Find the coefficient of correlation from the following data: x 2 3 5 7 3 y 15 17 4 5 4

Given that, the data is x 2 3 5 7 3 y 15 17 4 5 4 We know that, correlation coefficient r = ∑ x i - x ¯ y i - y ¯ ∑ x i - x ¯ 2 ∑ y i - y ¯ 2 ⇒ Mean of x values is x ¯ = 2 + 3 + 5 + 7 + 3 5 = 4 ⇒ Mean of y values is y ¯ = 15 + 17 + 4 + 5 + 4 5 = 9 x y x i - x ¯ y i - y ¯ x i - x ¯ 2 y i - y ¯ 2 x i - x ¯ y i - y ¯ 2 15 - 2 6 4 36 - 12 3 17 - 1 8 1 64 - 8 5 4 1 - 5 1 25 - 5 7 5 3 - 4 9 16 - 12 3 4 0 - 5 0 25 0 Σ x = 20 Σ y = 45 Σ x i - x ¯ = 1 Σ y i - y ¯ = 0 Σ x i - x ¯ 2 = 15 Σ y i - y ¯ 2 = 166 Σ x i - x ¯ y i - y ¯ = - 37 ⇒ r = - 37 15 166 = - 0 . 74

An article costs Rs. 70 at Gwalior, find the corresponding most appropriate vale at Bhopal using the following data: Gwalior Bhopal Mean Value 65 67 Standard Deviation 2 . 5 3 . 5 The correlation coefficient between the values of the two cities is 0 . 8 .

The line of regression of y on x is y = m x + c , where m is slope and c is y -intercept. Gwalior x Bhopal y Mean Value x ¯ = 65 y ¯ = 67 Standard Deviation s x = 2 . 5 s y = 3 . 5 Here, slope m = r s y s x = 0 . 8 × 3 . 5 2 . 5 = 1 . 12 and y -intercept is y ¯ - m x ¯ = 67 - 1 . 12 × 65 = - 5 . 8 The line of regression of y on x is y = 1 . 12 x - 5 . 8 When x = 70 , then y = 1 . 12 × 70 - 5 . 8 = 72 . 6 Thus, the approximate value at Bhopal is 72 . 6 .

Two lines of regression are x + 2 y = 5 and 2 x + 3 y = 8 , find: (I) x ¯ and y ¯ , (II) b y x and b x y and (III) ρ x y

Two lines of regression intersect at the point (mean of ' x ', mean of ' y ') where x and y are the variables under consideration. Given two lines of regression : x + 2 y - 5 = 0 ........(i) 2 x + 3 y - 8 = 0 ......(ii) Solving the given two regression equations, we get the point of intersection x ¯ and y ¯ . Multiplying equation (i) by 2 and subtracting equation(i) and (ii), ( 2 x + 4 y - 10 ) - ( 2 x + 3 y - 8 ) = 0 ⇒ y - 2 = 0 ⇒ y = 2 Substituting in equation (i), we get, ⇒ x = 1 Hence, x ¯ = 1 , y ¯ = 2 Now, from equation (i), x + 2 y = 5 ⇒ 2 y = - x - 5 ⇒ y = - 1 2 x - 5 2 On comparing with the regression equation of y on x i.e., y = m x + c where m = b y x Thus, b y x = - 1 2 Similarly from equation (i), x + 2 y = 5 ⇒ x = - 2 y + 5 On comparing with regression equation of x on y i.e., x = m y + c , where m = b x y Thus, b x y = - 2 Now, correlation coefficient ρ x y = b x y × b y x = - 1 2 × - 2 = 1

Two lines of regression are x + 2 y = 5 and 2 x + 3 y = 8 , find: (I) x ¯ and y ¯ , (II) b y x and b x y and (III) ρ x y

Two lines of regression intersect at the point (mean of ' x ', mean of ' y ') where x and y are the variables under consideration. Given two lines of regression : x + 2 y - 5 = 0 ........(i) 2 x + 3 y - 8 = 0 ......(ii) Solving the given two regression equations, we get the point of intersection x ¯ and y ¯ . Multiplying equation (i) by 2 and subtracting equation(i) and (ii), ( 2 x + 4 y - 10 ) - ( 2 x + 3 y - 8 ) = 0 ⇒ y - 2 = 0 ⇒ y = 2 Substituting in equation (i), we get, ⇒ x = 1 Hence, x ¯ = 1 , y ¯ = 2 Now, from equation (i), x + 2 y = 5 ⇒ 2 y = - x - 5 ⇒ y = - 1 2 x - 5 2 On comparing with the regression equation of y on x i.e., y = m x + c where m = b y x Thus, b y x = - 1 2 Similarly from equation (i), x + 2 y = 5 ⇒ x = - 2 y + 5 On comparing with regression equation of x on y i.e., x = m y + c , where m = b x y Thus, b x y = - 2 Now, correlation coefficient ρ x y = b x y × b y x = - 1 2 × - 2 = 1

An article costs Rs. 70 at Gwalior, find the corresponding most appropriate vale at Bhopal using the following data: Gwalior Bhopal Mean Value 65 67 Standard Deviation 2 . 5 3 . 5 The correlation coefficient between the values of the two cities is 0 . 8 .

The line of regression of y on x is y = m x + c , where m is slope and c is y -intercept. Gwalior x Bhopal y Mean Value x ¯ = 65 y ¯ = 67 Standard Deviation s x = 2 . 5 s y = 3 . 5 Here, slope m = r s y s x = 0 . 8 × 3 . 5 2 . 5 = 1 . 12 and y -intercept is y ¯ - m x ¯ = 67 - 1 . 12 × 65 = - 5 . 8 The line of regression of y on x is y = 1 . 12 x - 5 . 8 When x = 70 , then y = 1 . 12 × 70 - 5 . 8 = 72 . 6 Thus, the approximate value at Bhopal is 72 . 6 .

Find the coefficient of correlation from the following data: x 2 3 5 7 3 y 15 17 4 5 4

Given that, the data is x 2 3 5 7 3 y 15 17 4 5 4 We know that, correlation coefficient r = ∑ x i - x ¯ y i - y ¯ ∑ x i - x ¯ 2 ∑ y i - y ¯ 2 ⇒ Mean of x values is x ¯ = 2 + 3 + 5 + 7 + 3 5 = 4 ⇒ Mean of y values is y ¯ = 15 + 17 + 4 + 5 + 4 5 = 9 x y x i - x ¯ y i - y ¯ x i - x ¯ 2 y i - y ¯ 2 x i - x ¯ y i - y ¯ 2 15 - 2 6 4 36 - 12 3 17 - 1 8 1 64 - 8 5 4 1 - 5 1 25 - 5 7 5 3 - 4 9 16 - 12 3 4 0 - 5 0 25 0 Σ x = 20 Σ y = 45 Σ x i - x ¯ = 1 Σ y i - y ¯ = 0 Σ x i - x ¯ 2 = 15 Σ y i - y ¯ 2 = 166 Σ x i - x ¯ y i - y ¯ = - 37 ⇒ r = - 37 15 166 = - 0 . 74