Karl Pearson’s Coefficient of Correlation

Correlation is a statistical measure that shows the degree of relationship between two variables. Karl Pearson’s coefficient of correlation is one of the most widely used methods for determining the strength and direction of this relationship. It plays an important role in statistics, data science, and also in competitive exams like JEE Main, JEE Advanced, and other entrance tests.

1.0Karl Pearson’s Coefficient of Correlation – Definition and Theory

• Karl Pearson’s coefficient of correlation measures the linear relationship between two variables.
• It is denoted by ‘r’.
• The value of Karl Pearson’s coefficient of correlation lies between -1 and +1.
• • If r = +1 → Perfect positive correlation 
  • If r = -1 → Perfect negative correlation
  • If r = 0 → No correlation
• Karl Pearson’s coefficient of correlation method uses covariance and standard deviation.

Formula:

$r=\frac{\text{Cov}(X,Y)}{{\sigma }_X{\sigma }_Y}$

Where,

• Cov(X,Y) = Covariance between variables X and Y
• ${\sigma }_X,{\sigma }_Y$ = Standard deviations of X and Y

2.0Methods to Calculate Karl Pearson’s Coefficient of Correlation

1. Direct Method

This is the standard method using raw data.

$r=\frac{n\sum xy-(\sum x)(\sum y)}{\sqrt{[n\sum x^2-(\sum x{)}^2][n\sum y^2-(\sum y{)}^2]}}$

• n: Number of pairs of observations
• x, y: Data values of two variables

Use this method when numbers are small and simple.

2. Assumed Mean Method (Short-cut Method)

When data values are large, calculations can be simplified by taking deviations from assumed means.

Let:

$d_x=x-A_x,d_y=y-A_y$

Where $A_x,A_y$ are assumed means.

Formula:

$r=\frac{\sum d_x{d}_y}{\sqrt{\sum d_x^2\cdot \sum d_y^2}}$

This reduces calculation effort.

3. Step Deviation Method (Coding Method)

When data is very large or has a common difference (like in grouped data), divide deviations by a factor.

$d_x=\frac{x-A_x}{h_x},d_y=\frac{y-A_y}{h_y}$

Formula becomes:

$r=\frac{\sum d_x{d}_y}{\sqrt{\sum d_x^2\cdot \sum d_y^2}}$

This method is useful for grouped frequency distributions.

4. Using Covariance Formula

Another approach is to use:

$r=\frac{\text{Cov}(X,Y)}{{\sigma }_X{\sigma }_Y}$

• First, calculate the covariance of X and Y.
• Then divide by the product of their standard deviations.

Very common in statistics and probability questions.

So, the main methods are:

1. Direct Method
2. Assumed Mean Method
3. Step Deviation Method
4. Covariance Method

3.0Examples Based on Methods to Calculate Karl Pearson’s Coefficient of Correlation

Method 1 — Direct (raw-sum) formula

Example 1 (Direct method)

Data: (X,Y) = (1, 2), (2, 3), (3, 5), (4, 4). Compute r.

Solution

Compute sums:

$n=4,\sum x=10,\sum y=14,\sum xy=39,\sum x^2=30,\sum y^2=54.$

Use raw-sum formula

$r=\frac{n\sum xy-(\sum x)(\sum y)}{\sqrt{(n\sum x^2-(\sum x{)}^2)(n\sum y^2-(\sum y{)}^2)}}.$

Numerator: $4\cdot 39-10\cdot 14=156-140=16.$

Denominator: $\sqrt{(120-100)(216-196)}=\sqrt{20\cdot 20}=20$_.

So

$r=\frac{16}{20}=0.8$

Method 2 — Assumed-mean (shortcut) method

Example 2 (Assumed mean)

Data: X = {98, 102, 100, 105, 95}, Y = {200, 210, 205, 215, 195}.

Take assumed means $A_x=100,A_y=205$.

Solution

Compute deviations $d_x=x-A_x,d_y=y-A_y$ :

Now

$r=\frac{\sum d_x{d}_y}{\sqrt{(\sum d_x^2)(\sum d_y^2)}}=\frac{120}{\sqrt{58\cdot 250}}.$

Compute denominator: $\sqrt{14500}\approx 120.415$_. Hence

$r\approx \frac{120}{120.415}\approx 0.9965$

(very strong positive linear association).

Note: assumed-mean reduces arithmetic — you only work with small deviations.

