Question

Given the measurements.

$\begin{array}{l}\mathbf{x}\mathbf{:}\mathbf{5}\mathbf{.8}\mathbf{,}\mathbf{6}\mathbf{.1}\mathbf{,}\mathbf{6}\mathbf{.4}\mathbf{,}\mathbf{5}\mathbf{.9}\mathbf{,}\mathbf{5}\mathbf{.7}\mathbf{,}\mathbf{6}\mathbf{.2}\mathbf{,}\mathbf{5}\mathbf{.9}\\ \mathbf{y}\mathbf{:}\mathbf{2}\mathbf{.7}\mathbf{,}\mathbf{3}\mathbf{.0}\mathbf{,}\mathbf{2}\mathbf{.9}\mathbf{,}\mathbf{3}\mathbf{.3}\mathbf{,}\mathbf{3}\mathbf{.1}\end{array}$

Find the mean value and the probable error of $\mathbf{2}\mathit{x}\mathbf{-}\mathit{y}\mathbf{,}{\mathit{y}}^{\mathbf{2}}\mathbf{-}\mathit{x}\mathbf{,}{\mathit{e}}^{\mathbf{y}}$and $\frac{\mathbf{x}}{{\mathbf{y}}^{\mathbf{2}}}$.

Short answer

Required expressions are,

$\begin{array}{c}\overline{x}=6,\\ r_x=0.0925\\ \overline{y}=3,\\ r_y=0.0624\end{array}$

$\begin{array}{c}\overline{2x-y}=9,\\ r_{\text{w}}=0.14\\ \overline{y^2-x}=3,\\ r=0.409\end{array}$

$\begin{array}{c}\overline{e^y}=20,\\ r=1.35\\ \overline{x/y^2}=\frac{2}{3},\\ r=0.03\end{array}$

Step-by-step solution

Given Information

Measurements are mentioned below:

$\begin{array}{l}x:5.8,6.1,6.4,5.9,5.7,6.2,5.9\\ y:2.7,3.0,2.9,3.3,3.1\end{array}$

Definition of Probable error. 

The amount by which a sample's arithmetic mean is predicted to vary solely due to chance.

Calculate mean and probable error value of x .

Calculate expected value.

$\begin{array}{c}\overline{x}=\frac{{\displaystyle \sum _{i=1}^n{x}_i}}{n}\\ =\frac{5.7+5.8+2\times 5.9+6.1+6.2+6.4}{7}\\ =6\end{array}$

Calculate variance.

$\begin{array}{c}\mathrm{Var}\left(x\right)=\frac{1}{n_x-1}\sum _{i=1}^n{(x_i-\overline{x})}^2\\ =\frac{1}{6}[{\left(0.3\right)}^2+2{\left(0.2\right)}^2+3{\left(0.1\right)}^2+{\left(0.4\right)}^2]\\ =0.06\end{array}$

Calculate error of values.

$\begin{array}{c}{\sigma }_{mx}=\sqrt{\frac{\mathrm{Var}\left(x\right)}{n}}\\ =\sqrt{\frac{0.06}{7}}\\ =0.0925\end{array}$

Calculate probable error.

$\begin{array}{c}{r}_x=0.6745\left(0.0925\right)\\ =0.0624\end{array}$

Calculate mean and probable error value of y 

Calculate expected value.

$\begin{array}{c}\overline{y}=\frac{{\displaystyle \sum _{i=1}^n{y}_i}}{n}\\ =\frac{2.7+2.9+3.0+3.1+3.3}{5}\\ =3\end{array}$

Calculate variance.

$\begin{array}{c}{\sigma }_y^2=\frac{1}{n_i-1}\sum _{i=1}^n{(y_i-\overline{y})}^2\\ =\frac{1}{4}[2\times {\left(0.3\right)}^2+2\times {\left(0.1\right)}^2]\\ =0.05\end{array}$

Calculate error of values.

$\begin{array}{c}{\sigma }_{my}=\sqrt{\frac{\mathrm{Var}\left(y\right)}{n_y}}\\ =\sqrt{\frac{0.05}{5}}\\ =0.1\end{array}$

Calculate probable error.

$\begin{array}{c}{r}_y=0.6745\left(0.1\right)\\ =0.0674\end{array}$

Calculate mean and probable error value of  .2x−y

Calculate expected value.

