Rank correlation is the measure of association or strength between the ranked variables. For example: the rank of this  numerical data 65, 25, 75, 69 would be 3, 4, 1, 2 respectively.

Spearman's rank cofficient : $R=1-\dfrac { 6\sum { { d } _{ i }^{ 2 } }  }{ n({ n }^{ 2 }-1) } $

Fact. If the variables are uncorrelated or independent then covariance
and coefficient of correlation between the variable both are equal to 0
i.e. $r _{xy}=Cov\left ( x, y \right )=0$ 

Correlation coefficient 
${ r } _{ x,y }=\dfrac { n\sum { xy } -\sum { x } \sum { y }  }{ \sqrt { \left[ n\sum { { x }^{ 2 }-{ \left( \sum { x }  \right)  }^{ 2 } }  \right] \left[ n\sum { { y }^{ 2 }-{ \left( \sum { y }  \right)  }^{ 2 } }  \right]  }  } $

$=\displaystyle\frac { 30-12 }{ \sqrt { 64\times 81 }  } $
$\Rightarrow r _{x,y}=\dfrac{1}{4}$

Rank Difference $(d)$ | $d^2$ | | --- | --- | --- | | 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. | -2 -4 -1 3 2 0 -2 3 3 -2 | 4 16 1 9 4 0 4 9 9 4 |

 $\sum d^2=60,\quad n=10$

$r=1-\cfrac{6\sum d^2}{n(n^2-1)}$

$r=1-\cfrac{6(60)}{10(10^2-1)}$

$r=1-\cfrac{360}{990}$

$r=0.6363....\approx 0.64$

Descending order arranged data will be as follows:

Rank | $Y$ | Rank | $|d|$ | $d^2$ | | --- | --- | --- | --- | --- | --- | | 48 33 40 9 16 16 65 25 16 57 | 3 5 4 10 7 7 1 6 7 2 | 13 13 24 6 15 4 20 9 6 19 | 5 5 1 8 4 10 2 7 8 3 | 2 0 3 2 3 3 1 1 1 1 | 4 0 9 4 9 9 1 1 1 1 |

$n=10,\quad \sum d^2=39$

$r=1-\cfrac{6\sum d^2}{n(n^2-1)}=1-\cfrac{6\times 39}{10(10^2-1)}=1-\cfrac{234}{990}=0.76$

Attendance | Rank | Position | $|d|$ | $d^2$ | | --- | --- | --- | --- | --- | --- | | A B C D E F G H I J K L | 27 30 18 25 32 12 19 11 32 12 12 15 | 4 3 7 5 1 9 6 12 1 9 9 8 | 1 2 3 4 5 6 7 8 9 10 11 12 | 3 1 4 1 4 3 1 4 8 1 2 4 | 9 1 16 1 16 9 1 16 64 1 4 16 |

$n=12,\quad \sum d^2=154$

$r=1-\cfrac{6\sum d^2}{n(n^2-1)}=1-\cfrac{6\times 154}{12(12^2-1)}=1-\cfrac{924}{1716}=0.48$

Rank | Height(Son) | Rank | $|d|$ | $d^2$ | | --- | --- | --- | --- | --- | --- | | 65 66 67 67 68 69 70 72 | 8 7 5 5 4 3 2 1 | 67 68 65 68 72 72 69 71 | 7 5 8 5 1 1 4 3 | 1 2 3 0 3 2 2 2   | 1 4 9 0 9 4 4 4 |

$n=08,\quad \sum d^2=35$

$r=1-\cfrac{6\sum d^2}{n(n^2-1)}=1-\cfrac{6\times 35}{8(8^2-1)}=1-\cfrac{210}{504}=0.58$

History $(H)$ | Mathematics$(M)$ | Rank $(H)$ | Rank$(M)$ | $|d|$ | $d^2$ | | --- | --- | --- | --- | --- | --- | --- | | A B C D E F G H I J K L | 69 36 39 71 67 76 40 20 85 65 55 34   | 33 52 71 25 79 22 83 81 24 35 46 64 | 4 10 9 3 5 2 8 12 1 6 7 11 | 9 6 4 10 3 12 1 2 11 8 7 5 | 5 4 5 7 2 10 7 10 10 2 0 6   | 25 6 25 49 4 100 49 100 100 4 0 36   |

$n=12,\quad \sum d^2=508$

$r=1-\cfrac{6\sum d^2}{n(n^2-1)}=1-\cfrac{6\times 508}{12(12^2-1)}=1-\cfrac{3048}{1716}=-0.77$

Mathematics $(Y)$ | $XY$ | $X^2$ | $Y^2$ | | --- | --- | --- | --- | --- | | 1 3 3 4 5 6 7 8 9 10 | 2 4 1 5 3 9 7 10 6 8 | 2 12 3 20 15 54 49 80 54 80 | 1 9 9 16 25 6 49 64 81 100 | 4 16 1 25 9 81 49 100 36 64 |

