Maharashtra Board Class 11 Maths Part 2 Chapter 5 Correlation 5.1 Solutions

Question 1. Draw a scatter diagram for the data given below and interpret it.

Solution:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक स्कैटर डायग्राम दिखाता है जहाँ बिंदु बाएं से दाएं ऊपर की ओर बढ़ते हुए एक बैंड में फैले हुए हैं। यह दर्शाता है कि जैसे-जैसे X का मान बढ़ता है, वैसे-वैसे Y का मान भी बढ़ता है, जो एक सकारात्मक संबंध को इंगित करता है। Since all the points lie in a band rising from left to right. Therefore, there is a positive correlation between the values of X and Y respectively.
Answer: The scatter diagram shows a positive correlation as the points form an upward-sloping band from left to right, indicating that as X values increase, Y values also tend to increase.
In simple words: The diagram displays scattered points forming an upward trend, which means as one variable goes up, the other tends to go up too, showing a positive connection.

🎯 Exam Tip: When interpreting a scatter diagram, observe the direction (upward/downward) and spread (tight/scattered) of the points to determine the type and strength of correlation. Clearly label axes and scales.

Question 2. For the following data of marks of 7 students in Physics (x) and Mathematics (y), draw scatter diagram and state the type of correlation.

Solution: We take marks in Physics on X-axis and marks in Mathematics on Y-axis and plot the points as below.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह स्कैटर डायग्राम भौतिकी और गणित में छात्रों के अंकों के बीच संबंध दर्शाता है। बिंदु बाएं से दाएं ऊपर की ओर एक सीधी रेखा के आसपास क्लस्टर किए गए हैं, जो दोनों विषयों के अंकों के बीच एक मजबूत सकारात्मक सहसंबंध का सुझाव देता है। We get a band of points rising from left to right. This indicates the positive correlation between marks in Physics and marks in Mathematics.
Answer: The scatter diagram shows a positive correlation because the points generally move upwards from left to right, suggesting that students who score well in Physics also tend to score well in Mathematics.
In simple words: The graph shows that as Physics scores increase, Math scores also tend to increase, indicating a positive relationship between the two subjects.

🎯 Exam Tip: When drawing scatter diagrams, ensure accurate plotting of points. A clear upward trend from left to right signifies positive correlation, while a downward trend indicates negative correlation. No clear trend means no correlation.

Question 3. Draw a scatter diagram for the data given below. Is there any correlation between Aptitude score and Grade points?

Solution:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह स्कैटर डायग्राम एप्टीट्यूड स्कोर और ग्रेड पॉइंट्स के बीच संबंध को दर्शाता है। बिंदु पूरे ग्राफ पर बेतरतीब ढंग से बिखरे हुए हैं, जिससे कोई स्पष्ट पैटर्न या प्रवृत्ति नहीं दिख रही है। यह एप्टीट्यूड स्कोर और ग्रेड पॉइंट्स के बीच सहसंबंध की अनुपस्थिति को इंगित करता है। The points are completely scattered i.e., no trend is observed.
\( \therefore \) there is no correlation between Aptitude score (X) and Grade point (Y).
Answer: The scatter diagram shows that the points are randomly scattered without any discernible pattern or trend. Therefore, there is no correlation between Aptitude score and Grade points.
In simple words: The graph shows no clear pattern between aptitude scores and grade points, meaning there's no relationship or correlation linking them.

🎯 Exam Tip: A scattered distribution of points on a scatter diagram indicates no correlation. Clearly state this conclusion and justify it by mentioning the absence of any trend (upward or downward).

