Question

In an experiment the refractive index of glass was observed to be

1.45, 1.56 , 1.54, 1.44, 1.54 & 1.53

Calculate :
(i) Mean value of refractive index
(ii) Mean absolute error
(iii) fractional error
(iv) Percentage error
(v) express the result in terms of absolute error and percentage error....


Steps needed....​

Answer 1

I hope the answers are right. l have a doubt in (v) the answer can also be 1.51 + 3%

Answer 2

Given,

The refractive index of glass was observed to be :

1.45 , 1.56 , 1.54 , 1.44 , 1.54 , 1.53

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(i) Mean value of refractive index

$\sf \mu_m=\dfrac{1.45+1.56+1.54+1.44+1.54+1.53}{6}$

$\implies\sf \mu_m=\dfrac{9.06}{6}$

$\implies\sf \mu_m=1.51$

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(ii) Mean absolute error

• $\sf \Delta\mu_1=|1.45-1.51|=0.06$

• $\sf \Delta\mu_2=|1.56-1.51|=0.5$

• $\sf \Delta\mu_3=|1.54-1.51|=0.3$

• $\sf \Delta\mu_4=|1.44-1.51|=0.07$

• $\sf \Delta\mu_5=|1.54-1.51|=0.3$

• $\sf \Delta\mu_6=|1.53-1.51|=0.2$

$\sf\Delta\mu_m=\dfrac{0.06+0.5+0.3+0.070.3+0.2}{6}$

$\implies\sf \Delta\mu_m=\dfrac{0.26}{6}$

$\implies\sf \Delta\mu_m=0.0433$

$\implies\sf \Delta\mu_m\approx 0.04$

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(iii) fractional error

$\sf \dfrac{\Delta\mu_m}{\mu_m}=\dfrac{0.04}{1.51}$

$\implies\sf \dfrac{\Delta\mu_m}{\mu_m}=0.02649$

$\implies\sf \dfrac{\Delta\mu_m}{\mu_m}=0.03$

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(iv) Percentage error

$\sf\dfrac{\Delta\mu_m}{\mu_m}\times 100=0.03\times 100$

$\implies\sf\dfrac{\Delta\mu_m}{\mu_m}\times 100=3\%$

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(v)

Result in terms of Absolute Error

$\sf Absolute\:Error=\mu_m\pm \Delta\mu_m$

$\implies\sf Absolute\:Error=1.51\pm 0.04$

Result in terms of Percentage Error

$\sf Percentage\:Error={\mu_m}\pm\dfrac{\Delta\mu_m}{\mu}\times100$

$\implies\sf Percentage\:Error=1.51\pm3\%$

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Answers

(i) Mean value of refractive index = 1.51

(ii) Mean absolute error  = 0.04

(iii) Fractional error  = 0.03

(iv) Percentage error  = 3%

(v) Result in terms of absolute error = 1.51 ± 0.04

    Result in terms of percentage error = 1.51 ± 3%

$\huge\boxed{\rm Hope\:it\:Helps}$