$\begin{array}{c}\overline{w}=\overline{2x-y}\\ =2\overline{x}-\overline{y}\\ =2\left(6\right)-3\\ =9\end{array}$$\begin{array}{c}\overline{w}=\overline{2x-y}\\ =2\overline{x}-\overline{y}\\ =2\left(6\right)-3\\ =9\end{array}$

Calculate error of values.

$\begin{array}{c}{\sigma }_{mw}=\sqrt{4{\sigma }_{mu}^2+{\sigma }_{my}^2}\\ =\sqrt{4{\left(0.0925\right)}^2+{\left(0.1\right)}^2}\\ =0.21\end{array}$

Calculate probable error.

$\begin{array}{c}{r}_w=0.6745\left(0.21\right)\\ =0.14\end{array}$

Calculate mean and probable error value of.y2−x 

Calculate expected value.

$\begin{array}{c}w=y^2-x\\ E\left(w\right)={\mu }_y^2-{\mu }_x\\ =3^2-6\\ =3\end{array}$

Calculate error of values.

$\begin{array}{c}{\sigma }_{\mathrm{m}w}={\left[\sqrt{{\left(\frac{\partial w}{\partial x}\right)}^2{\sigma }_{mx}^2+{\left(\frac{\partial w}{\partial y}\right)}^2{\sigma }_{my}^2}\right]}_{(x,y)=(\overline{x},\overline{y})}\\ {\sigma }_{mw}=\sqrt{{(-1)}^2{\left(0.0925\right)}^2+{(2\times 3)}^2{\left(0.1\right)}^2}\\ =0.607\end{array}$

Calculate probable error.

$\begin{array}{c}{r}_w=\left(0.6745\right)\left(0.607\right)\\ =0.409\end{array}$

Calculate mean and probable error value of  ey. 

Calculate expected value.

$\begin{array}{c}w=e^y\\ E\left(w\right)=e^{\mu }\\ =e^3\\ =20\end{array}$

Calculate error of values.

$\begin{array}{c}{\sigma }_{\mathrm{m}w}={\left[\sqrt{{\left(\frac{\partial w}{\partial x}\right)}^2{\sigma }_{me}^2+{\left(\frac{\partial w}{\partial y}\right)}^2{\sigma }_{\mathrm{my}}^2}\right]}_{(x,y)=(\overline{x},\overline{y})}\\ {\sigma }_{mw}=\sqrt{{\left(e^3\right)}^2{\left(0.1\right)}^2}=2.0\end{array}$

Calculate probable error.

$\begin{array}{c}{r}_w=\left(0.6745\right)\left(2.0\right)\\ =1.35\end{array}$

Calculate mean and probable error value of .xy2

Calculate expected value.

$\begin{array}{c}w=\frac{x}{y^2}\\ E\left(w\right)=\frac{x}{y^2}\\ =\frac{{\mu }_x}{{\mu }_y^2}\\ \equiv \frac{6}{3^2}\end{array}$

$E\left(w\right)=\frac{2}{3}$

Calculate error of values.

$\begin{array}{c}{\sigma }_{mw}={\left[\sqrt{{\left(\frac{\partial w}{\partial x}\right)}^2{\sigma }_{mx}^2+{\left(\frac{\partial w}{\partial y}\right)}^2{\sigma }_{my}^2}\right]}_{(x,y)=(\overline{x},\overline{y})}\\ {\sigma }_{\text{mw}}=\sqrt{{\left(\frac{1}{3^2}\right)}^2{\left(0.0925\right)}^2+{(-\frac{2\times 6}{3^3})}^2{\left(0.1\right)}^2}\\ =0.046\end{array}$

Calculate probable error.

$\begin{array}{c}{r}_w=\left(0.6745\right)\left(0.046\right)\\ =0.03\end{array}$

Hence, required expressions are,

$\begin{array}{c}\overline{x}=6,\\ r_x=0.0925\\ \overline{y}=3,\\ r_y=0.0624\end{array}$

$\begin{array}{c}\overline{2x-y}=9,\\ r_{\text{w}}=0.14\\ \overline{y^2-x}=3,\\ r=0.409\end{array}$

$\begin{array}{c}\overline{e^y}=20,\\ r=1.35\\ \overline{x/y^2}=\frac{2}{3},\\ r=0.03\end{array}$