91影视

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.

$\begin{array}{c}\stackrel{炉}{x}=\frac{1}{n}\underset{i=1}{\overset{n}{鈭�}}{x}_i\\ =\frac{2.3+2.1+1.8+1.7+2.1}{5}\\ =2\end{array}$

The standard deviation for x is given below.

$\begin{array}{c}{蟽}_n^2=\frac{{\displaystyle \underset{i=1}{\overset{n}{鈭�}}{(x鈭�\stackrel{炉}{x})}^2}}{n_a鈭�1}\\ =\frac{2{\left(0.3\right)}^2+2{\left(0.1\right)}^2+{\left(0.2\right)}^2}{4}\\ =0.06\end{array}$

The probable error for x is given below.

$\begin{array}{c}{蟽}_{ms}=\sqrt{\frac{{蟽}_z^2}{n}}\\ =\sqrt{\frac{0.06}{5}}\\ =0.11\end{array}$

The mean for y is given below.

$\begin{array}{c}\stackrel{炉}{y}=\frac{1}{n}\underset{i=1}{\overset{n}{鈭�}}{y}_i\\ =\frac{1+1.1+0.9}{3}\\ =1\end{array}$

The variance for y is given below.

$\begin{array}{c}{蟽}_n^2=\frac{{\displaystyle \underset{i=1}{\overset{n}{鈭�}}{(y鈭�\stackrel{炉}{y})}^2}}{n_y鈭�1}\\ =\frac{2{\left(0.1\right)}^2}{2}\\ =0.01\end{array}$

The standard deviation for y is given below.

$\begin{array}{c}{蟽}_{\text{y}}=\sqrt{\frac{{蟽}_y^2}{n}}\\ =\sqrt{\frac{0.01}{3}}\\ =0.0577\end{array}$

The probable error for y is given below.

$\begin{array}{c}{r}_y={蟽}_{y}I\\ =\left(0.6745\right)\left(0.0577\right)\\ =0.039\end{array}$

Let us assume the equations mentioned below.

$\begin{array}{c}E\left(w\right)=w({渭}_v,{渭}_{\text{v}})\\ w=x鈭�y\end{array}$

The mean for $x鈭�y$ is given below.

$\begin{array}{c}E\left(w\right)={渭}_e鈭�{渭}_v\\ =2鈭�1\\ =1\end{array}$

The standard deviation for $x鈭�y$ is given below.

$\begin{array}{c}{蟽}_{ma}={\left[\sqrt{{\left(\frac{鈭�w}{鈭�x}\right)}^2{蟽}_{mz}^2+{\left(\frac{鈭�w}{鈭�y}\right)}^2{蟽}_{m,}^2}\right]}_{(x,y)=(\stackrel{炉}{x},\stackrel{炉}{y})}\\ {蟽}_{ma}=\sqrt{{\left(1\right)}^2{\left(0.11\right)}^2+{\left(1\right)}^2{\left(0.0577\right)}^2}\\ =0.124\end{array}$

The probable error for $x鈭�y$is given below.

$\begin{array}{c}{r}_{\text{w}}=\left(0.6745\right)\left(0.124\right)\\ =0.083\end{array}$

Let us assume the equations mentioned below.

$\begin{array}{c}E\left(w\right)=w({渭}_v,{渭}_{\text{v}})\\ w=xy\end{array}$

The mean for $xy$ is given below.

$\begin{array}{c}E\left(w\right)={渭}_x{渭}_y\\ =2脳1\\ =2\end{array}$

The standard deviation for $xy$is given below.

$\begin{array}{c}{蟽}_v=\sqrt{{\stackrel{炉}{y}}^2{蟽}_{me}^2+{\stackrel{炉}{x}}^2{蟽}_{nvy}^2}\\ =\sqrt{{\left(0.11\right)}^2+2^2{\left(0.0577\right)}^2}\\ =0.16\end{array}$

The probable error for $xy$ is given below.

$\begin{array}{c}{r}_{\text{w}}=0.16脳0.6745\\ =0.108\end{array}$

Let us assume the equations mentioned below.

$\begin{array}{c}E\left(w\right)=w({渭}_v,{渭}_{\text{v}})\\ w=\frac{x}{y^3}\end{array}$

The mean for $\frac{x}{y^3}$ is given below.

$\begin{array}{c}E\left(w\right)=\frac{{渭}_z}{{渭}_v^3}\\ =\frac{2}{1}\\ =2\end{array}$

The standard deviation for $\frac{x}{y^3}$ is given below.

$\begin{array}{c}{蟽}_w=\sqrt{{\left(\frac{1}{y^3}\right)}^2{蟽}_{me}^2+{(鈭�\frac{3x}{y^4})}^2{蟽}_m^2}\\ =\sqrt{\left(1\right){\left(0.11\right)}^2+36{\left(0.0577\right)}^2}\\ =0.363\end{array}$

The probable error for$\frac{x}{y^3}$ is given below.

$\begin{array}{c}{r}_v=0.363脳0.6745\\ =0.245\end{array}$

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

$\begin{array}{c}E\left(w\right)=1\\ {蟽}_{ma}=0.124\\ r_{\text{w}}=0.083\end{array}$

The values of the function for $xy$ is given below.

$\begin{array}{c}E\left(w\right)=2\\ {蟽}_v=0.16\\ r_{\text{w}}=0.108\end{array}$

Thvalues of the function for $\frac{x}{y^3}$ is given below.