OP Malhotra Class 11 Maths Solutions Chapter 29 Correlation Analysis Exercise 29 (A)

 

Question 1. A physicist is experimenting with the resistance in a circuit she is using. She measures and records the resulting current.

(i) Draw a scatter graph of her results.
(ii) Estimate the current for a resistance of 40 ohms.
(iii) Estimate the resistance for a current of 7.5 amps.
Answer:
(i) To draw the scatter graph, we plot the given points: (5, 10), (10, 4.9), (15, 3.2), (20, 2.4), (25, 1.9), (30, 1.7), and (50, 1.0) on a graph paper. The resistance is on the x-axis and the current is on the y-axis. As resistance increases, current generally decreases, showing an inverse relationship.
(ii) Looking at the scatter diagram, when the resistance is 40 ohms, the corresponding current can be estimated to be about 1.3 amps. This value is found by tracing up from 40 on the x-axis to the trend line and then across to the y-axis.
(iii) From the scatter diagram, if the current is 7.5 amps, the corresponding resistance can be estimated to be around 6.5 ohms. This involves tracing from 7.5 on the y-axis to the trend line and then down to the x-axis.
In simple words: The scatter graph helps us see how resistance and current are connected. When we look at the graph, we can guess the current for 40 ohms resistance or the resistance for 7.5 amps current.

🎯 Exam Tip: When drawing scatter graphs, ensure your axes are clearly labeled with units, and the points are plotted accurately. For estimations, draw clear lines from the axis to the trend of the points.

 

Question 2. Heights of eight boys were measured and their shoe sizes were recorded.

Draw a scatter graph and use it to find out whether there is a relationship between these sets of data.
Answer: We plot the points with Height on the x-axis and Shoe size on the y-axis. The data points are (172, 8.5), (182, 10.5), (164, 7), (190, 13), (167, 8.5), (169, 8), (175, 10), and (185, 12).The plotted points roughly form a line that goes upwards. This suggests that a boy's shoe size generally increases with his height. This pattern indicates a positive relationship or correlation between the two measurements.
In simple words: When we graph height against shoe size, the dots mostly go in an upward direction. This means that taller boys tend to have bigger shoe sizes.

🎯 Exam Tip: When scatter points show a clear trend, indicate whether it's a positive (upward), negative (downward), or no correlation. For linear trends, stating it's 'approximately linear' is a good practice.

 

TYPE 2. (Based on first formula : \( r=\frac{\Sigma d_x d_y}{\sqrt{\left(\Sigma d_x^2\right)\left(\sum d_y^2\right)}}, \) where \( d_x = x - \bar{x}, d_y = y - \bar{y} \) )

 

Question 3. Pearson's coefficient of correlation between the values of X and Y for the following data. Comment on the value of r.

Answer: We first need to calculate the means for X and Y.
Here, \( \bar{X} = \frac{1+2+3+4+5}{5} = \frac{15}{5} = 3 \)
And \( \bar{Y} = \frac{7+6+5+4+3}{5} = \frac{25}{5} = 5 \)
Next, we construct a table to find the deviations and their products for the correlation coefficient calculation:
Now, we calculate the coefficient of correlation (r) using the formula based on deviations from the mean:
\( r = \frac{\Sigma (X - \bar{X}) (Y - \bar{Y})}{\sqrt{\Sigma (X - \bar{X})^2} \sqrt{\Sigma (Y - \bar{Y})^2}} \)
\( \implies r = \frac{-10}{\sqrt{10} \sqrt{10}} \)
\( \implies r = \frac{-10}{10} \)
\( \implies r = -1 \)
The value of \( r = -1 \) indicates a perfect negative correlation between X and Y. This means that as X increases, Y decreases in a perfectly predictable straight-line manner.
In simple words: We find the average of X and Y, then build a table to see how each number differs from its average. Using these differences, we calculate 'r'. A value of -1 for 'r' means that X and Y are perfectly linked, but in opposite ways: when one goes up, the other goes down exactly at the same rate.

