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Answer 1

The degree of dispersion in your data collection is indicated by variance.
It is given by the formula:
$\overline{d}= \sum_{i=1}^{i=n}\frac{|x_i-\overline{x}|^2}{n}$

To determine the values of a certain set of scattered data, the standard deviation formula is utilized. The dispersion of the quantities or data from the mean is how the standard deviation is simply defined. The conclusion drawn from a lower standard deviation is that the results are fairly near to the average. Higher values, on the other hand, indicate a significant dispersion from the mean. The standard deviation number can never be negative, it should be positive.

It is given by the formula:

$s_d = \sqrt{\sum_{i=1}^{i=n}\frac{|x_i-\overline{x}|^2}{n}} =\sqrt{\overline{d}}$

We have that the mean of the values is:  

at 8 AM $\overline{x}=\frac{97.9+99.3+97.9+97.6+97.4}{5}=98.02$

at 12 AM $\overline{x}=\frac{98.5+99.7+98+97.4+97.6}{5}=98.24$

Using the above definitions we have that

at 8 AM $\overline{d}=\frac{(98.02-97.9)^2+(98.02-99.3)^2+(98.02-97.9)^2+(98.02-97.6)^2+(98.02-97.4)^2}{5}=0.4456$

$s_d=\sqrt{\overline{d}}=\sqrt{0.4456}=0.6675$

at 12 AM

$\overline{d}=\frac{(98.24-98.5)^2+(98.24-99.7)^2+(98.24-98)^2+(98.24-97.4)^2+(98.24-97.6)^2}{5}=0.6744$

$s_d=\sqrt{\overline{d}}=\sqrt{0.6744}=0.8212$

In general, $\mu_d$ represents the mean of the differences from the population of matched data.

Learn more about the variance and standard deviation here-

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