Question

The radius of a sphere measured

repeatedly yields values 5.63 m, 5.54 m, 5.44

m, 5.40 m and 5.35 m. Determine the most

probable value of radius and the mean absolute,

relative and percentage errors. ​

Answer 1

To Find:

The most probable value of radius and the mean absolute, relative and percentage errors

Solution:

Values = 5.63 m, 5.54 m, 5.44 m, 5.40 m, 5.35 m.

The mean value = (5.63 + 5.54 + 5.44 + 5.40 + 5.35)/5

= 27.36/5

= 5.472 m.

Absolute errors:-

5.63 - 5.472 = 0.158  

5.54 - 5.472 = 0.068

5.56 - 5.472 = 0.088

5.40 - 5.472 = |-0.072| = 0.074

5.30 - 5.472 = |-0.172| = 0.172

Absolute mean error = (0.158 + 0.068 + 0.088 + 0.074 + 0.172)/5

= 0.56/5

= 0.112 m.

So, the Relative error:- ∆a_mean/a_mean

= 0.112/5.472

= 0.02046783625 m.

Hence, the percentage error = 0.02046783625 * 100 =2.05 %.To Find:

The most probable value of radius and the mean absolute, relative and percentage errors

Solution:

Values = 5.63 m, 5.54 m, 5.44 m, 5.40 m, 5.35 m.

The mean value = (5.63 + 5.54 + 5.44 + 5.40 + 5.35)/5

= 27.36/5

= 5.472 m.

Absolute errors:-

5.63 - 5.472 = 0.158  

5.54 - 5.472 = 0.068

5.56 - 5.472 = 0.088

5.40 - 5.472 = |-0.072| = 0.074

5.30 - 5.472 = |-0.172| = 0.172

Absolute mean error = (0.158 + 0.068 + 0.088 + 0.074 + 0.172)/5

= 0.56/5

= 0.112 m.

So, the Relative error:- ∆a_mean/a_mean

= 0.112/5.472

= 0.02046783625 m.

Hence, the percentage error = 0.02046783625 * 100 =2.05 %.

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