91Ó°ÊÓ

Measurements

Frequency distribution of the number of successful outcomes that can be achieved in a given number of trials, each with an equal chance of success.

The mean for x is given below.

$\begin{array}{c}\stackrel{¯}{x}=\frac{1}{n}\underset{i=1}{\overset{n}{∑}}{x}_i\\ =\frac{5.7+4.5+4.8+5.1+4.9}{5}\\ =5\end{array}$

The variance for x is given below.

$\begin{array}{c}{\mathrm{σ}}_x^2=\frac{{\mathrm{Î\pounds }}_{i=1}^n{(x−\stackrel{¯}{x})}^2}{n_x−1}\\ =\frac{2{\left(0.1\right)}^2+{\left(0.7\right)}^2+{\left(0.2\right)}^2+{\left(0.5\right)}^2}{4}\\ =0.02\end{array}$

The standard deviation for x is given below.

$\begin{array}{c}{\mathrm{σ}}_{ma}=\sqrt{\frac{{\mathrm{σ}}_z^2}{n}}I\\ =\sqrt{\frac{0.02}{5}}\end{array}$

The probable error for x is given below.

$\begin{array}{c}{r}_x={\mathrm{σ}}_{mx}I\\ =\left(0.6745\right)\left(0.02\right)\\ =0.135\end{array}$

The mean for y is given below.

$\begin{array}{c}\stackrel{¯}{y}=\frac{1}{n}\underset{i=1}{\overset{n}{∑}}{y}_i\\ =\frac{61.5+60.1+59.7+60.3+58.4}{5}\\ =60\end{array}$

The variance for y is given below.

$\begin{array}{c}{\mathrm{σ}}_z^2=\frac{{\mathrm{Î\pounds }}_{i=1}^n{(y−\stackrel{¯}{y})}^2}{n_y−1}\\ =\frac{{\left(1.5\right)}^2+{\left(0.1\right)}^2+2{\left(0.3\right)}^2+{\left(1.6\right)}^2}{4}\\ =1.25\end{array}$

The standard deviation for y is given below.

$\begin{array}{c}{\mathrm{σ}}_{my}=\sqrt{\frac{{\mathrm{σ}}_y^2}{n}}I\\ =\sqrt{\frac{1.25}{5}}\\ =0.5\end{array}$

The probable error for y is given below.

$\begin{array}{c}{r}_y={\mathrm{σ}}_{y}I\\ =\left(0.6745\right)\left(0.5\right)\\ =0.337\end{array}$

Let us assume the equations mentioned below.

$\begin{array}{c}E\left(w\right)=w({\mathrm{μ}}_v,{\mathrm{μ}}_{\text{v}})\\ w=x+y\end{array}$

The mean for $x+y$is given below.

$\begin{array}{c}\mathbf{E}\left(w\right)={\mathrm{μ}}_v+{\mathrm{μ}}_y\\ =5+60\\ =65\end{array}$

The standard deviation for$x+y$ is given below.

$\begin{array}{c}\mathrm{σ}={\left[\sqrt{{\left(\frac{∂w}{∂x}\right)}^2{\mathrm{σ}}_{mx}^2+{\left(\frac{∂w}{∂y}\right)}^2{\mathrm{σ}}_{my}^2}\right]}_{(x,y)=(x,y)}\\ \mathrm{σ}=\sqrt{{\left(1\right)}^2{\left(0.2\right)}^2+{\left(1\right)}^2{\left(0.5\right)}^2}\\ =0.538\end{array}$

The probable error for$x+y$ is given below.

$\begin{array}{c}{r}_w=\left(0.6745\right)\left(0.538\right)\\ =0.363\end{array}$

Let us assume the equations mentioned below.

$\begin{array}{c}E\left(w\right)=w({\mathrm{μ}}_v,{\mathrm{μ}}_{\text{v}})\\ w=\frac{y}{x}\end{array}$

The mean for $\frac{y}{x}$ is given below.

$\begin{array}{c}E\left(w\right)=\frac{{\mathrm{μ}}_y}{{\mathrm{μ}}_z}\\ =\frac{60}{5}\\ =12\end{array}$

The standard deviation for $\frac{y}{x}$is given below.

$\begin{array}{c}{\mathrm{σ}}_w=\sqrt{{\left(\frac{1}{\stackrel{¯}{x}}\right)}^2{\mathrm{σ}}_{ma}^2+{(−\frac{\stackrel{¯}{y}}{\stackrel{¯}{x}})}^2{\mathrm{σ}}_{me}^2}\\ =\sqrt{{\left(\frac{1}{5}\right)}^2{\left(0.5\right)}^2+{\left(\frac{60}{5^2}\right)}^2{\left(0.2\right)}^2}\\ =0.49\end{array}$

The probable error for$\frac{y}{x}$ is given below.

$\begin{array}{c}{r}_w=0.49×0.6745\\ =0.33\end{array}$

Let us assume the equations mentioned below.

$\begin{array}{c}E\left(w\right)=w({\mathrm{μ}}_v,{\mathrm{μ}}_{\text{v}})\\ w=x^2\end{array}$

The mean for $x^2$is given below.

$\begin{array}{c}E\left(w\right)={\mathrm{μ}}^2\\ =5^2\\ =25\end{array}$

The standard deviation for $x^2$is given below.

$\begin{array}{c}{\mathrm{σ}}_w=\sqrt{{\left(2x\right)}^2{\mathrm{σ}}_m^2}\\ =\sqrt{{(2×5)}^2{\left(0.2\right)}^2}\\ =2\end{array}$

The probable error for$x^2$ is given below.

$\begin{array}{c}{r}_v=2×0.6745\\ =1.35\end{array}$

The values of the function for $x+y$ is given below.

$\begin{array}{c}\mathbf{E}\left(w\right)=65\\ \mathrm{σ}=0.538\\ r_w=0.363\end{array}$

The values of the function for$\frac{y}{x}$ is given below.

$\begin{array}{c}E\left(w\right)=12\\ {\mathrm{σ}}_w=0.49\\ r_w=0.33\end{array}$

The values of the function for$x^2$ is given below.

$\begin{array}{c}E\left(w\right)=25\\ {\mathrm{σ}}_w=2\\ r_v=1.35\end{array}$