 $\sum X=56,\quad \sum Y=55,\quad \sum XY=369,\quad \sum X^2=390,\quad \sum Y^2=385$

$N=10$

Cov$(x,y)=\cfrac{\sum XY}{N}-\cfrac{\sum X}{N}.\cfrac{\sum Y}{N}=\cfrac{369}{10}-\cfrac{56}{10}.\cfrac{55}{10}=6.1$

$\sigma _x=\sqrt{\cfrac{\sum X^2}{N}-\left(\cfrac{\sum X^2}{N}\right)^2}=\sqrt{\cfrac{390}{10}-\left(\cfrac{56}{10}\right)^2}=2.76$

$\sigma _y=\sqrt{\cfrac{\sum Y^2}{N}-\left(\cfrac{\sum Y^2}{N}\right)^2}=\sqrt{\cfrac{385}{10}-\left(5.5\right)^2}=2.87$

$r=\cfrac{Cov(x,y)}{\sigma _x.\sigma _y}=\cfrac{6.1}{2.87\times 2.76}=0.77$

The formula for speraman's rank coefficient is

Let the ranks of students obtained in Physics be $x$ and the ranks of students obtained in Chemistry be $y$.

Rank Correlation Coefficient,$r _s=1-\cfrac{6\sum{\rho _i^2}}{n(n^2-1)}$

The term tie is used in connection with rank order statistics. 

Rank correlation is the technique in which ranks are provided to the data after sorting it so sometimes, it becomes very difficult to assign ranks if the variables are large in numbers and also in this, ranks are taken inspite of original values.

Rank correlation coefficient is developed by Charles Spearman, a renowned statistician that measures the strength between the ranked variables.

Rank correlation technique measures the strength and direction between two variables by providing suitable ranks to the concerned variables, e.g. marks of students in a class can be easily ranked.

$1st$ Judge | Rank | $2nd$ Judge | Rank | $|d|$ | $d^2$ | | --- | --- | --- | --- | --- | --- | --- | | A B C D E | 5.7 5.8 5.9 5.6 5.5 | 3 2 1 4 5 | 5.6 5.7 6.0 5.5 5.8 | 4 3 1 5 2 | 1 1 0 1 3 | 1 1 0 1 9 |

$n=5,\quad \sum d^2=12$

$r=1-\cfrac{6\sum d^2}{n(n^2-1)}=1-\cfrac{6\times 12}{5(5^2-1)}=1-\cfrac{72}{240}=0.44$

Marks in Maths$(Y)$ | $XY$ | $X^2$ | $Y^2$ | | --- | --- | --- | --- | --- | | 35 23 47 17 10 43 9 6 28 | 30 33 45 23 8 49 12 4 31 | 1050 759 2115 391 80 2107 108 24 868 | 1225 529 2209 289 100 1849 81 35 784 | 900 1089 2025 529 64 2401 144 16 961 |

 $\sum X=218,\quad \sum Y=235,\quad \sum XY=7502,\quad \sum X^2=7102,\quad \sum Y^2=8129$

$N=09$

Cov$(x,y)=\cfrac{\sum XY}{N}-\cfrac{\sum X}{N}.\cfrac{\sum Y}{N}=\cfrac{7502}{09}-\cfrac{218}{9}.\cfrac{235}{9}=201.08$

$\sigma _x=\sqrt{\cfrac{\sum X^2}{N}-\left(\cfrac{\sum X^2}{N}\right)^2}=\sqrt{\cfrac{7102}{09}-\left(\cfrac{218}{09}\right)^2}=14.22$

$\sigma _y=\sqrt{\cfrac{\sum Y^2}{N}-\left(\cfrac{\sum Y^2}{N}\right)^2}=\sqrt{\cfrac{8129}{09}-\left(\cfrac{235}{9}\right)^2}=14.88$

$r=\cfrac{Cov(x,y)}{\sigma _x.\sigma _y}=\cfrac{201.08}{14.22\times 14.88}=0..95$

The correction factor in the rank correlation coefficient is given by $\sum m(m^2-1)$

Let the ranks of students obtained in Physics be $x$ and the ranks of students obtained in Chemistry be $y$.

$Correlation \:coefficient = \dfrac{cov (x,y)}{std\: deviation (x) \times std\: deviation (y)}$

Let std deviation of $x$ be $x$.

We have $correlation \:coefficient=0.6, cov(x,y)=27, std\:deviation(y)=\sqrt{25}=5$

Substituting respective values

$0.6=\dfrac{27}{5\times x}$

$\Rightarrow x=9$

So variance of $x$ is $9^2=81$