sqrt{(1)/(6)*0.0132}`

`|E| = √(0.0022)`

`|E| = 0.04`

Next, calculate the relative error (R)

`R = (|E|)/(\bar x)`

`R = (0.04)/(1.51)`

`R = 0.026`

Lastly, the percentage error is calculated as:

`\% Error = R * 100\%`

`\% Error = 0.026 * 100\%`

`\% Error = 2.6\%`