Question

the radius of a sphere gives values 5.63m,5.54m,5.44m,5.40m,and5.35m.find the most probable values of radius .means absolute error ,Relative error and percentage error.​

Answer 1

Answer:firstly, find the mean =5.63+5.54+5.44+5.40+5.35/5

=27.36/5

a(mean)=5.472m

NOW SEE THE PIC OF WHOLE ANS ..

Answer 2

Concept:

The figure with the highest likelihood of being accurate is referred to as the most probable value. Simply the arithmetic means of the supplied data.

The mean of the absolute departure from the mean value is known as the mean absolute error.

The ratio of mean absolute error to mean value is known as a relative error.

Data percentage error can be calculated by multiplying the relative error by 100.

Given:

The radius of a sphere are 5.63m, 5.54m, 5.44m, 5.40m, and 5.35m.

Find:

The probable values of radius means, absolute error, relative error and percentage error.

Solution:

The given values of the radius of a sphere is 5.63m, 5.54m, 5.44m, 5.40m, and 5.35m.

The mean of the radius is given as:

a = sum of observations/ the total number of observations

$\text{mean}=\frac{5.63+5.54+5.44+5.40+5.35}{5}=5.472 m$

The mean absolute error is:

$\Delta a_{mean}=\frac{1}{n}\sum_{i=1}^{n}\Delta a_1$

Now, $\Delta a_n=|a_{mean}-a_n}|$

$=\frac{1}{5}[|5.472-5.63|+|5.472-5.54|+|5.472-5.44|+|5.472-5.40|+|5.472-5.35|$

$=\frac{1}{5}[0.158+0.068+0.032+0.072+0.122]$

$=\frac{0.452}{2}=0.0904 m$

The relative error is given as:

Relative error = Mean absolute error/ mean value

$=\frac{\Delta a_{mean}}{a}$

$=\frac{0.0904}{5.472}=0.01652 m$

The percentage error is given as:

Percentage error = relative error × 100

= 0.01652 × 100 = 1.652 %

The mean, mean absolute error, relative error, and the percentage error of the radius of the sphere is 5.472 m, 0.0904 m, 0.01652 m, and 1.652% respectively.

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