Maharashtra Board Class 11 Maths Part 2 Chapter 5 Correlation Miscellaneous Solutions

Question 1. Two series of x and y with 50 items each have standard deviations of 4.8 and 3.5 respectively. If the sum of products of deviations of x and y series from respective arithmetic means is 420, then find the correlation coefficient between x and y.
Answer: Solution:
Given, n = 50, \(\sigma_x\) = 4.8, \(\sigma_y\) = 3.5, \(\sum(x-\bar{x})(y-\bar{y})\) = 420
Cov (X, Y) = \(\frac{1}{n} \sum(x_i-\bar{x})(y_i-\bar{y})\)
\( = \frac{1}{50} \times 420 \)
\( = 8.4 \)
\(\therefore\) Cov (X, Y) = 8.4
\(r = \frac{\text{Cov (X, Y)}}{\sigma_x \sigma_y}\)
\( = \frac{8.4}{(4.8)(3.5)}\)
\( = \frac{84 \times 10}{48 \times 35}\)
\( = \frac{1}{2} \)
\( = 0.5 \)
In simple words: The correlation coefficient of 0.5 indicates a moderate positive relationship between the two series. This means as one series increases, the other tends to increase, but not perfectly.

🎯 Exam Tip: Remember the formula for the correlation coefficient (r) involving covariance and standard deviations. Accurately calculate covariance first, then substitute values to find r.

Question 2. Find the number of pairs of observations from the following data, r = 0.15, \(\sigma_y\) = 4, \(\sum (x_i – \bar{x}) (y_i – \bar{y})\) = 12, \(\sum(x_i – \bar{x})^2\) = 40.
Answer: Solution:
Given, r = 0.15, \(\sigma_y\) = 4, \(\sum (x_i – \bar{x}) (y_i – \bar{y})\) = 12, \(\sum(x_i – \bar{x})^2\) = 40
Since, \(\sigma_x = \sqrt{\frac{1}{n}\sum(x_i-\bar{x})^2} = \sqrt{\frac{40}{n}}\)
Cov (X, Y) = \(\frac{1}{n} \sum(x_i-\bar{x})(y_i-\bar{y})\)
\( = \frac{1}{n} \times 12 \)
\( = \frac{12}{n} \)
\(\therefore\) Cov (X, Y) = \(\frac{12}{n}\)
Since, \(r = \frac{\text{Cov (X, Y)}}{\sigma_x \sigma_y}\)
\( 0.15 = \frac{\frac{12}{n}}{\sqrt{\frac{40}{n}} \times 4} \)
\( 0.15 = \frac{\frac{12}{n}}{\frac{\sqrt{40}}{\sqrt{n}} \times 4} \)
\( 0.15 = \frac{12}{n} \times \frac{\sqrt{n}}{4\sqrt{40}} \)
\( 0.15 = \frac{3}{\sqrt{n}\sqrt{40}} \)
\( 0.05 = \frac{1}{\sqrt{n}\sqrt{40}} \)
Squaring on both the sides, we get
\( 0.0025 = \frac{1}{n \times 40} \)
\( n = \frac{1}{0.0025 \times 40} \)
\( n = \frac{10000}{25 \times 40} \)
\( n = \frac{10000}{1000} \)
\( n = 10 \)
In simple words: By using the formula for the correlation coefficient and substituting the given values, we can isolate and solve for 'n', which represents the number of pairs of observations. In this case, there are 10 pairs.

🎯 Exam Tip: When given `r`, `\(\sigma_y\)`, sum of product of deviations, and sum of squared deviations of x, remember to express `\(\sigma_x\)` and `Cov(X,Y)` in terms of 'n' before plugging into the correlation formula.

Question 3. Given that r = 0.4, \(\sigma_y\) = 3, \(\sum(x_i - \bar{x}) (y_i - \bar{y})\) = 108, \(\sum (x_i – \bar{x})^2\) = 900. Find the number of pairs of observations.
Answer: Solution:
Given, r = 0.4, \(\sigma_y\) = 3, \(\sum (x_i – \bar{x}) (y_i – \bar{y})\) = 108, \(\sum(x_i – \bar{x})^2\) = 900
Cov (X, Y) = \(\frac{1}{n} \sum(x_i-\bar{x})(y_i-\bar{y})\)
\( = \frac{1}{n} \times 108 \)
Cov (X, Y) = \(\frac{108}{n}\)
\(\sigma_x = \sqrt{\frac{1}{n}\sum(x_i-\bar{x})^2}\)
\( = \sqrt{\frac{1}{n} \times 900} \)
\( = \sqrt{\frac{900}{n}} \)
\( = \frac{30}{\sqrt{n}} \)
Since, \(r = \frac{\text{Cov (X, Y)}}{\sigma_x \sigma_y}\)
\( 0.4 = \frac{\frac{108}{n}}{\frac{30}{\sqrt{n}} \times 3} \)
\( 0.4 = \frac{\frac{108}{n}}{\frac{90}{\sqrt{n}}} \)
\( 0.4 = \frac{108}{n} \times \frac{\sqrt{n}}{90} \)
\( 0.4 = \frac{108}{90\sqrt{n}} \)
\( 0.4 = \frac{12}{10\sqrt{n}} \)
\( 4\sqrt{n} = 12 \)
\( \sqrt{n} = \frac{12}{4} \)
\( \sqrt{n} = 3 \)
\( n = 9 \)
In simple words: To find the number of observations 'n', we first calculate the covariance and standard deviation of X using 'n' in their formulas. Then, substitute these into the correlation coefficient formula and solve the resulting equation for 'n'. Here, 'n' is found to be 9.

