Introduction to Correlation Coefficient and Previous Year JEE Main Problems on Statistics

Correlation Coefficient

A statistical tool that is used to measure the strength of the relationship between any two given variables is a correlation coefficient. The range of a correlation coefficient is from – 1 to +1. In this article, the correlation coefficient and previous year JEE Main problems on statistics are discussed.

• – 1 indicates a negative correlation. The variables will be moving in opposite directions: if one variable shows a positive increase, the second variable decreases.
• + 1 indicates a positive correlation. A positive increase in one variable results in a positive increase in the second variable.
• 0 indicates no relationship between the variables

The most commonly used correlation coefficient is the Pearson correlation coefficient.

Steps to determine the Pearson correlation coefficient are as follows:

• Determine the covariance of the two variables.
• Find the standard deviation of each variable. The measure of the spread of the numbers from the mean is called the standard deviation.
• The correlation coefficient is the ratio of the covariance and the product of the two variables’ standard deviations.

⍴_xy = cov (x, y) / _{x y}

where: ⍴_xy =Pearson product-moment correlation coefficient

Cov (x, y) = covariance of variables x and y

σ_x = standard deviation of x

σ_y = standard deviation of y

Inverse correlation is the relationship between the two variables such that one of the variables is high and the other is low.

Previous Year JEE Main Problems on Statistics

Below are the previous year JEE Main problems on statistics. It will help the students to ace the JEE examination.

Example 1: The sum of squares of deviations for 10 observations taken from mean 50 is 250. The coefficient of variation is

Solution:

S.D. (σ) = √250 / 10 = √25 = 5

Hence, coefficient of variation = [σ / mean] × 100                                                                                       = [5 / 50] × 100

= 10%

Example 2: The means of five observations is 4 and their variance is 5.2. If three of these observations are 1, 2 and 6, then the other two are

Solution:

Let the two unknown items be x and y, then

Mean = 4 ⇒ [1 + 2 + 6 + x + y] / 5 = 4

x + y = 11   ….(i)  and variance = 5. 2

41 + x² + y² = 5 [5.2 + (4)²]

41 + x² + y² = 106

x² + y² = 65  …..(ii)

Solving (i) and (ii) for x and y, we get

x = 4, y = 7 or x = 7, y = 4.

Example 3: The quartile deviation for the following data is

Solution:

N = (Σf) = 20

Q₁ = [(N + 1)/ 4]^th observation = (21 / 4)^th observation = 3

Similarly, Q₃ = [3 (N + 1) / 4)]^th observation

= (63 / 4)^th observation = 5

Quartile Deviation = (1 / 2) (Q₃ − Q₁)

= (1 / 2) (5 − 3)

= 1

Example 4: The variance of the data 2, 4, 6, 8, 10 is

Solution:

Mean = [2 + 4 + 6 + 8 + 10] / 5 = 6

Variance = (1 / n) Σ (x_i − x {bar})²

= (1 / 5) {(2 − 6)² + (4 − 6)² + (6 − 6)² + (8 − 6)² + (10 − 6)²}

= (1 / 5) {(16 + 4 + 0 + 4 + 16}

= (1 / 5) {40}

= 8

Example 5: The variance of the first n natural numbers is

Solution:

Variance = (S.D.)²  = (1 / n) Σx² − (Σx / n)²    {mean = Σx / n}

= [n (n + 1) (2n + 1)] / [6n] − (n (n + 1) / 2n)²

= [n² − 1] / 12

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