Question

find mean absolute, relative and percentage error of 1.55,1.54,1.53,1.52,1.49​

Answer 1

Let ,

A = 1.55

B =1.54

C =1.53

D = 1.52

E = 1.49

True Value = $\large{ \frac{1.55 + 1.54 + 1.53 + 1.52 + 1.49}{5} = 1.52}$

Absolute Error in A = Measured value - True value = 1.55 - 1.52 = 0.3

Absolute Error in B = Measured value - True value = 1.54 - 1.52 = 0.2

Absolute Error in C = Measured value - True value = 1.53 - 1.52 = 0.1

Absolute Error in D = Measured value - True value = 1.52 - 1.52 = 0

Absolute Error in E = Measured value - True value = 1.49 - 1.52 = 0.3

Mean Absolute Error = Sum of Absolute errors / Total number of absolute error

Mean Absolute Error = 0.9 / 5

Mean Absolute Error = 9 / 50

Mean Absolute Error = 0.18

Relative Error = Mean Absolute error / True Value

Relative error = 0.18 / 1.52 = 0.11

Percentage Error = Relative error × 100 = 11 %

Answer 2

Answer:

The values given :- 1.55, 1.54, 1.53, 1.52 and 1.49.

So, here we have a1 = 1.55, a2 = 1.54, a3 = 1.53, a4 = 1.52 and a5 = 1.49.

Hence, a_mean = (1.55 + 1.54 + 1.53 + 1.52 + 1.49)/5

= 7.63/5 = 1.52

Now, Finding the absolute errors :-

For a1 : 1.55 - 1.52 = 0.03

For a2 : 1.54 - 1.52 = 0.02

For a3 : 1.53 - 1.52 = 0.01

For a4 : 1.52 - 1.52 = -0.00 = 0.00

For a5 : 1.49 - 1.52 = -0.03 = 0.03

Now, Finding the mean absolute error :-

∆a_mean = (∆a1 + ∆a2 +.....+ ∆a5)/5

= (0.03 + 0.02 + 0.001 + 0.00 + 0.03)/5

= (0.09)/5

= 9/500

= 0.018

Now, Finding the relative error :-

= ∆a_mean/a_mean

= 0.018/1.52

= 0.011

Also, the percentage error :- 0.011 * 100% = 1.1%