Math Problem Statement

Consider the following data and find Co variance and Coefficient of Correlation. X 15 29 87 11 43 Y 34 10 30 22 50

Solution

To find the Covariance and Coefficient of Correlation between XX and YY, let's proceed step by step:


Given Data:

X=[15,29,87,11,43]X = [15, 29, 87, 11, 43] Y=[34,10,30,22,50]Y = [34, 10, 30, 22, 50]


Step 1: Compute Mean of X and Y

The formula for the mean: Xˉ=Xin,Yˉ=Yin\bar{X} = \frac{\sum X_i}{n}, \quad \bar{Y} = \frac{\sum Y_i}{n} where nn is the number of data points (5 in this case).

Xˉ=15+29+87+11+435=1855=37\bar{X} = \frac{15 + 29 + 87 + 11 + 43}{5} = \frac{185}{5} = 37 Yˉ=34+10+30+22+505=1465=29.2\bar{Y} = \frac{34 + 10 + 30 + 22 + 50}{5} = \frac{146}{5} = 29.2


Step 2: Compute Deviations

Calculate the deviations from the mean for both XX and YY: dXi=XiXˉ,dYi=YiYˉdX_i = X_i - \bar{X}, \quad dY_i = Y_i - \bar{Y}

XiX_iYiY_idXi=XiXˉdX_i = X_i - \bar{X}dYi=YiYˉdY_i = Y_i - \bar{Y}dXidYidX_i \cdot dY_idXi2dX_i^2dYi2dY_i^2
15341537=2215 - 37 = -223429.2=4.834 - 29.2 = 4.8224.8=105.6-22 \cdot 4.8 = -105.648423.04
29102937=829 - 37 = -81029.2=19.210 - 29.2 = -19.2819.2=153.6-8 \cdot -19.2 = 153.664368.64
87308737=5087 - 37 = 503029.2=0.830 - 29.2 = 0.8500.8=4050 \cdot 0.8 = 4025000.64
11221137=2611 - 37 = -262229.2=7.222 - 29.2 = -7.2267.2=187.2-26 \cdot -7.2 = 187.267651.84
43504337=643 - 37 = 65029.2=20.850 - 29.2 = 20.8620.8=124.86 \cdot 20.8 = 124.836432.64

Step 3: Compute Covariance

The formula for covariance is: Cov(X,Y)=(dXidYi)n\text{Cov}(X, Y) = \frac{\sum (dX_i \cdot dY_i)}{n}

(dXidYi)=105.6+153.6+40+187.2+124.8=400\sum (dX_i \cdot dY_i) = -105.6 + 153.6 + 40 + 187.2 + 124.8 = 400

Cov(X,Y)=4005=80\text{Cov}(X, Y) = \frac{400}{5} = 80


Step 4: Compute Coefficient of Correlation (r)

The formula for the correlation coefficient is: r=Cov(X,Y)Var(X)Var(Y)r = \frac{\text{Cov}(X, Y)}{\sqrt{\text{Var}(X) \cdot \text{Var}(Y)}}

Var(X)=dXi2n,Var(Y)=dYi2n\text{Var}(X) = \frac{\sum dX_i^2}{n}, \quad \text{Var}(Y) = \frac{\sum dY_i^2}{n}

dXi2=484+64+2500+676+36=3760,Var(X)=37605=752\sum dX_i^2 = 484 + 64 + 2500 + 676 + 36 = 3760, \quad \text{Var}(X) = \frac{3760}{5} = 752 dYi2=23.04+368.64+0.64+51.84+432.64=876.8,Var(Y)=876.85=175.36\sum dY_i^2 = 23.04 + 368.64 + 0.64 + 51.84 + 432.64 = 876.8, \quad \text{Var}(Y) = \frac{876.8}{5} = 175.36

r=80752175.36=80131936.32=80363.20.22r = \frac{80}{\sqrt{752 \cdot 175.36}} = \frac{80}{\sqrt{131936.32}} = \frac{80}{363.2} \approx 0.22


Final Results:

  1. Covariance: Cov(X,Y)=80\text{Cov}(X, Y) = 80
  2. Correlation Coefficient: r0.22r \approx 0.22

Do you want me to explain any specific part or further verify these calculations?

Related Questions:

  1. How is the correlation coefficient affected by outliers in data?
  2. Can covariance be negative, and what does it imply in practical terms?
  3. What is the difference between covariance and variance?
  4. How would adding a constant to all XX values affect covariance and correlation?
  5. What are the limitations of using the correlation coefficient?

Tip: The correlation coefficient rr ranges between 1-1 and +1+1. An rr value close to 0 indicates little to no linear relationship between the variables.

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Math Problem Analysis

Mathematical Concepts

Statistics
Covariance
Correlation Coefficient

Formulas

Mean: X̄ = ΣXi / n, Ȳ = ΣYi / n
Covariance: Cov(X, Y) = Σ((Xi - X̄) * (Yi - Ȳ)) / n
Correlation Coefficient: r = Cov(X, Y) / √(Var(X) * Var(Y))
Variance: Var(X) = Σ(Xi - X̄)^2 / n, Var(Y) = Σ(Yi - Ȳ)^2 / n

Theorems

Properties of Covariance
Properties of Pearson's Correlation Coefficient

Suitable Grade Level

Grades 10-12