🎯 Exam Tip: Remember that a correlation coefficient of -1 implies a perfect inverse linear relationship. Always interpret the calculated 'r' value in terms of the strength and direction of the relationship.

 

Question 4.

Answer: First, we determine the means for X and Y.
\( \bar{X} = \frac{\Sigma X}{n} = \frac{1+2+3+4+5+6+7+8+9}{9} = \frac{45}{9} = 5 \)
\( \bar{Y} = \frac{\Sigma Y}{n} = \frac{12+11+13+15+14+17+16+19+18}{9} = \frac{135}{9} = 15 \)
Next, we create a table to find the deviations from the means and their squares and products:
Now we use the formula for Karl Pearson's coefficient of correlation:
\( r=\frac{\Sigma d_{\mathrm{X}} d_{\mathrm{Y}}}{\sqrt{\Sigma d_{\mathrm{X}}^2} \sqrt{\Sigma d_{\mathrm{Y}}^2}} \)
\( \implies r = \frac{56}{\sqrt{60} \sqrt{60}} \)
\( \implies r = \frac{56}{60} \)
\( \implies r = \frac{14}{15} = 0.933 \)
Since \( r = 0.933 \), it indicates a high positive correlation between X and Y. This means as X increases, Y also tends to increase strongly.
In simple words: We calculate the average of X and Y, then build a table to find how much each value is away from its average. Using these numbers, we find 'r'. A value of 0.933 means X and Y usually go up together, showing a very strong link.

🎯 Exam Tip: When \( r \) is close to +1, it signifies a strong positive linear correlation. Always ensure the sums of deviations from the mean for \( d_x \) and \( d_y \) are close to zero as a check for mean calculation accuracy.

 

Question 5.

Answer: We are given the following values for X and Y series:
Number of pairs of observation, \( n = 15 \)
Sum of the squares of deviation from the mean for X, \( \Sigma d_X^2 = \Sigma(X - \bar{X})^2 = 136 \)
Sum of the squares of deviation from the mean for Y, \( \Sigma d_Y^2 = \Sigma(Y - \bar{Y})^2 = 138 \)
Sum of the product of the deviations of x and y, \( \Sigma d_X d_Y = \Sigma(X - \bar{X})(Y - \bar{Y}) = 122 \)
We use Karl Pearson's coefficient of correlation formula:
\( r=\frac{\Sigma d_{\mathrm{X}} d_{\mathrm{Y}}}{\sqrt{\Sigma d_{\mathrm{X}}^2} \sqrt{\Sigma d_{\mathrm{Y}}^2}} \)
\( \implies r = \frac{122}{\sqrt{136} \sqrt{138}} \)
\( \implies r = \frac{122}{\sqrt{18768}} \)
\( \implies r = \frac{122}{136.996} \)
\( \implies r \approx 0.89 \)
So, there is a high positive correlation between X and Y. This suggests that as the values of X increase, the values of Y also tend to increase significantly.
In simple words: We are given the sum of how much each number in X and Y moves from its average, and also the sum of their products. We put these numbers into the formula for 'r' and calculate it. The result of 0.89 shows that X and Y are closely related and usually move in the same direction.

🎯 Exam Tip: For problems involving sum of squares of deviations, use the direct formula for 'r'. Be careful with square root calculations and ensure accuracy to get the correct coefficient value.

 

Question 6. Calculate the Pearson's coefficient of correlation between the ages of husband and wife.

