Overview
Basically correlation is the measurement of the strength of a linear relationship between two variables.
In other words, we define it as- if the change in one variable affects a change in other variable, then there will be a corr. between two variables
Example
1. A person’s income and expenditures.
2. As the temperature goes up, the demand of ice cream also go up.
Classification–
Positive correlation- When both variables move in the same direction, or if the increase in one variable results in a corresponding increase in the other. this is the condition of positive corr.
Negative correlation:
When one variable increases and other decreases or vise-versa.
No correlation:
When two variables are independent and do not affect each other then there will be no corr. between the two and this ia the condition of no-correlation.
Note- (Perfect correlation)– When a variable changes constantly with the other variable, then there will be perfect corr.
Scatter plots or dot diagrams
Scatter or dot diagram is used to check the corr. between the two variables.
It is the simplest method to represent a bivariate data.
When the dots in diagram are very close to each other, then we can say that there is a fairly good corr.
If the dots are scattered then we get a poor corr.
Karl Pearson’s method for correlation
We also call Karl Person’s coefficient of corr. as product moment correlation coefficient.
It is denoted by ‘r’, and defined as-
Here 
Alternate formula-
Note-
1. The value of ‘r’ lies between -1 and +1.
2. ‘r’ is independent of change of origin and scale.
3. If the two variables are independent then the value of r will be zero.
| Value of correlation coefficient (r) | Type of correlation |
| +1 | Perfect positive corr. |
| -1 | Perfect negative corr. |
| 0.25 | Weak positive corr. |
| 0.75 | Strong positive corr. |
| -0.25 | Weak negative corr. |
| -0.75 | Strong negative corr. |
| 0 | No corr. |
Solved exampled of correlation coefficient
Example: The data given below is about the marks obtained by a student and hours she studied.
Find the corr. coefficient between hours and marks obtained.
| Hours | 1 | 3 | 5 | 7 | 8 | 10 |
| marks | 8 | 12 | 15 | 17 | 18 | 20 |
Solution:
Let hours = x and marks = y
Karl Person’s formula is given by-
The correlation coefficient between hours and marks obtained is- 0.98
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