A distance AB is observed repeatedly using the same equipment and procedures, and the results, in meters, are listed below:69.401, 69.400, 69.402, 69.396, 69.406, 69.401, 69.396, 69.401, 69.405, and 69.404Calculate (a) the line's most probable length, (b) the standard deviation.

Answer :

Answer:

The line's most probable length is 69.4012 meters.

The standard deviation is 0.0032 meters.

Step-by-step explanation:

We are given the following in the question:

69.401, 69.400, 69.402, 69.396, 69.406, 69.401, 69.396, 69.401, 69.405, 69.404      

Formula:

[tex]\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n}}[/tex]  

where [tex]x_i[/tex] are data points, [tex]\bar{x}[/tex] is the mean and n is the number of observations.  

[tex]Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}[/tex]

[tex]Mean =\displaystyle\frac{694.012}{10} = 69.4012[/tex]

The line's most probable length is 69.4012 meters.

Sum of squares of differences = 0.0001016

[tex]\sigma = \sqrt{\dfrac{0.0001016}{10}} = 0.0032[/tex]

The standard deviation is 0.0032 meters.

(a)The line's most probable length is 69.4012 meters.

(b)The standard deviation is 0.0032 meters.

We have the data set:

69.401, 69.400, 69.402, 69.396, 69.406, 69.401, 69.396, 69.401, 69.405, 69.404      

Mean = sum of all observations divide by total number of observations

(a).Mean

[tex]=\frac{69.401+69.400+ 69.402+69.396+ 69.406+ 69.401+ 69.396+ 69.401+69.405+ 69.404 }{10} \\=69.412[/tex]  

 The line's most probable length is 69.4012 meters.

Sum of squares of differences =

(b).Standard deviation:

[tex]=\frac{\sqrt{(x_i-x')^{2} } }{n} \\=\frac{\sqrt{0.0001016 } }{10}\\=0.0032[/tex]

The standard deviation is 0.0032 meters.

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