🎯 Exam Tip: This problem is similar to Question 2. Ensure careful algebraic manipulation to solve for 'n', especially when dealing with square roots in the denominator. Double-check all calculations.

Question 4. Given the following information, \(\sum x_i^2\) = 90, \(\sum x_i y_i\) = 60, r = 0.8, \(\sigma_y\) = 2.5, where \(x_i\) and \(y_i\) are the deviations from their respective means, find the number of items.
Answer: Solution:
Here, \(\sum x_i^2\) = 90, \(\sum x_i y_i\) = 60, r = 0.8, \(\sigma_y\) = 2.5
Here, \(x_i\) and \(y_i\) are the deviations from their respective means.
\(\therefore\) If \(X_i, Y_i\) are elements of x and y series respectively, then \(X_i – \bar{x} = x_i\) and \(Y_i – \bar{y} = y_i\)
\(\therefore \sum x_i y_i = \sum (X_i – \bar{x}) (Y_i – \bar{y})\) = 60, \(\sum x_i^2 = \sum (X_i – \bar{x})^2\) = 90
Now, \(\sigma_x^2 = \frac{\sum(X_i-\bar{x})^2}{n}\)
\(\therefore \sigma_x^2 = \frac{90}{n}\)
\(\therefore \sigma_x = \sqrt{\frac{90}{n}}\)
Also, Cov (X, Y) = \(\frac{1}{n} \sum(X_i-\bar{x})(Y_i-\bar{y})\)
Cov (X, Y) = \(\frac{60}{n}\)
\(r = \frac{\text{Cov(X,Y)}}{\sigma_x \sigma_y}\)
\( 0.8 = \frac{\frac{60}{n}}{\sqrt{\frac{90}{n}} \times 2.5} \)
\( 0.8 \times 2.5 \times \sqrt{\frac{90}{n}} = \frac{60}{n} \)
\( 2 \times \frac{\sqrt{90}}{\sqrt{n}} = \frac{60}{n} \)
\( \frac{2\sqrt{90}}{\sqrt{n}} = \frac{60}{\sqrt{n}\sqrt{n}} \)
\( 2\sqrt{90} = \frac{60}{\sqrt{n}} \)
\( \sqrt{n} = \frac{60}{2\sqrt{90}} \)
\( \sqrt{n} = \frac{30}{\sqrt{90}} \)
\( \sqrt{n} = \frac{30}{\sqrt{90}} \times \frac{\sqrt{30}}{\sqrt{30}} \) (Note: The OCR shows \(\sqrt{30} \times \sqrt{30}\) but \(\sqrt{90}\) is not \(\sqrt{30} \times \sqrt{30}\). It should be rationalizing \(\frac{30}{\sqrt{90}}\)
Let's re-evaluate: \( \sqrt{n} = \frac{30}{\sqrt{90}} = \frac{30}{3\sqrt{10}} = \frac{10}{\sqrt{10}} = \sqrt{10} \)
So, \( \sqrt{n} = \sqrt{10} \)
\( n = 10 \)
In simple words: Since \(x_i\) and \(y_i\) are already deviations from the mean, the covariance and variance calculations are simplified. We then use the given correlation coefficient and standard deviation of Y to solve for 'n', the number of items, which is 10.

🎯 Exam Tip: Pay close attention when `x_i` and `y_i` are defined as deviations from the mean; this simplifies the calculation of covariance and standard deviations, as `\(\sum x_i = 0\)` and `\(\sum y_i = 0\)`. Ensure correct algebraic simplification, especially with square roots.