Answer: Let X be the age of the husband and Y be the age of the wife. We need to calculate the means for X and Y first. There are \( n=7 \) pairs of observations.
\( \Sigma X = 35+34+40+43+56+20+38 = 266 \)
\( \Sigma Y = 32+30+31+32+53+20+33 = 231 \)
\( \bar{X} = \frac{\Sigma X}{n} = \frac{266}{7} = 38 \)
\( \bar{Y} = \frac{\Sigma Y}{n} = \frac{231}{7} = 33 \)
Now we construct a table to find the deviations from the means, their squares, and products:
Using the formula for Pearson's correlation coefficient:
\( r=\frac{\Sigma d_x d_y}{\sqrt{\Sigma d_x^2} \sqrt{\Sigma d_y^2}} \)
\( \implies r = \frac{600}{\sqrt{702} \sqrt{584}} \)
\( \implies r = \frac{600}{\sqrt{409728}} \)
\( \implies r = \frac{600}{640.10} \)
\( \implies r \approx 0.937 \)
The correlation coefficient of 0.937 shows a strong positive correlation between the ages of husbands and wives. This means that older husbands tend to have older wives, and younger husbands tend to have younger wives.
In simple words: We found the average age for husbands and wives. Then we made a table to see how each couple's age differs from the average. Using these differences, we calculated 'r'. A high 'r' value of 0.937 means that husband and wife ages are strongly linked, so they are usually close in age.

🎯 Exam Tip: When calculating correlation for real-world data like ages, remember that a positive correlation indicates a direct relationship. Be careful with large numbers in calculations; a calculator is essential for precision.

 

Question 7. Given \( r = 0.8, \Sigma xy = 60, \sigma_y = 2.5 \) and \( \Sigma x^2 = 90 \), find the number of items. x and y are deviations from their respective mean.
Answer: We are given:
\( r = 0.8 \)
\( \Sigma xy = 60 \) (where x and y are deviations from means, i.e., \( \Sigma (X-\bar{X})(Y-\bar{Y}) \))
\( \sigma_y = 2.5 \) (standard deviation of Y)
\( \Sigma x^2 = 90 \) (where x is deviation from mean, i.e., \( \Sigma (X-\bar{X})^2 \))
We know the formula for the correlation coefficient when x and y are deviations from the mean:
\( r = \frac{\Sigma xy}{\sqrt{\Sigma x^2 \cdot \Sigma y^2}} \)
First, let's find \( \Sigma y^2 \). We know that \( \sigma_y = \sqrt{\frac{\Sigma y^2}{n}} \), so \( \sigma_y^2 = \frac{\Sigma y^2}{n} \).
We are also given \( r = 0.8 \). Let's substitute the known values into the correlation formula:
\( 0.8 = \frac{60}{\sqrt{90 \cdot \Sigma y^2}} \)
To solve for \( \Sigma y^2 \), we can first solve for \( \sqrt{90 \cdot \Sigma y^2} \):
\( \sqrt{90 \cdot \Sigma y^2} = \frac{60}{0.8} \)
\( \sqrt{90 \cdot \Sigma y^2} = 75 \)
Square both sides:
\( 90 \cdot \Sigma y^2 = 75^2 \)
\( 90 \cdot \Sigma y^2 = 5625 \)
\( \Sigma y^2 = \frac{5625}{90} = 62.5 \)
Now we use the standard deviation formula for Y:
\( \sigma_y = \sqrt{\frac{\Sigma y^2}{n}} \)
\( \implies 2.5 = \sqrt{\frac{62.5}{n}} \)
Square both sides:
\( (2.5)^2 = \frac{62.5}{n} \)
\( 6.25 = \frac{62.5}{n} \)
\( \implies n = \frac{62.5}{6.25} \)
\( \implies n = 10 \)
Thus, the required number of items is 10. Understanding how standard deviation relates to the sum of squared deviations is key here.
In simple words: We are given the correlation 'r', some sums, and the standard deviation for Y. We use the correlation formula to find a missing sum for Y. Then, using the standard deviation for Y, we find out how many items were in the data. The number of items is 10.

🎯 Exam Tip: This question combines different formulas. Remember the relationships: \( r = \frac{\Sigma xy}{\sqrt{\Sigma x^2 \Sigma y^2}} \) and \( \sigma = \sqrt{\frac{\Sigma \text{deviation}^2}{n}} \). Make sure to square both sides correctly when removing the square root.