Question 4. Find correlation coefficient between x and y series for the following data: n = 15, \( \bar{x} \) = 25, \( \bar{y} \) = 18, \( \sigma_x \) = 3.01, \( \sigma_y \) = 3.03, \( \sum (x_i - \bar{x}) (y_i - \bar{y}) \) = 122 Solution: Here, n = 15, \( \bar{x} \) = 25, \( \bar{y} \) = 18, \( \sigma_x \) = 3.01, \( \sigma_y \) = 3.03, \( \sum (x_i - \bar{x}) (y_i - \bar{y}) \) = 122 Since, Cov (X, Y) = \( \frac{1}{n} \sum (x_i - \bar{x}) (y_i - \bar{y}) \)
\( \implies \) Cov (X, Y) = \( \frac{1}{15} \times 122 \)
\( \implies \) = 8.13 Since, r = \( \frac{\text{Cov(X,Y)}}{\sigma_x \sigma_y} \)
\( \therefore \) r = \( \frac{8.13}{3.01 \times 3.03} \)
\( \implies \) = \( \frac{8.13}{9.1203} \)
\( \therefore \) r = 0.89
Answer: The correlation coefficient, r, is calculated as follows: First, calculate Covariance (X, Y): Cov (X, Y) = \( \frac{1}{n} \sum (x_i - \bar{x}) (y_i - \bar{y}) = \frac{1}{15} \times 122 = 8.13 \) Then, calculate the correlation coefficient r: r = \( \frac{\text{Cov(X,Y)}}{\sigma_x \sigma_y} = \frac{8.13}{3.01 \times 3.03} = \frac{8.13}{9.1203} \approx 0.89 \) The correlation coefficient between x and y is 0.89.
In simple words: We calculated the covariance and then used it along with the given standard deviations to find the correlation coefficient, which turned out to be 0.89, indicating a strong positive relationship.

🎯 Exam Tip: Remember the formula for correlation coefficient \( r = \frac{\text{Cov(X,Y)}}{\sigma_x \sigma_y} \). Ensure accurate substitution of values and careful calculation, especially when dealing with decimals.

Question 5. The correlation coefficient between two variables x and y are 0.48. The covariance is 36 and the variance of x is 16. Find the standard deviation of y. Solution: Given, r = 0.48, Cov(X, Y) = 36 Since \( \sigma_x^2 \) = 16
\( \therefore \sigma_x \) = 4 Since, r = \( \frac{\text{Cov(X,Y)}}{\sigma_x \sigma_y} \)
\( \implies \) 0.48 = \( \frac{36}{4 \times \sigma_y} \)
\( \implies \) 0.48 = \( \frac{9}{\sigma_y} \)
\( \implies \sigma_y \) = \( \frac{9}{0.48} \)
\( \implies \sigma_y \) = \( \frac{900}{48} \)
\( \implies \sigma_y \) = 18.75
\( \therefore \) the standard deviation of y is 18.75.
Answer: Given r = 0.48, Cov(X, Y) = 36, and variance of x (\( \sigma_x^2 \)) = 16. First, find \( \sigma_x \): \( \sigma_x = \sqrt{16} = 4 \). Using the formula for correlation coefficient: \( r = \frac{\text{Cov(X,Y)}}{\sigma_x \sigma_y} \) \( 0.48 = \frac{36}{4 \times \sigma_y} \) \( 0.48 = \frac{9}{\sigma_y} \) \( \sigma_y = \frac{9}{0.48} = 18.75 \) The standard deviation of y is 18.75.
In simple words: Given the correlation coefficient, covariance, and standard deviation of x, we rearranged the correlation formula to solve for the standard deviation of y, which we found to be 18.75.

🎯 Exam Tip: Make sure to correctly extract the standard deviation from the variance (square root). Rearranging the correlation coefficient formula to solve for an unknown variable is a common application; practice these algebraic manipulations.

Question 6. In the following data, one of the values of y is missing. Arithmetic means of x and y series are 6 and 8 respectively. (√2 = 1.4142)

Solution: (i) Estimate missing observation. (ii) Calculate correlation coefficient. Solution: (i) Let X = \( x_i \), Y = \( y_i \) and missing observation be 'a'. Given, \( \bar{x} \) = 6, \( \bar{y} \) = 8, n = 5
\( \therefore \) 8 = \( \frac{35+a}{5} \)
\( \therefore \) 40 = 35 + a
\( \therefore \) a = 5 (ii) We construct the following table:

