Question

the refractive index of water is found to have values 1.29 1.33 1.34 1.35 1.32 1.36 1.30 and 1.33 calculate mean value absolute error relative error and percentage error

Answer 1

Given values ⇒

1.29 1.33 1.34 1.35 1.32 1.36 1.30 and 1.33

Now, there are total of the 8 term. 

∵ Mean value = Sum of all the observations/No. of observations. 
∴ Mean value = (1.29 + 1.33 + 1.34 + 1.35 + 1.32 + 1.36  + 1.30 + 1.33)/8
   = 10.62/8
   = 1.3275 
   ≈ 1.33

Now, for absolute error in each term, 

x₁ = |1.33- 1.29| = 0.04
x₂ = |1.33 - 1.33| = 0.00
x₃ = |1.33 - 1.34| = 0.01
x₄ = |1.33 - 1.35| = 0.02
x₅ = |1.33 - 1.32| = 0.01
x₆ = |1.33 - 1.36| = 0.03
x₇ = |1.33 - 1.30| = 0.03
x₈ = |1.33 - 1.33| = 0.00

∴ Mean Absolute error = (0.04 + 0.00 + 0.01 + 0.02 + 0.01 + 0.03 + 0.03 + 0.00)/8
= 0.14/8
= 0.0175
≈ 0.018 


∴ Relative Error = Mean absolute error/Mean Value

   = 0.018/1.33
   = 0.135
   ≈ 0.12 

∴ % Error = 0.12 × 100
   % Error = 12 %


Hope it helps. :-)

Answer 2

Explanation:

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