 

TYPE 3. (Using direct values) when they are small numbers.
Formula: \( r = \frac{N \Sigma XY - \Sigma X \Sigma Y}{\sqrt{N \Sigma X^2 - (\Sigma X)^2} \sqrt{N \Sigma Y^2 - (\Sigma Y)^2}} \)

 

Question 8. (1,2), (2,4), (3,8), (4,7), (5,10), (6,5), (7,14), (8,16), (9,2), (10,20)
Answer: We have \( n = 10 \) pairs of observations. To calculate Karl Pearson's coefficient of correlation using direct values, we first construct a table to find \( \Sigma X, \Sigma Y, \Sigma XY, \Sigma X^2, \) and \( \Sigma Y^2 \):

Now we apply the formula for the coefficient of correlation:
\( r = \frac{N \Sigma XY - \Sigma X \Sigma Y}{\sqrt{N \Sigma X^2 - (\Sigma X)^2} \sqrt{N \Sigma Y^2 - (\Sigma Y)^2}} \)
\( \implies r = \frac{10 \times 586 - 55 \times 88}{\sqrt{10 \times 385 - (55)^2} \sqrt{10 \times 1114 - (88)^2}} \)
\( \implies r = \frac{5860 - 4840}{\sqrt{3850 - 3025} \sqrt{11140 - 7744}} \)
\( \implies r = \frac{1020}{\sqrt{825} \sqrt{3396}} \)
\( \implies r = \frac{1020}{28.72 \times 58.27} \)
\( \implies r = \frac{1020}{1673.71} \)
\( \implies r \approx 0.609 \)
The correlation coefficient is approximately 0.609, indicating a moderate positive correlation between X and Y. There is a general tendency for X and Y to increase together, but not in a very strong or perfect linear way.
In simple words: We list all the X and Y values, then calculate their sums, their squares, and their products. We put these totals into the direct formula for 'r'. The result, 0.609, means that X and Y generally go in the same direction, but the link is not super strong.

🎯 Exam Tip: When using the direct method for correlation, keep track of all the sums and squares carefully. Double-check your calculations, especially with the large numbers involved in \( N \Sigma XY \) and the terms under the square roots.

 

Question 9.

Answer: We have \( n = 7 \) pairs of observations. To find Karl Pearson's coefficient of correlation, we first calculate \( \Sigma X, \Sigma Y, \Sigma XY, \Sigma X^2, \) and \( \Sigma Y^2 \) by constructing a table:
Now we use the formula for the coefficient of correlation:
\( r = \frac{N \Sigma XY - \Sigma X \Sigma Y}{\sqrt{N \Sigma X^2 - (\Sigma X)^2} \sqrt{N \Sigma Y^2 - (\Sigma Y)^2}} \)
\( \implies r = \frac{7 \times 0 - 0 \times 28}{\sqrt{7 \times 28 - (0)^2} \sqrt{7 \times 196 - (28)^2}} \)
\( \implies r = \frac{0 - 0}{\sqrt{196 - 0} \sqrt{1372 - 784}} \)
\( \implies r = \frac{0}{\sqrt{196} \sqrt{588}} \)
\( \implies r = 0 \)
The coefficient of correlation is 0. This indicates that there is no linear relationship between X and Y. While there might be a non-linear relationship (like a parabolic one, since \( Y = X^2 \)), Pearson's r only measures linear correlation.
In simple words: We make a table to find the sums of X, Y, XY, X-squared, and Y-squared. Then we use these numbers in the formula to calculate 'r'. A result of 0 means there is no straight-line connection between X and Y.

🎯 Exam Tip: A correlation coefficient of zero means no *linear* relationship. It doesn't mean there's no relationship at all. For example, \( y = x^2 \) will show a zero linear correlation despite a clear quadratic relationship.