From the table, we have \( \sum x_i = 30, \sum y_i = 40, \sum x_i^2 = 220, \sum y_i^2 = 340, \sum x_i y_i = 214 \) Since, Cov (X, Y) = \( \frac{1}{n} \sum x_i y_i - \bar{x} \bar{y} \)
\( \therefore \) Cov (X, Y) = \( \frac{1}{5} \times 214 - 6 \times 8 \)
\( \implies \) = 42.8 - 48
\( \implies \) = -5.2 \( \sigma_x^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2 \)
\( \implies \) = \( \frac{220}{5} - (6)^2 = 44 - 36 \)
\( \therefore \sigma_x^2 \) = 8
\( \therefore \sigma_x \) = \( \sqrt{8} = 2\sqrt{2} = 2(1.4142) = 2.83 \) \( \sigma_y^2 = \frac{\sum y_i^2}{n} - (\bar{y})^2 \)
\( \implies \) = \( \frac{340}{5} - (8)^2 = 68 - 64 \)
\( \therefore \sigma_y^2 \) = 4
\( \therefore \sigma_y \) = \( \sqrt{4} = 2 \) Thus, the correlation coefficient between X and Y is r = \( \frac{\text{Cov(X,Y)}}{\sigma_x \sigma_y} \)
\( \implies \) = \( \frac{-5.2}{2.83 \times 2} \)
\( \implies \) = \( \frac{-5.2}{5.66} \)
\( \implies \) = -0.9187...
\( \implies \) = -0.92 (approx)
Answer:
(i) Estimate missing observation: Given \( \bar{y} \) = 8 and n = 5. \( \bar{y} = \frac{\sum y_i}{n} \implies 8 = \frac{9+11+a+8+7}{5} \) \( 8 = \frac{35+a}{5} \implies 40 = 35+a \implies a = 5 \). The missing observation is 5.
(ii) Calculate correlation coefficient: Using the completed data:

\( \bar{x} = \frac{30}{5} = 6 \), \( \bar{y} = \frac{40}{5} = 8 \) Cov (X, Y) = \( \frac{\sum x_i y_i}{n} - \bar{x} \bar{y} = \frac{214}{5} - (6 \times 8) = 42.8 - 48 = -5.2 \) \( \sigma_x^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2 = \frac{220}{5} - (6)^2 = 44 - 36 = 8 \) \( \sigma_x = \sqrt{8} = 2\sqrt{2} = 2 \times 1.4142 = 2.8284 \approx 2.83 \) \( \sigma_y^2 = \frac{\sum y_i^2}{n} - (\bar{y})^2 = \frac{340}{5} - (8)^2 = 68 - 64 = 4 \) \( \sigma_y = \sqrt{4} = 2 \) Correlation coefficient, r = \( \frac{\text{Cov(X,Y)}}{\sigma_x \sigma_y} = \frac{-5.2}{2.83 \times 2} = \frac{-5.2}{5.66} \approx -0.92 \)
In simple words: First, we used the given mean of y to find the missing value. Then, we calculated the covariance and the standard deviations of both x and y. Finally, we used these values to compute the correlation coefficient, which indicates a strong negative relationship.

🎯 Exam Tip: When a value is missing, always use the given mean to find it first. Organize your calculations in a table to avoid errors, and double-check standard deviation calculations. Remember that a negative correlation coefficient means an inverse relationship.

Question 7. Find correlation coefficient from the following data. [Given: √3 = 1.732]

Solution:

From the table, we have n = 5, \( \sum x_i = 25, \sum y_i = 30, \sum x_i^2 = 155, \sum y_i^2 = 190, \sum x_i y_i = 147 \) \( \bar{x} = \frac{\sum x_i}{n} \)
\( \implies \) = \( \frac{25}{5} \)
\( \implies \) = 5 \( \bar{y} = \frac{\sum y_i}{n} \)
\( \implies \) = \( \frac{30}{5} \)
\( \implies \) = 6 Since, Cov (X, Y) = \( \frac{1}{n} \sum x_i y_i - \bar{x} \bar{y} \) Cov (X, Y) = \( \frac{1}{5} \times 147 - (5 \times 6) \)
\( \implies \) = 29.4 - 30
\( \implies \) = -0.6 \( \sigma_x^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2 \)
\( \implies \) = \( \frac{155}{5} - (5)^2 \)
\( \implies \) = 31 - 25
\( \therefore \sigma_x^2 \) = 6
\( \therefore \sigma_x \) = \( \sqrt{6} \) \( \sigma_y^2 = \frac{\sum y_i^2}{n} - (\bar{y})^2 \)
\( \implies \) = \( \frac{190}{5} - (6)^2 \)
\( \implies \) = 38 - 36
\( \therefore \sigma_y^2 \) = 2
\( \therefore \sigma_y \) = \( \sqrt{2} \)
\( \therefore \sigma_x \sigma_y \) = \( \sqrt{6} \sqrt{2} = \sqrt{12} \)
\( \implies \) = \( 2\sqrt{3} \)
\( \implies \) = \( 2(1.732) = 3.464 \) Thus, the correlation coefficient between X and Y is r = \( \frac{\text{Cov(X,Y)}}{\sigma_x \sigma_y} \)
\( \implies \) = \( \frac{-0.6}{3.464} \)
\( \implies \) = -0.1732
Answer:

