Chapter 3.2, Problem 19E

a.

To determine

Find the uncertainty in the average of ten independent measurements of length of a component.

Answer to Problem 19E

The uncertainty in the average of ten independent measurements of length of a component is $\underset{\_}{{\sigma }_{\overline{X}}=\text{0}\text{.016 mm}}$.

Explanation of Solution

Given info:

Ten independent measurements are made on the length of a component using an instrument. The uncertainty in the instrument is ${\sigma }_X=0.05\text{mm}$.

Justification:

Uncertainty:

The uncertainty of a process is determined by the standard deviation of the measurements. In other words it can be said that, measure of variability of a process is known as uncertainty of the process.

Therefore, it can be said that uncertainty is simply $\left(\sigma \right)$.

Standard deviation:

The standard deviation is based on how much each observation deviates from a central point represented by the mean. In general, the greater the distances between the individual observations and the mean, the greater the variability of the data set.

The general formula for standard deviation is,

$s=\sqrt{\frac{{\displaystyle \sum x_i{}^2-\frac{{\left({\displaystyle \sum x_i}\right)}^2}{n}}}{n-1}}$.

From the properties of linear combinations of measurements it is known that,

• If $X_1,X_2,\mathrm{...},X_n$ are independent measurements with mean $\mu$ and uncertainty $\sigma$ then the mean and uncertainty of sample mean measurement $\overline{X}$ is are ${\mu }_{\overline{X}}=\mu$ and ${\sigma }_{\overline{X}}=\frac{\sigma }{\sqrt{n}}$.

Here, the number of measurements made on the length of a component using the instrument is $n=10$ and uncertainty in the instrument is ${\sigma }_X=0.05\text{mm}$.

The uncertainty in the average of ten measurements of length of a component is,

$\begin{array}{c}{\sigma }_{\overline{X}}=\frac{{\sigma }_X}{\sqrt{n}}\\ =\frac{0.05}{\sqrt{10}}\\ =0.016\end{array}$

Hence, the uncertainty in the average of ten measurements of length of a component is $\underset{\_}{{\sigma }_{\overline{X}}=\text{0}\text{.016 mm}}$.

b.

To determine

Find the uncertainty in the average of five independent measurements of length of a component using the new measuring device.

Answer to Problem 19E

The uncertainty in the average of five independent measurements of length of a component using the new measuring device is $\underset{\_}{{\sigma }_{\overline{Y}}=\text{0}\text{.0089 mm}}$.

Explanation of Solution

Given info:

Five independent measurements are made on the length of a component using a new measuring device. The uncertainty in the new measuring device is ${\sigma }_Y=0.02\text{mm}$.

Justification:

From the properties of linear combinations of measurements it is known that,

• If $X_1,X_2,\mathrm{...},X_n$ are independent measurements with mean $\mu$ and uncertainty $\sigma$ then the mean and uncertainty of sample mean measurement $\overline{X}$ is are ${\mu }_{\overline{X}}=\mu$ and ${\sigma }_{\overline{X}}=\frac{\sigma }{\sqrt{n}}$.

Here, the number of measurements made on the length of a component using the new measuring device is $n=5$ and uncertainty in the new measuring device is ${\sigma }_Y=0.02\text{mm}$.

The uncertainty in the average of five measurements of length of a component is,

$\begin{array}{c}{\sigma }_{\overline{Y}}=\frac{{\sigma }_Y}{\sqrt{n}}\\ =\frac{0.02}{\sqrt{5}}\\ =0.0089\end{array}$

Hence, the uncertainty in the average of five measurements of length of a component is $\underset{\_}{{\sigma }_{\overline{Y}}=\text{0}\text{.0089 mm}}$.

c.

To determine

Find the uncertainty in $\frac{\overline{X}+\overline{Y}}{2}$.

Find the uncertainty in $\left(\frac{10}{15}\right)\overline{X}+\left(\frac{5}{15}\right)\overline{Y}$.

Answer to Problem 19E

The uncertainty in $\frac{\overline{X}+\overline{Y}}{2}$ is $\underset{\_}{{\sigma }_{\frac{\overline{X}+\overline{Y}}{2}}=\text{0}\text{.0091 mm}}$.