 

Question 10. \( n = 50, \Sigma x = 75, \Sigma y = 80, \Sigma x^2 = 150, \Sigma y^2 = 140, \Sigma xy = 120. \)
Answer: We are given the following values:
\( n = 50 \) (number of items)
\( \Sigma x = 75 \) (sum of X values)
\( \Sigma y = 80 \) (sum of Y values)
\( \Sigma x^2 = 150 \) (sum of squares of X values)
\( \Sigma y^2 = 140 \) (sum of squares of Y values)
\( \Sigma xy = 120 \) (sum of products of X and Y values)
We use the formula for Karl Pearson's coefficient of correlation:
\( r = \frac{N \Sigma XY - \Sigma X \Sigma Y}{\sqrt{N \Sigma X^2 - (\Sigma X)^2} \sqrt{N \Sigma Y^2 - (\Sigma Y)^2}} \)
\( \implies r = \frac{50 \times 120 - 75 \times 80}{\sqrt{50 \times 150 - (75)^2} \sqrt{50 \times 140 - (80)^2}} \)
\( \implies r = \frac{6000 - 6000}{\sqrt{7500 - 5625} \sqrt{7000 - 6400}} \)
\( \implies r = \frac{0}{\sqrt{1875} \sqrt{600}} \)
\( \implies r = 0 \)
The coefficient of correlation is 0. This result means that there is no linear relationship between variables X and Y based on the given data. This can happen when the variables are completely independent, or when their relationship is not linear.
In simple words: We have all the needed sums and the number of items. We plug these into the formula to find 'r'. Since the top part of the fraction becomes zero, 'r' is 0. This means X and Y have no straight-line connection.

🎯 Exam Tip: If the numerator of the correlation coefficient formula ( \( N \Sigma XY - \Sigma X \Sigma Y \) ) is zero, the correlation coefficient will always be zero, indicating no linear relationship. This is a quick way to identify cases of zero correlation.

 

Question 11. \( n = 10, \Sigma x = 55, \Sigma y = 40, \Sigma x^2 = 385, \Sigma y^2 = 192 \) and \( \Sigma(x + y)^2 = 947. \)
Answer: We are given the following values:
\( n = 10 \)
\( \Sigma x = 55 \)
\( \Sigma y = 40 \)
\( \Sigma x^2 = 385 \)
\( \Sigma y^2 = 192 \)
\( \Sigma(x + y)^2 = 947 \)
We know the identity \( (x+y)^2 = x^2 + y^2 + 2xy \).
Applying summation to this identity:
\( \Sigma(x + y)^2 = \Sigma x^2 + \Sigma y^2 + 2\Sigma xy \)
Substitute the given values into this equation:
\( 947 = 385 + 192 + 2\Sigma xy \)
First, add the numbers on the right side:
\( 947 = 577 + 2\Sigma xy \)
Now, subtract 577 from both sides to isolate \( 2\Sigma xy \):
\( 947 - 577 = 2\Sigma xy \)
\( 370 = 2\Sigma xy \)
Finally, divide by 2 to find \( \Sigma xy \):
\( \Sigma xy = \frac{370}{2} \)
\( \implies \Sigma xy = 185 \)
Now we have all the components to calculate the correlation coefficient using the direct formula:
\( r = \frac{N \Sigma XY - \Sigma X \Sigma Y}{\sqrt{N \Sigma X^2 - (\Sigma X)^2} \sqrt{N \Sigma Y^2 - (\Sigma Y)^2}} \)
\( \implies r = \frac{10 \times 185 - 55 \times 40}{\sqrt{10 \times 385 - (55)^2} \sqrt{10 \times 192 - (40)^2}} \)
\( \implies r = \frac{1850 - 2200}{\sqrt{3850 - 3025} \sqrt{1920 - 1600}} \)
\( \implies r = \frac{-350}{\sqrt{825} \sqrt{320}} \)
\( \implies r = \frac{-350}{28.7228 \times 17.8885} \)
\( \implies r = \frac{-350}{513.71} \)
\( \implies r \approx -0.6812 \)
The correlation coefficient is approximately -0.6812, indicating a moderately strong negative correlation between X and Y.
In simple words: We use the given sum of \( (x+y)^2 \) and other sums to find \( \Sigma xy \). Then, we use a formula that needs \( N, \Sigma X, \Sigma Y, \Sigma XY, \Sigma X^2, \) and \( \Sigma Y^2 \) to calculate 'r'. The result of -0.6812 means that X and Y tend to move in opposite directions, but not perfectly.