Method 3 — Step-deviation / coding method (useful when data have constant step)

Example 3 (Step-deviation)

Data: X = {100, 110, 120, 130}, Y = {45, 50, 60, 65}.

Use h = 10. Let $u=(x-Ax)/h,v=(y-Ay)/hwith{A}_x=115,A_y=55.$

Solution

Compute coded deviations:

Then

$r=\frac{\sum uv}{\sqrt{(\sum u^2)(\sum v^2)}}=\frac{3.5}{\sqrt{5\cdot 2.5}}=\frac{3.5}{\sqrt{12.5}}\approx \frac{3.5}{3.5355}\approx \mathrm{0.990.}$

Because we scaled both variables by constants, r is unchanged — coding just simplifies arithmetic.

Method 4 — Covariance and standard deviations (conceptual form)

Example 4 (Covariance method, same data as Example 1)

Data: (1, 2), (2, 3), (3, 5), (4, 4). We already found r = 0.8. Show via covariance.

Solution

Means: $\bar{x}=2.5,\bar{y}=3.5$_.

Compute sums of deviations:

Population covariance: $\text{Cov}(X,Y)=\frac{\sum (x-\bar{x})(y-\bar{y})}{n}=\frac{4}{4}=1.$

Population standard deviations: 

${\sigma }_x=\sqrt{\frac{5}{4}}=\sqrt{1.25},{\sigma }_y=\sqrt{1.25}.$
Thus

$r=\frac{\text{Cov}(X,Y)}{{\sigma }_X{\sigma }_Y}=\frac{1}{(\sqrt{1.25}{)}^2}=\frac{1}{1.25}=0.8,$ same result as Example 1.

4.0Solved Examples on Karl Pearson’s Coefficient of Correlation 

Example 1: Find Karl Pearson’s coefficient of correlation for the following data:

Solution:

Here, Y = 2X.

Clearly, the relation is perfectly linear.

Thus, r = +1.

Example 2: Given the following data, calculate Karl Pearson’s coefficient of correlation:

Solution:

Step 1: Compute mean of X and Y. 

$\bar{X}=\frac{10+12+13+16+17+18+19}{7}=15$

$\bar{Y}=\frac{40+38+43+45+37+43+49}{7}=42.14$

Step 2: Find deviations

$dx=X-\bar{X},dy=Y-\bar{Y}$.

Step 3: Apply formula:

$r=\frac{\sum dx\cdot dy}{\sqrt{\sum d{x}^2\cdot \sum d{y}^2}}$

After calculations:

$r\approx 0.82$

So, there is a strong positive correlation.

Example 3: If the correlation coefficient between X and Y is r = 0.6, standard deviation of X = 3, standard deviation of Y = 4, and covariance of (X, Y) = ?

Solution:
We know,

$r=\frac{\text{Cov}(X,Y)}{{\sigma }_X{\sigma }_Y}$$0.6=\frac{\text{Cov}(X,Y)}{3\times 4}$$\text{Cov}(X,Y)=0.6\times 12=7.2$

Example 4: For the data X = {1, 2, 3}, Y = {6, 4, 2}, find r.

Solution:

Here Y = -2X + 8 (perfect linear with negative slope). Hence correlation is perfect negative:

r=−1. 

Example 5: For data (X, Y): (1, 2), (2, 3), (3, 5), (4, 4). Compute Karl Pearson’s r.

Solution:

Compute sums (careful arithmetic):

  n = 4

$$\begin{gathered} \sum x=1+2+3+4=10 \\ \sum y=2+3+5+4=14 \\ \sum xy=1\cdot 2+2\cdot 3+3\cdot 5+4\cdot 4=2+6+15+16=39 \\ \sum x^2=1+4+9+16=30 \\ \sum y^2=4+9+25+16=54 \end{gathered}$$

Use formula

$r=\frac{n\sum xy-(\sum x)(\sum y)}{\sqrt{(n\sum x^2-(\sum x{)}^2)(n\sum y^2-(\sum y{)}^2)}}.$

Numerator: $4\cdot 39-10\cdot 14=156-140=16.$

Denominator: $\sqrt{(4\cdot 30-100)(4\cdot 54-196)}=\sqrt{(120-100)(216-196)}=\sqrt{20\cdot 20}=20.$

So r = 16/20 = 0.8.

r = 0.8 — strong positive linear correlation.