The uncertainty in $\left(\frac{10}{15}\right)\overline{X}+\left(\frac{5}{15}\right)\overline{Y}$ is $\underset{\_}{{\sigma }_{\left(\frac{10}{15}\right)\overline{X}+\left(\frac{5}{15}\right)\overline{Y}}=\text{0}\text{.0111 mm}}$.

Explanation of Solution

Justification:

Here, the number of measurements made on the length of a component using the instrument is $n=10$ and uncertainty in the instrument is ${\sigma }_X=0.05\text{mm}$.

The uncertainty in the average of ten independent measurements of length of a component is ${\sigma }_{\overline{X}}=\text{0}\text{.016 mm}$.

Here, the number of measurements made on the length of a component using the new measuring device is $n=5$ and uncertainty in the new measuring device is ${\sigma }_Y=0.02\text{mm}$.

The uncertainty in the average of five independent measurements of length of a component using the new measuring device is ${\sigma }_{\overline{Y}}=\text{0}\text{.0089 mm}$.

From the properties of linear combinations of measurements it is known that,

• If $X_1,X_2,\mathrm{...},X_n$ are independent measurements and $c_1,c_2,\mathrm{...},c_n$ are constants then the uncertainty of the random variable $c_1{X}_1+c_2{X}_2+\mathrm{...}+c_n{X}_n$ is ${\sigma }_{c_1{X}_1+c_2{X}_2+\mathrm{...}+c_n{X}_n}=\sqrt{c_1^2{\sigma }_{X_1}^2+c_2^2{\sigma }_{X_2}^2+\mathrm{...}+c_n^2{\sigma }_{X_n}^2}$.

Uncertainty in$\frac{\overline{X}+\overline{Y}}{2}$:

The uncertainty of the random variable $\frac{\overline{X}+\overline{Y}}{2}$ is,

$\begin{array}{c}{\sigma }_{\frac{\overline{X}+\overline{Y}}{2}}=\sqrt{{\left(\frac{1}{2}\right)}^2\times {\sigma }_{\overline{X}}^2+{\left(\frac{1}{2}\right)}^2\times {\sigma }_{\overline{Y}}^2}\\ =\sqrt{\frac{1}{4}\times {0.016}^2+\frac{1}{4}\times \times {0.0089}^2}\\ =\sqrt{0.000825}\\ =0.0091\end{array}$

Thus, the standard deviation of random variable $\frac{\overline{X}+\overline{Y}}{2}$ is 0.0091.

Hence, the uncertainty of the random variable $\frac{\overline{X}+\overline{Y}}{2}$ is $\underset{\_}{{\sigma }_{\frac{\overline{X}+\overline{Y}}{2}}=\text{0}\text{.0091 mm}}$.

Uncertainty in $\left(\frac{10}{15}\right)\overline{X}+\left(\frac{5}{15}\right)\overline{Y}$:

The uncertainty of the random variable $\left(\frac{10}{15}\right)\overline{X}+\left(\frac{5}{15}\right)\overline{Y}$ is,

$\begin{array}{c}{\sigma }_{\left(\frac{10}{15}\right)\overline{X}+\left(\frac{5}{15}\right)\overline{Y}}=\sqrt{{\left(\frac{10}{15}\right)}^2\times {\sigma }_{\overline{X}}^2+{\left(\frac{5}{15}\right)}^2\times {\sigma }_{\overline{Y}}^2}\\ =\sqrt{\frac{4}{9}\times {0.016}^2+\frac{1}{9}\times \times {0.0089}^2}\\ =\sqrt{0.00012258}\\ =0.0111\end{array}$

Thus, the standard deviation of random variable $\left(\frac{10}{15}\right)\overline{X}+\left(\frac{5}{15}\right)\overline{Y}$ is 0.0111.

Hence, the uncertainty of the random variable $\left(\frac{10}{15}\right)\overline{X}+\left(\frac{5}{15}\right)\overline{Y}$ is $\underset{\_}{{\sigma }_{\left(\frac{10}{15}\right)\overline{X}+\left(\frac{5}{15}\right)\overline{Y}}=\text{0}\text{.0111 mm}}$.

d.

