Question

A distance AB is observed repeatedly using the same equipment and procedures, and the results, in meters, are listed below: 67.401, 67.400, 67.402, 67.396, 67.406, 67.401, 67.396, 67.401, 67.405, and 67.404Calculate (a) the line's most probable length, (b) the standard deviation.

Answer 1

Answer:

a) $\bar X =\frac{67.401+67.400+67.402+67.396+67.406+67.401+67.396+67.401+67.405+67.404}{10}=67.401$

b) The sample deviation is calculated from the following formula:

$s=\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

And for this case after replace the values and with the sample mean already calculated we got:

$s=0.0036$

If we assume that the data represent a population then the standard deviation would be given by:

$\sigma=\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n}}$

And then the deviation would be:

$\sigma =0.00319$

Step-by-step explanation:

For this case we have the following dataset:

67.401, 67.400, 67.402, 67.396, 67.406, 67.401, 67.396, 67.401, 67.405, and 67.404

Part a: Determine the most probable value.

For this case the most probably value would be the sample mean given by this formula:

$\bar X =\frac{\sum_{i=1}^n X_i}{n}$

And if we replace we got:

$\bar X =\frac{67.401+67.400+67.402+67.396+67.406+67.401+67.396+67.401+67.405+67.404}{10}=67.401$

Part b: Determine the standard deviation

The sample deviation is calculated from the following formula:

$s=\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

And for this case after replace the values and with the sample mean already calculated we got:

$s=0.0036$

If we assume that the data represent a population then the standard deviation would be given by:

$\sigma=\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n}}$

And then the deviation would be:

$\sigma =0.00319$