\( n = 5, \sum x_i = 25, \sum y_i = 30, \sum x_i^2 = 155, \sum y_i^2 = 190, \sum x_i y_i = 147 \) \( \bar{x} = \frac{25}{5} = 5 \) \( \bar{y} = \frac{30}{5} = 6 \) Cov (X, Y) = \( \frac{\sum x_i y_i}{n} - \bar{x} \bar{y} = \frac{147}{5} - (5 \times 6) = 29.4 - 30 = -0.6 \) \( \sigma_x^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2 = \frac{155}{5} - (5)^2 = 31 - 25 = 6 \) \( \sigma_x = \sqrt{6} \) \( \sigma_y^2 = \frac{\sum y_i^2}{n} - (\bar{y})^2 = \frac{190}{5} - (6)^2 = 38 - 36 = 2 \) \( \sigma_y = \sqrt{2} \) \( \sigma_x \sigma_y = \sqrt{6} \times \sqrt{2} = \sqrt{12} = 2\sqrt{3} = 2 \times 1.732 = 3.464 \) Correlation coefficient, r = \( \frac{\text{Cov(X,Y)}}{\sigma_x \sigma_y} = \frac{-0.6}{3.464} \approx -0.1732 \)
In simple words: We calculated the means, sums of squares, and products from the data, then used these to find the covariance and standard deviations. Finally, we applied the formula to get the correlation coefficient, which is approximately -0.1732, indicating a weak negative relationship.

🎯 Exam Tip: Always construct a table to calculate the necessary sums (\( \sum x_i, \sum y_i, \sum x_i^2, \sum y_i^2, \sum x_i y_i \)). Be meticulous with calculations and decimal places, especially when using given root values like √3.

Question 8. The correlation coefficient between x and y is 0.3 and their covariance is 12. The variance of x is 9, find the standard deviation of y. Solution: Given, r = 0.3, Cov(X, Y) = 12, \( \sigma_x^2 \) = 9
\( \therefore \sigma_x \) = 3 Since, r = \( \frac{\text{Cov(X,Y)}}{\sigma_x \sigma_y} \)
\( \implies \) 0.3 = \( \frac{12}{(3) (\sigma_y)} \)
\( \implies \) 0.3 = \( \frac{4}{\sigma_y} \)
\( \therefore \sigma_y \) = \( \frac{4}{0.3} \)
\( \therefore \sigma_y \) = 13.33
\( \therefore \) the standard deviation of y is 13.33.
Answer: Given r = 0.3, Cov(X, Y) = 12, and variance of x (\( \sigma_x^2 \)) = 9. First, find \( \sigma_x \): \( \sigma_x = \sqrt{9} = 3 \). Using the formula for correlation coefficient: \( r = \frac{\text{Cov(X,Y)}}{\sigma_x \sigma_y} \) \( 0.3 = \frac{12}{3 \times \sigma_y} \) \( 0.3 = \frac{4}{\sigma_y} \) \( \sigma_y = \frac{4}{0.3} \approx 13.33 \) The standard deviation of y is 13.33.
In simple words: We used the given correlation coefficient, covariance, and standard deviation of x to calculate the standard deviation of y by rearranging the correlation formula, yielding approximately 13.33.

🎯 Exam Tip: Pay close attention to distinguishing between variance and standard deviation. Always take the square root of the variance to get the standard deviation. Practice solving for different variables within the correlation coefficient formula.