Example 6: Paired grouped data (midpoints used as scores): $(x_i,f_i,y_i):(5,4,50),(15,6,60),(25,5,70)$. 

Compute r treating $f_i$ as weights.

Solution:

First compute weighted sums (N = total frequency = 4 + 6 + 5 = 15).

= 1000 + 5400 + 8750 = 15150.

$\sum fx=5\cdot 4+15\cdot 6+25\cdot 5=20+90+125=235.$$\sum fy=50\cdot 4+60\cdot 6+70\cdot 5=200+360+350=910.$$\sum f{x}^2=5^2\cdot 4+15^2\cdot 6+25^2\cdot 5=100+1350+3125=4575.$$\sum f{y}^2=50^2\cdot 4+60^2\cdot 6+70^2\cdot 5=10000+21600+24500=56100.$$\sum fxy=(5\cdot 50)\cdot 4+(15\cdot 60)\cdot 6+(25\cdot 70)\cdot 5$= 1000 + 5400 + 8750 = 15150.

Now use weighted form:

$r=\frac{N\sum fxy-(\sum fx)(\sum fy)}{\sqrt{(N\sum f{x}^2-(\sum fx{)}^2)(N\sum f{y}^2-(\sum fy{)}^2)}}.$

Numerator: $15\cdot 15150-235\cdot 910=227250-213850=13400.$

First variance term: $15\cdot 4575-235^2=68625-55225=13400.$

Second variance term: $15\cdot 56100-910^2=841500-828100=13400.$

Denominator: $\sqrt{13400\cdot 13400}=13400$_.

Hence r = 13400/13400 = 1.

r = 1 — perfect positive linear correlation (weighted).

Example 7: Given regression coefficients $b_{xy}=0.8$ (regression of Y on X) and $b_{yx}=0.45$ (regression of X on Y), find r.

Solution:

Property: $r^2=b_{xy}{b}_{yx}$ Sign of r equals sign of both coefficients (if both positive → r > 0).

Compute $r^2=0.8\times 0.45=\mathrm{0.36.}$

So $r=\sqrt{0.36}=\mathrm{0.6.}$

r = 0.6. 

Example 8: If r = -0.5, ${\sigma }_X=4,{\sigma }_Y=3,$ find Cov(X, Y).

Solution:

$Cov(X,Y)=-6.$

$\text{Cov}(X,Y)=r{\sigma }_X{\sigma }_Y=-0.5\times 4\times 3=-6.$$\text{Cov}(X,Y)=-6.$

Example 9: For a sample of n = 10 pairs you observe r = 0.8. Test significance at 5% level (two-tailed). Use $t=\frac{r\sqrt{n-2}}{\sqrt{1-r^2}}$_.

Solution:

Degrees of freedom = n – 2 = 8. Compute

$t=0.8\sqrt{\frac{8}{1-0.64}}=0.8\sqrt{\frac{8}{0.36}}=0.8\sqrt{22.22\overline{2}}\approx 0.8\times 4.714=\mathrm{3.771.}$

Critical two-tailed $t_{0.025,8}\approx \mathrm{2.306.}Since3.771>2.306$, the correlation is statistically significant at 5% level.

$\text{Reject }{H}_0:r=0;\text{ significant correlation.}$

5.0Practice Questions on Karl Pearson’s Coefficient of Correlation

1. Calculate Karl Pearson’s coefficient of correlation for the following:

2. If the correlation coefficient r = -0.75, ${\sigma }_X=5,{\sigma }_Y=4$_, find covariance.
3. For two variables, the regression coefficients are , .b_{xy} = 0.8, \quad b_{yx} = 0.5 Find the value of Karl Pearson’s coefficient of correlation.
4. Compute r for (X, Y) = (1, 1), (2, 2), (3, 2), (4, 3).

Answer hint: use raw-sum formula.

5. For a sample n = 6, you get r = 0.45. Is it significant at 5%? (df = 4, $t_{0.025,4}\approx 2.776$ ).
   Answer hint: compute “t” as in Example 6.
6. Given $\sum x=40,\sum y=50,\sum x^2=350,\sum y^2=520,\sum xy=430$ for n = 10. Find r.

Answer hint: apply the raw-sum formula.

7. If r=0.3, ${\sigma }_X=5,{\sigma }_Y=2$, compute covariance.
8. Paired grouped data: (x, f, y) = (2, 3, 10), (6, 2, 20), (10, 5, 30). Compute weighted r.