TS EAMCET 6-Aug-2021 Shift 1 Question Paper

Given that, number of observations, n=5
‌a1=2.4m,a2=2.5m,a3=2.6m,a4=2.8m‌, ‌
‌a5=3.0m
Mean value of observations,
a‌‌=‌

a1+a2+a3+a4+a5

5

a‌‌=‌

2.4+2.5+2.6+2.8+3.0

5

‌‌=‌

13.3

5

=2.66
Absolute errors in individual observed value
‌|∆a1|=|a1−a|=|2.4−2.66|=|−0.26|=0.26
‌|∆a2|=|a2−a|=|2.5−2.66|=|−0.16|=0.16
‌|∆a3|=|a3−a|=|2.6−2.66|=|−0.06|=0.06
‌|∆a4|=|a4−a|=|2.8−2.66|=0.14
‌|∆a5|=|a5−a|=|3.0−2.66|=0.34
Mean absolute error,
‌∆a=‌

|∆a1|+|∆a2|+|∆a3|+|∆a4|+|∆a5|

n

‌∆a=‌

0.26+0.16+0.06+0.14+0.34

5

=‌

0.96

5

‌∆a=0.192
‌ Relative error ‌‌‌=‌

‌ Mean absolute error ‌

‌ Mean value ‌

‌‌=‌

∆a

a

=‌

0.192

2.66

=0.072