Question 5. A sample of 5 items is taken from the production of a firm. The length and weight of 5 items are given below. [Given: \(\sqrt{0.8823} = 0.9393\)] Calculate the correlation coefficient between length and weight and interpret the result.
Answer: Solution:
Let length = \(x_i\) (in cm), Weight = \(y_i\) (in gm)

From the table, we have
n = 5, \(\sum x_i\) = 30, \(\sum y_i\) = 65, \(\sum x_i^2\) = 210, \(\sum y_i^2\) = 879, \(\sum x_i y_i\) = 420
\(\bar{x} = \frac{\sum x_i}{n} = \frac{30}{5} = 6\), \(\bar{y} = \frac{\sum y_i}{n} = \frac{65}{5} = 13\)
Cov (X, Y) = \(\frac{1}{n} \sum x_i y_i - \bar{x} \bar{y}\)
\( = \frac{1}{5} \times 420 - 6 \times 13 \)
\( = 84 - 78 \)
\( = 6 \)
\(\sigma_x^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2 = \frac{210}{5} - (6)^2 = 42 - 36 \)
\(\sigma_x^2 = 6 \)
\(\sigma_x = \sqrt{6}\)
\(\sigma_y^2 = \frac{\sum y_i^2}{n} - (\bar{y})^2\)
\( = \frac{879}{5} - (13)^2 \)
\( = 175.8 - 169 \)
\(\sigma_y^2 = 6.8\)
\(\sigma_y = \sqrt{6.8}\)
Thus, the coefficient of correlation between X and Y is
\(r = \frac{\text{Cov(X,Y)}}{\sigma_x \sigma_y} = \frac{6}{\sqrt{6}\sqrt{6.8}} \)
\( = \frac{6}{\sqrt{6 \times 6.8}} \)
\( = \frac{6}{\sqrt{40.8}} \)
\( = \frac{6}{6.3875} \)
\( r = 0.9393 \approx 0.94 \)
\(\therefore\) the value of r indicates a high degree of positive correlation between length and weight of items.
In simple words: We calculated the correlation coefficient 'r' using the given length and weight data. The result, approximately 0.94, shows a very strong positive correlation, meaning that as the length of the items increases, their weight tends to increase significantly.

🎯 Exam Tip: For problems involving raw data, create a comprehensive calculation table for `\(\sum x_i, \sum y_i, \sum x_i^2, \sum y_i^2, \sum x_i y_i\)`. Ensure correct calculation of means, covariance, standard deviations, and finally, the correlation coefficient. Don't forget to interpret the `r` value.

Question 6. Calculate the correlation coefficient from the following data and interpret it.
Answer: Solution:

From the table, we have
n = 7, \(\sum x_i\) = 49, \(\sum y_i\) = 42, \(\sum x_i^2\) = 455, \(\sum y_i^2\) = 364, \(\sum x_i y_i\) = 182.
\(\bar{x} = \frac{\sum x_i}{n} = \frac{49}{7} = 7\)
\(\bar{y} = \frac{\sum y_i}{n} = \frac{42}{7} = 6\)
Cov (X, Y) = \(\frac{1}{n} \sum x_i y_i - \bar{x} \bar{y}\)
\( = \frac{1}{7} \times 182 - (7 \times 6) \)
\( = 26 - 42 \)
Cov (X, Y) = -16
\(\sigma_x^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2\)
\( = \frac{455}{7} - (7)^2 = 65 - 49 \)
\(\sigma_x^2 = 16 \)
\(\sigma_x = 4 \)
\(\sigma_y^2 = \frac{\sum y_i^2}{n} - (\bar{y})^2\)
\( = \frac{364}{7} - (6)^2 = 52 - 36 \)
\(\sigma_y^2 = 16 \)
\(\sigma_y = 4 \)
Thus, the coefficient of correlation between X and Y is
\(r = \frac{\text{Cov (X, Y)}}{\sigma_x \sigma_y} = \frac{-16}{4 \times 4} \)
\( = \frac{-16}{16} \)
\( = -1 \)
\(\therefore\) the value of r indicates a perfect negative correlation between x and y.
In simple words: The calculated correlation coefficient is -1, which signifies a perfect negative correlation. This means that as the value of x increases, the value of y decreases in a perfectly linear and consistent manner.

🎯 Exam Tip: A correlation coefficient of -1 implies a perfect inverse linear relationship. Ensure all summation and mean calculations are accurate, as a small error can significantly change the final 'r' value and its interpretation.

