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OLEGan [10]
3 years ago
15

If I measure a set of times with a mean of 1.4 s, a standard deviation of 0.6 s, and an SDOM of 0.2 s, how should I report my me

asurement?
Mathematics
1 answer:
NNADVOKAT [17]3 years ago
7 0

Answer:

The measurements should be reported as 1.4s ± 0.2s.

Step-by-step explanation:

The mean of a data set consisting of <em>n</em> independent values is:

\bar X=\frac{1}{n}\sum (x_{1}+x_{2}+x_{3}...+x_{n})

The standard deviation of a data set consisting of <em>n</em> independent values is the measure of the spread of the data. It is computed as follows:

\sigma=\sqrt{\frac{1}{n}\sum(X_{i}-\bar X)^{2}}

The standard deviation of mean (SDOM) is also known as the standard error.

The standard error is:

\delta X=\frac{\sigma}{\sqrt {n}}

If we want to report these measurements the format is: \bar X\pm \delta X

Given:

\bar X=1.4s\\\delta =0.2s

These measurements can be reported as:

\bar X\pm \delta X=1.4s\pm 0.2s

Thus, the measurements should be reported as 1.4s ± 0.2s.

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