To determine

Find the value of c that minimizes the uncertainty in the weighted average $c\overline{X}+\left(1-c\right)\overline{Y}$.

Find the uncertainty the weighted average $c\overline{X}+\left(1-c\right)\overline{Y}$.

Answer to Problem 19E

The uncertainty in the weighted average $c\overline{X}+\left(1-c\right)\overline{Y}$ is minimum for $\underset{\_}{c=\text{0}\text{.24}}$.

The uncertainty the weighted average $c\overline{X}+\left(1-c\right)\overline{Y}$. is $\underset{\_}{{\sigma }_{c\overline{X}+\left(1-c\right)\overline{Y}}=\text{0}\text{.007778 mm}}$.

Explanation of Solution

Justification:

The value of c that minimizes the weighted average of $\overline{X}and\overline{Y}$:

From the properties of linear combinations of measurements it is known that,

• If $X\text{and}Y$ are independent measurements of the same quantity with uncertainties ${\sigma }_X$ and ${\sigma }_Y$ then the weighted average of the independent measurements $X\text{and}Y$ with minimum uncertainty is $cX+\left(1-c\right)Y$. Here, the value of c is $c=\frac{{\sigma }_Y^2}{{\sigma }_X^2+{\sigma }_Y^2}$ and $1-c=\frac{{\sigma }_X^2}{{\sigma }_X^2+{\sigma }_Y^2}$.

Here, the uncertainty in the average of ten independent measurements of length of a component is ${\sigma }_{\overline{X}}=\text{0}\text{.016 mm}$ and the uncertainty in the average of five independent measurements of length of a component using the new measuring device is ${\sigma }_{\overline{Y}}=\text{0}\text{.0089 mm}$.

Here, $\overline{X}\text{and}\overline{Y}$ are two independent measurements made on same quantity.

Therefore, the weighted average of $\overline{X}\text{and}\overline{Y}$ is $c\overline{X}+\left(1-c\right)\overline{Y}$.

The value of c that minimizes the weighted average $c\overline{X}+\left(1-c\right)\overline{Y}$ is,

$\begin{array}{c}c=\frac{{\sigma }_Y^2}{{\sigma }_X^2+{\sigma }_Y^2}\\ =\frac{{0.0089}^2}{{0.0089}^2+{0.016}^2}\\ =0.2363\\ \simeq 0.24\end{array}$

Thus, the uncertainty in the weighted average $c\overline{X}+\left(1-c\right)\overline{Y}$ is minimum for $\underset{\_}{c=\text{0}\text{.24}}$.

Uncertainty in the weighted average $c\overline{X}+\left(1-c\right)\overline{Y}$:

From the properties of linear combinations of measurements it is known that,

• If $X_1,X_2,\mathrm{...},X_n$ are independent measurements and $c_1,c_2,\mathrm{...},c_n$ are constants then the uncertainty of the random variable $c_1{X}_1+c_2{X}_2+\mathrm{...}+c_n{X}_n$ is ${\sigma }_{c_1{X}_1+c_2{X}_2+\mathrm{...}+c_n{X}_n}=\sqrt{c_1^2{\sigma }_{X_1}^2+c_2^2{\sigma }_{X_2}^2+\mathrm{...}+c_n^2{\sigma }_{X_n}^2}$.

The uncertainty in the weighted average $c\overline{X}+\left(1-c\right)\overline{Y}$ is,

$\begin{array}{c}{\sigma }_{c\overline{X}+\left(1-c\right)\overline{Y}}=\sqrt{{\left(0.24\right)}^2\times {\sigma }_{\overline{X}}^2+{\left(1-0.24\right)}^2\times {\sigma }_{\overline{Y}}^2}\\ =\sqrt{{\left(0.24\right)}^2\times {0.016}^2+{\left(0.76\right)}^2\times \times {0.0089}^2}\\ =0.007778\end{array}$

Thus, the standard deviation of random variable $c\overline{X}+\left(1-c\right)\overline{Y}$ is 0.007778.

Hence, the uncertainty the weighted average $c\overline{X}+\left(1-c\right)\overline{Y}$. is $\underset{\_}{{\sigma }_{c\overline{X}+\left(1-c\right)\overline{Y}}=\text{0}\text{.007778 mm}}$.