Question 7. Calculate the correlation coefficient from the following data and interpret it.
Answer: Solution:

From the table, we have
n = 7, \(\sum x_i\) = 52, \(\sum y_i\) = 122, \(\sum x_i^2\) = 396, \(\sum y_i^2\) = 2136, \(\sum x_i y_i\) = 916
\(\bar{x} = \frac{\sum x_i}{n} = \frac{52}{7}\)
\(\bar{y} = \frac{\sum y_i}{n} = \frac{122}{7}\)
\(\bar{x}\bar{y} = \frac{52 \times 122}{49} = \frac{6344}{49}\)
Cov (X, Y) = \(\frac{1}{n} \sum x_i y_i - \bar{x} \bar{y}\)
\( = \frac{916}{7} - \frac{6344}{49} \)
\( = \frac{916 \times 7 - 6344}{49} \)
\( = \frac{6412 - 6344}{49} = \frac{68}{49} \)
\(\sigma_x^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2\)
\( = \frac{396}{7} - \left(\frac{52}{7}\right)^2 \)
\( = \frac{396 \times 7 - 52^2}{49} \)
\( = \frac{2772 - 2704}{49} = \frac{68}{49} \)
\(\sigma_y^2 = \frac{\sum y_i^2}{n} - (\bar{y})^2\)
\( = \frac{2136}{7} - \left(\frac{122}{7}\right)^2 \)
\( = \frac{2136 \times 7 - 122^2}{49} \)
\( = \frac{14952 - 14884}{49} = \frac{68}{49} \)
\(\sigma_x \sigma_y = \sqrt{\sigma_x^2 \sigma_y^2} \)
\( = \sqrt{\frac{68}{49} \times \frac{68}{49}} \)
\( = \frac{68}{49} \)
\(r = \frac{\text{Cov (X, Y)}}{\sigma_x \sigma_y}\)
\( = \frac{\frac{68}{49}}{\frac{68}{49}} \)
\( = 1 \)
\(\therefore\) the value of r indicates a perfect positive correlation between x and y.
In simple words: The correlation coefficient for this dataset is 1, which means there is a perfect positive linear relationship between x and y. As x increases, y also increases in a perfectly consistent and proportional manner.

🎯 Exam Tip: When the correlation coefficient is 1, it indicates that all data points lie on a straight line with a positive slope. Accurate calculation of means, covariance, and standard deviations is crucial to arrive at this specific result.

Question 8. If the correlation coefficient between X and Y is 0.8, what is the correlation coefficient between
(i) 2X and Y
(ii) \(\frac{X}{2}\) and Y
(iii) X and 3Y
(iv) X – 5 and Y – 3
(v) X + 7 and Y + 9
(vi) \(\frac{X-5}{7}\) and \(\frac{Y-3}{8}\)
Answer: Solution:
The correlation coefficient remains unaffected by the change of origin and scale. i.e., if \(u_i = \frac{x_i-a}{h}\) and \(v_i = \frac{y_i-b}{k}\), then Corr(U, V) = \(\pm\)Corr(X, Y). according to the same or opposite signs of h and k.
(i) \(u_i = \frac{2(x_i-0)}{1}\), \(v_i = \frac{y_i-0}{1}\)
Here, h = 1 and k = 1 are of the same signs.
\(\therefore\) Corr (U, V) = Corr (X, Y) = 0.8
(ii) \(u_i = \frac{x_i-0}{2}\), \(v_i = \frac{y_i-0}{1}\)
Here, h = 2 and k = 1 are of the same signs.
\(\therefore\) Corr (U, V) = Corr (X, Y) = 0.8
(iii) Corr (X, 3Y) = Corr (X, Y) = 0.8
(iv) Corr (X – 5, Y – 3) = Corr(X, Y) = 0.8
(v) Corr (X + 7, Y + 9) = Corr(X, Y) = 0.8
(vi) Corr\(\left(\frac{X-5}{7}, \frac{Y-3}{8}\right)\) = Corr(X, Y) = 0.8
In simple words: The correlation coefficient is a pure number unaffected by changes in the origin (adding/subtracting constants) or scale (multiplying/dividing by positive constants) of the variables. Therefore, for all these linear transformations, the correlation coefficient remains 0.8.

🎯 Exam Tip: This is a key theoretical concept. Remember that `r` is invariant to linear transformations \(u = (x-a)/h\) and \(v = (y-b)/k\) as long as \(h\) and \(k\) have the same sign. If \(h\) and \(k\) have opposite signs, the sign of `r` flips.

Question 9. In the calculation of the correlation coefficient between the height and weight of a group of students of a college, one investigator took the measurements in inches and pounds while the other investigator took the measurements in cm. and kg. Will they get the same value of the correlation coefficient or different values? Justify your answer.
Answer: Solution:
The coefficient of correlation is a ratio of covariance and standard deviations.
Since covariance and standard deviations are independent of units of measurement.
\(\therefore\) coefficient of correlation is also independent of units of measurement.
\(\therefore\) values of coefficient of correlation obtained by first and second investigators are the same.
In simple words: Both investigators will get the same correlation coefficient. This is because the correlation coefficient measures the strength and direction of the linear relationship between two variables, regardless of the units used for measurement.

🎯 Exam Tip: Understand that the correlation coefficient is a unitless measure. Changing units (e.g., inches to cm, pounds to kg) is a change of scale, and as established in previous questions, correlation is invariant to changes in scale and origin when the scaling factors have the same sign. This is a common conceptual question.