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 calculate Covariance and the Coefficient of Correlation, let's proceed step by step:

Given data:

X1529871143
Y3410302250

1. Formulas:

Covariance:

Cov(X,Y)=(XiXˉ)(YiYˉ)n\text{Cov}(X, Y) = \frac{\sum{(X_i - \bar{X})(Y_i - \bar{Y})}}{n} where:

  • Xi,YiX_i, Y_i are individual data points.
  • Xˉ,Yˉ\bar{X}, \bar{Y} are the means of XX and YY, respectively.
  • nn is the number of data points.

Coefficient of Correlation (rr):

r=Cov(X,Y)σXσYr = \frac{\text{Cov}(X, Y)}{\sigma_X \cdot \sigma_Y} where:

  • σX,σY\sigma_X, \sigma_Y are the standard deviations of XX and YY.

2. Calculate Means (Xˉ\bar{X} and Yˉ\bar{Y}):

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


3. Calculate Deviations and Cross Products:

XiX_iYiY_iXiXˉX_i - \bar{X}YiYˉY_i - \bar{Y}(XiXˉ)(YiYˉ)(X_i - \bar{X})(Y_i - \bar{Y})
1534-224.8-105.6
2910-8-19.2153.6
8730500.840
1122-26-7.2187.2
4350620.8124.8

(XiXˉ)(YiYˉ)=105.6+153.6+40+187.2+124.8=400\sum (X_i - \bar{X})(Y_i - \bar{Y}) = -105.6 + 153.6 + 40 + 187.2 + 124.8 = 400


4. Covariance:

Cov(X,Y)=(XiXˉ)(YiYˉ)n=4005=80\text{Cov}(X, Y) = \frac{\sum (X_i - \bar{X})(Y_i - \bar{Y})}{n} = \frac{400}{5} = 80


5. Standard Deviations (σX\sigma_X and σY\sigma_Y):

σX=(XiXˉ)2nandσY=(YiYˉ)2n\sigma_X = \sqrt{\frac{\sum{(X_i - \bar{X})^2}}{n}} \quad \text{and} \quad \sigma_Y = \sqrt{\frac{\sum{(Y_i - \bar{Y})^2}}{n}}

For XX: (XiXˉ)2=(22)2+(8)2+502+(26)2+62=484+64+2500+676+36=3760\sum{(X_i - \bar{X})^2} = (-22)^2 + (-8)^2 + 50^2 + (-26)^2 + 6^2 = 484 + 64 + 2500 + 676 + 36 = 3760 σX=37605=75227.42\sigma_X = \sqrt{\frac{3760}{5}} = \sqrt{752} \approx 27.42

For YY: (YiYˉ)2=(4.8)2+(19.2)2+(0.8)2+(7.2)2+(20.8)2\sum{(Y_i - \bar{Y})^2} = (4.8)^2 + (-19.2)^2 + (0.8)^2 + (-7.2)^2 + (20.8)^2 =23.04+368.64+0.64+51.84+432.64=876.8= 23.04 + 368.64 + 0.64 + 51.84 + 432.64 = 876.8 σY=876.85=175.3613.24\sigma_Y = \sqrt{\frac{876.8}{5}} = \sqrt{175.36} \approx 13.24


6. Coefficient of Correlation:

r=Cov(X,Y)σXσY=8027.4213.2480363.120.22r = \frac{\text{Cov}(X, Y)}{\sigma_X \cdot \sigma_Y} = \frac{80}{27.42 \cdot 13.24} \approx \frac{80}{363.12} \approx 0.22


Final Results:

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

Do you have any questions or want a deeper explanation?
Here are 5 follow-up questions:

  1. How do you interpret the positive covariance value?
  2. Why is the correlation coefficient only 0.22 despite the positive covariance?
  3. Can covariance be compared across different datasets?
  4. How would the values change if all XX and YY values were doubled?
  5. What does the sign of rr indicate about the relationship between XX and YY?

Tip: Covariance measures the directional relationship, while the correlation coefficient standardizes it for comparison.

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Math Problem Analysis

Mathematical Concepts

Statistics
Covariance
Correlation
Mean
Standard Deviation

Formulas

Cov(X, Y) = (Σ(X_i - X̄)(Y_i - Ȳ)) / n
r = Cov(X, Y) / (σ_X * σ_Y)

Theorems

Covariance theorem
Coefficient of correlation theorem

Suitable Grade Level

Grades 11-12