Question

A distance AB is observed repeatedly using the same equipment and procedures, and the results, in meters, are listed in Problems 3.7 through 3.10. Calculate (a) the line's most probable length, (b) the standard deviation, and (c) the standard deviation of the mean for each set of results.

Answer 1

The most probable length is 65.4012

The standard deviation is 0.003359

The standard deviation of mean is  1.0626 x 10⁻³

How to solve for the most probable length

The most probable length for the group of data would be the mean of the data set. In order to solve for this, you have to add up all of the values that are in the data set and divide by the total number in the data set

65.400 + 65.402 + 65.396+ 65.406 + 65.401 + 65.396 + 65.401 + 65.405 + 65.404 / 10

= 65.4012

The most probable length is 65.401

How to solve for the standard deviation.

To solve for the standard deviation, we would have to make use of a standard deviation calculator

= (65.400 - 65.4012)² + ... + (65.404 - 65.4012)² / 9

= 0.00010155555555557 / 9

= 1.1283950617286E-5

σ = √1.1283950617286E-5

= 0.0033591592128516

= 0.003359

The standard deviation of the mean

σm = σ / √N

= 0.0036 / √10

= 0.00336 / 3.162

= 1.0626 x 10⁻³

Read more on standard deviation here:

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Complete question

distance AB is observed repeatedly using the same equipment and procedures, and the results, in meters, are listed below. Calculate 65.401, 65.400, 65.402, 65.396, 65.406, 65.401, 65.396, 65.401, 65.405, and 65.404.

Calculate:

(a) the line's most probable length,

(b) the standard deviation, and

(c) the standard deviation of the mean.