$$Measured \ Value = True \ Value + Bias + Random \ Error$$
The mean $\mu$ of the population represents the part of the measurement that is the same for every measurement. Hence, $\mu$ is the sum of the true value and the bias. The smaller the bias, the more accurate the measuring process is. If the mean is equal to the true value, the measuring process is said to be unbiased. The standard deviation $\sigma$ of the population is the the standard deviation of the random error. The precision of the measurement is determined by the standard deviation of the measurement process. The smaller the value of $\sigma$, the more precise the process. $\sigma$ is often referred to as the uncertainty in the measuring process.
For example, if $X_1, X_2, …, X_n$ are the independent measurements, all made on the same quantity by the same process, the sample standard deviation $s$ can be used to estimate the uncertainty in the process. If the true value of the measuring quantity is known, the sample mean $\overline{X}$ can be used to estimate the bias as $Bias \approx \overline{X} - True \ Value$. If the true value is unknown, the bias can not be estimated from the repeated measurements. If bias has been reduced to a negligible level, the measurements can be described as:
$$Measured \ Value \pm \sigma$$
#### Linear Combinations of Independent Measurements :If $X$ is a measurement and $c$ a constant, then the uncertainty in the measurement of $cX$ can be given as:
$$\sigma _{cX} = \left|c\right| \sigma _{X}$$
For the independet measurements $X_1,X_2, …, X_n$ and constants $c_1, c_2, …, c_n$, the uncertainty of the measurement $c_1X_1 + c_2X_2 + … + c_nX_n$ is given as:
$$\sigma _{c_1X_1 + c_2X_2 + … + c_nX_n} = \sqrt{c_1^2\sigma _{X_1}^2 + c_2^2\sigma _{X_2}^2 + … + c_n^2\sigma _{X_n}^2}$$
The above mentioned calculation of uncertainties is a simple implication of the linear combination of random variables.
#### Repeated Measurements :Taking repeated independent measurements of the same quantity is a good way to reduce the overall uncertainty or variance of the measurement. The average of these measurements will have the same mean but the standard deviation will be reduced by a factor of square root of number of measurements. Hence, for $n$ independent measurements $X_1,X_2, …, X_n$, each with mean $\mu$ and uncertainty $\sigma$, the sample mean $\overline{X}$ is a measurement with mean
$$\mu _{\overline{X}} = \mu$$
and with uncertainty
$$\sigma _{\overline{X}} = \frac{\sigma}{\sqrt{n}}$$
For the case of the repeated measurements $X_1,X_2, …, X_n$ with different uncertainties $\sigma _{X_1}, \sigma _{X_2}, …, \sigma _{X_n}$, the ovearall uncertainty of the average measurement can be given as:
$$\sigma _{avg} = \sqrt{\frac{1}{n^2}\sigma _{X_1}^2 + \frac{1}{n^2}\sigma _{X_2}^2 + … + \frac{1}{n^2}\sigma _{X_n}^2}$$
Instead, we can also take the weighted average by taking fractions $c_1, c_2, …, c_n$ measurements each individual one such that $c_1 + c_2 + … + c_n = 1$. The uncertainty of the final weighted average measurement will be:
$$\sigma _{weighted \ average} = \sqrt{c_1^2\sigma _{X_1}^2 + c_2^2\sigma _{X_2}^2 + … + c_n^2\sigma _{X_n}^2}$$
Example: An engineer measures the period of a pendulum (in seconds) to be 2.0 ± 0.2 s. Another independent measurement is made with a more precise clock, and the result is 2.2 ± 0.1 s. The average of these two measurements is 2.1 s. Find the uncertainty in this quantity.
Sol: The uncertainty will be:
$$\sigma _{avg} = \sqrt{\frac{1}{n^2}\sigma _{X}^2 + \frac{1}{n^2}\sigma _{Y}^2} = \sqrt{\frac{1}{4}(0.2)^2 + \frac{1}{4}(0.1)^2} = 0.11s$$
Example: In the above scenario, another engineer suggests that since Y is a more precise measurement than X, a weighted average in which Y is weighted more heavily than X might be more precise than the unweighted average. Express the uncertainty in the weighted average $cX + (1 − c)Y$ in terms of $c$, and find the value of c that minimizes the uncertainty
Sol: The uncertainty for the weighted average is given as:
$$\sigma _{weighted \ average} = \sqrt{c^2\sigma _{X}^2 + (1-c)^2\sigma _{Y}^2} = \sqrt{0.04c^2 + 0.01(1-c)^2} = \sqrt{0.05c^2 - 0.02c + 0.01}$$
Taking derivative with respect to $c$ and computing it to 0, we get
$$\frac{d\sigma _{weighted \ average}}{dc} = 0.10c - 0.02 = 0$$
and hence, the required value of $c$ is 0.2. The new mean and uncertainty is
$$\mu _{weighted \ average} = 0.2X + 0.8Y = 2.16$$
$$\sigma _{weighted \ average} = \sqrt{0.04(0.2)^2 + 0.01(0.8)^2} = 0.09s$$
#### Reference:https://www.mheducation.com/highered/product/statistics-engineers-scientists-navidi/M0073401331.html