Question

find absolute error, mean absolute error, percentage error of 2.63,2.56,2.42,2.71,2.80​

Answer 1

$\huge{ \blue{ \underline{ \underline{ \purple{ \rm{Solution : }}}}}}$

First, we have to find the true value i.e., actual value . It is the mean of all these values given :

$\blue{ \bf{a_i}}$

• 2.63

• 2.56

• 2.42

• 2.71

• 2.80

Here, $\sf{a_i}$ denotes measured (experimental) value. "i" is varying from 1 to 5.

The arithmetic mean, $\sf{ \overline{ a}}$ of period of oscillations, given as :

$\blue{ \sf{ \overline{a} = { \sum^{n}_{i =1}} \: a_i}}$

$\blue{ \sf = { \frac { 2.63 + 2.56 + 2.42 + 2.71 + 2.80} {5}}}$

$\blue{ \sf = \frac{13.12}{5} = 2.62}$

$\green{ \underline{ \boxed{ = 2.62 \: s{ \rm{ \small{ \: (approx)}}}}}}$

Now we will find errors from this true value (Arithmetic mean). Every measured value, we will subtract from a true value. Then, we find $\sf{\Delta {a_i}}$ i.e., Absolute Error.

Refer the attachment ⛈

In absolute error both negative ( - ) and positive ( + ) values are there, which means absolute errors can be positive or negative.

Now, we have to find mean absolute error.

$\blue{ \sf{ \overline { \Delta{a_i}} = \frac{1} {n} {\sum^{n}_{i = 1}} \: |{\Delta{a_i}|}}}$

$\blue{ \sf = { \frac {0.01 + 0.06 + 0.20 + 0.09 + 0.18} {5}}}$

$\green{ \underline{ \boxed{ \purple{ = 0.108 \: s \approx{0.11 \: s}}}}}$

Note,

$\sf{{\Delta{a_i} \: {\rightarrow \: {Absolute \: Error}}}}$

$\sf{ \overline{\Delta{a_i}} \: {\rightarrow \: Mean \: Absolute \: Error}}$

Here, we have taken only values, not signs. Hence, mean absolute error concentrates on magnitude not sign.....

Now, let's find out percentage error

$\blue{ \sf{Percentage \: error = \frac {Mean \: Absolute \: error}{Actual \: value} \times 100 }}$

$\blue{ \sf{ = \frac{0.11}{2.6} \times 100 }}$

$\green{ \underline{ \boxed{ \purple{ \sf{ = 4\%}}}}}$