Question

The diameter of a wire measured by a screw gauge is given by 0.012cm,1.015cm,1.018cm,1.013cm,1.017cm, respectively. Find the diameter of the wire, mean absolute error and percentage error.

Answer 1

$\sf{\gray{\underline{\underline{\purple{GIVEN:-}}}}}$

• The diameter of a wire measured by a screw gauge is given by “0.012cm” , “1.015cm” , “1.018cm” , “1.013cm” and “1.017cm” respectively .

$\sf{\gray{\underline{\underline{\purple{TO\: FIND:-}}}}}$

1. Mean Diameter of the wire .
2. Mean Absolute error .
3. Percentage error .

$\sf{\gray{\underline{\underline{\purple{SOLUTION:-}}}}}$

[1] MEAN DIAMETER :-

$\orange\star\:\bf{\pink{\boxed{\blue{Mean\:Diameter(\overline{D})\:=\:\dfrac{D_1\:+\:D_2\:+\:D_3\:+\:D_4\:+\:D_5}{5}\:}}}}$

Where,

• $\rm{D_1}$ = 0.012cm

• $\rm{D_2}$ = 1.015cm

• $\rm{D_3}$ = 1.018cm

• $\rm{D_4}$ = 1.013cm

• $\rm{D_5}$ = 1.017cm

$\rm{\implies\:Mean\:Diameter(\overline{D})\:=\:\dfrac{0.012\:+\:1.015\:+\:1.018\:+\:1.013\:+\:1.017}{5}\:}$

$\rm{\implies\:\overline{D}\:=\:\dfrac{4.075}{5}\:}$

$\rm{\implies\:\overline{D}\:=\:\dfrac{4.075}{5}\:}$

$\rm{\purple{\implies\:\overline{D}\:=\:0.815\:cm}}$

[2] MEAN ABSOLUTE ERROR :-

☃️ First of all, Calculate Absolute Error :-

$\red\star\:\bf{\pink{\boxed{\blue{\triangle{D_1}\:=\:\left|D_1\:-\:\overline{D}\right|\:}}}}$

$\rm{\implies\:\triangle{D_1}\:=\:\left|0.012\:-\:0.815\right|\:}$

$\rm{\implies\:\triangle{D_1}\:=\:\left|-0.803\right|\:}$

$\rm{\green{\implies\:\triangle{D_1}\:=\:0.803\:cm}}$

$\red\star\:\bf{\pink{\boxed{\blue{\triangle{D_2}\:=\:\left|D_2\:-\:\overline{D}\right|\:}}}}$

$\rm{\implies\:\triangle{D_2}\:=\:\left|1.015\:-\:0.815\right|\:}$

$\rm{\green{\implies\:\triangle{D_2}\:=\:0.200\:cm}}$

$\red\star\:\bf{\pink{\boxed{\blue{\triangle{D_3}\:=\:\left|D_3\:-\:\overline{D}\right|\:}}}}$

$\rm{\implies\:\triangle{D_3}\:=\:\left|1.018\:-\:0.815\right|\:}$

$\rm{\green{\implies\:\triangle{D_3}\:=\:0.818\:cm}}$

$\red\star\:\bf{\pink{\boxed{\blue{\triangle{D_4}\:=\:\left|D_4\:-\:\overline{D}\right|\:}}}}$

$\rm{\implies\:\triangle{D_4}\:=\:\left|1.013\:-\:0.815\right|\:}$

$\rm{\green{\implies\:\triangle{D_4}\:=\:0.198\:cm}}$

$\red\star\:\bf{\pink{\boxed{\blue{\triangle{D_5}\:=\:\left|D_5\:-\:\overline{D}\right|\:}}}}$

$\rm{\implies\:\triangle{D_5}\:=\:\left|1.017\:-\:0.815\right|\:}$

$\rm{\green{\implies\:\triangle{D_5}\:=\:0.202\:cm}}$

☃️ Now, to calculate Mean Absolute Error :-

$\orange\star\:\bf{\pink{\boxed{\blue{Mean\:Absolute\:Error\:=\:\dfrac{\triangle{D_1}\:+\:\triangle{D_2}\:+\:\triangle{D_3}\:+\:\triangle{D_4}\:+\:\triangle{D_5}}{5}\:}}}}$

$\rm{\implies\:Mean\:Absolute\:Error\:=\:\dfrac{0.803\:+\:0.200\:+\:0.818\:+\:0.198\:+\:0.202}{5}\:}$

$\rm{\implies\:Mean\:Absolute\:Error\:=\:\dfrac{2.221}{5}\:}$

$\rm{\purple{\implies\:Mean\:Absolute\:Error\:=\:0.444\:cm}}$

[3] PERCENTAGE ERROR :-

☃️ First of all, Calculate Relative Error :-

$\red\star\:\bf{\pink{\boxed{\blue{Relative\:Error\:=\:\dfrac{Mean\:Absolute\:Error}{\overline{D}}\:}}}}$

$\rm{\implies\:Relative\:Error\:=\:\dfrac{0.444}{0.815}\:}$

$\rm{\green{\implies\:Relative\:Error\:=\:0.544\:cm}}$

☃️ Now, to calculate Percentage Error :-

$\orange\star\:\bf{\pink{\boxed{\blue{Percentage\:Error\:=\:Relative\:Error\:\times{100}\:}}}}$

$\rm{\implies\:Percentage\:Error\:=\:0.544\:\times{100}\:}$

$\rm\purple{\implies\:Percentage\:Error\:=\:544\%\:}$