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juin [17]
2 years ago
13

A distance AB is observed repeatedly using the same equipment and procedures, and the results, in meters, are listed below:69.40

1, 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.
Mathematics
1 answer:
e-lub [12.9K]2 years ago
3 0

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:

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

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

Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}

Mean =\displaystyle\frac{694.012}{10} = 69.4012

The line's most probable length is 69.4012 meters.

Sum of squares of differences = 0.0001016

\sigma = \sqrt{\dfrac{0.0001016}{10}} = 0.0032

The standard deviation is 0.0032 meters.

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Solve the recurrence relation: hn = 5hn−1 − 6hn−2 − 4hn−3 + 8hn−4 with initial values h0 = 0, h1 = 1, h2 = 1, and h3 = 2 using (
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The required standard deviation : \sigma=0.2551

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