Question

X-50,54,56,58,59,60,61,62,65,75
Y-20,22,24,30,32,36,38,40,44,54
Calculate Karl Pearson's coefficient of correlation from the following data

Answer 1

Explanation:

50,54,56,58,59,60,61,62,65,75

Y-20,22,24,30,32,36,38,40,44,54

Answer 2

Answer :

Karl Pearson's coefficient of correlation between X and Y is 0.207.

Explanation :

Karl Pearson's coefficient of correlation (also known as Pearson's r) is a measure of the linear relationship between two variables, X and Y. Given the data for X and Y, we can calculate Pearson's coefficient of correlation as follows:

Step 1: Calculate the mean of the two variables X and Y.

The mean of X is (50 + 54 + 56 + 58 + 59 + 60 + 61 + 62 + 65 + 75) / 10 = 59

The mean of Y is (20 + 22 + 24 + 30 + 32 + 36 + 38 + 40 + 44 + 54) / 10 = 36

Step 2: Subtract the mean from each data point and calculate the deviations.

For X, the deviations would be (50-59), (54-59), (56-59), (58-59), (59-59), (60-59), (61-59), (62-59), (65-59), (75-59) = -9, -5, -3, -1, 0, 1, 2, 3, 6, 16

For Y, the deviations would be (20-36), (22-36), (24-36), (30-36), (32-36), (36-36), (38-36), (40-36), (44-36), (54-36) = -16, -14, -12, -6, -4, 0, 2, 4, 8, 18

Step 3: Multiply the deviations of X and Y together.

The product of deviations would be (-9) × (-16) = 144, (-5) × (-14) = 70, (-3) × (-12) = 36, (-1) × (-6) = 6, 0 × 0 = 0, 1 × 2 = 2, 2 × 4 = 8, 3 × 8 = 24, 6 × 18 = 108, 16 × 18 = 288

Step 4: Sum all the products from step 3.

The sum of all products would be 144 + 70 + 36 + 6 + 0 + 2 + 8 + 24 + 108 + 288 = 828

Step 5: Divide the sum from step 4 by the number of data points (n) and the product of the standard deviations of X and Y.

The standard deviation of X is sqrt( ( (50-59)^2 + (54-59)^2 + (56-59)^2 + (58-59)^2 + (59-59)^2 + (60-59)^2 + (61-59)^2 + (62-59)^2 + (65-59)^2 + (75-59)^2 ) / 10 ) = sqrt(368) = 19

The standard deviation of Y is sqrt( ( (20-36)^2 + (22-36)^2 + (24-36)^2 + (30-36)^2 + (32-36)^2 + (36-36)^2 + (38-36)^2 + (40-36)^2 + (44-36)^2 + (54-36)^2 ) / 10 ) = sqrt(420) = 21

The product of standard deviations would be 19 × 21 = 399

The Pearson's coefficient of correlation (r) would be 828 / (10 × 399) = 828 / 3990 = 0.207

So, the Karl Pearson's coefficient of correlation between X and Y is 0.936, which indicates a strong positive linear relationship between the two variables.

To know more about the concept please go through the links :

https://brainly.in/question/51415261

https://brainly.in/question/2113260

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