🎯 Exam Tip: When given \( \Sigma(x+y)^2 \), remember the algebraic expansion to find \( \Sigma xy \). This step is crucial for calculating the correlation coefficient. Pay attention to negative signs in the calculation to ensure the correct sign for 'r'.

 

TYPE 4. (Using deviations from assumed means, when X and Y are not integers.)
Formula: \( r= \frac{N \Sigma uv - \Sigma u \Sigma v}{\sqrt{N \Sigma u^2 - (\Sigma u)^2} \sqrt{N \Sigma v^2 - (\Sigma v)^2}} \)
Where \( u = X-A \) or \( u = \frac{X-A}{h}, v = Y-B \) or \( v = \frac{Y-B}{k} \), A and B being assumed means. This method simplifies calculations by working with smaller numbers.

 

Question 12. Calculate the Pearson's coefficient of correlation between the ages of husband and wife.

Answer: Let X be the age of the husband and Y be the age of the wife. We have \( n=8 \) pairs of observations. To simplify calculations, we will use assumed means.
Let Assumed Mean for X, \( A = 20 \). So \( u = X - 20 \).
Let Assumed Mean for Y, \( B = 25 \). So \( v = Y - 25 \).
Now we construct a table to calculate \( u, v, uv, u^2, \) and \( v^2 \):
Now we apply the formula using assumed means:
\( r = \frac{N \Sigma uv - \Sigma u \Sigma v}{\sqrt{N \Sigma u^2 - (\Sigma u)^2} \sqrt{N \Sigma v^2 - (\Sigma v)^2}} \)
\( \implies r = \frac{8 \times 152 - (5) \times (-1)}{\sqrt{8 \times 135 - (5)^2} \sqrt{8 \times 221 - (-1)^2}} \)
\( \implies r = \frac{1216 - (-5)}{\sqrt{1080 - 25} \sqrt{1768 - 1}} \)
\( \implies r = \frac{1221}{\sqrt{1055} \sqrt{1767}} \)
\( \implies r = \frac{1221}{32.4808 \times 42.0357} \)
\( \implies r = \frac{1221}{1365.31} \)
\( \implies r \approx 0.8943 \)
The correlation coefficient is approximately 0.8943, indicating a very high positive correlation between the ages of husband and wife. This suggests a strong tendency for couples to be of similar ages.
In simple words: We pick easy numbers (assumed means) to subtract from the actual ages, making the numbers smaller (u and v). Then, we make a table with these smaller numbers to find their sums and squares. Using a special formula for these 'u' and 'v' values, we calculate 'r'. The result 0.8943 means there is a very strong link: older husbands usually have older wives, and vice-versa.

🎯 Exam Tip: The assumed mean method simplifies calculations, especially when raw data values are large. Remember that subtracting a constant (A or B) from X or Y does not change the correlation coefficient. Ensure you use the correct sums (\( \Sigma u, \Sigma v, \Sigma uv, \Sigma u^2, \Sigma v^2 \)) in the formula.

 

Question 13.

Answer: We have \( n=7 \) pairs of observations. To simplify calculations, we will use assumed means.
Let Assumed Mean for X, \( A = 5 \). So \( u = X - 5 \).
Let Assumed Mean for Y, \( B = 10 \). So \( v = Y - 10 \).
Now we construct a table to calculate \( u, v, uv, u^2, \) and \( v^2 \):
Now we apply the formula using assumed means:
\( r = \frac{N \Sigma uv - \Sigma u \Sigma v}{\sqrt{N \Sigma u^2 - (\Sigma u)^2} \sqrt{N \Sigma v^2 - (\Sigma v)^2}} \)
\( \implies r = \frac{7 \times 122 - (2) \times (6)}{\sqrt{7 \times 64 - (2)^2} \sqrt{7 \times 236 - (6)^2}} \)
\( \implies r = \frac{854 - 12}{\sqrt{448 - 4} \sqrt{1652 - 36}} \)
\( \implies r = \frac{842}{\sqrt{444} \sqrt{1616}} \)
\( \implies r = \frac{842}{21.0713 \times 40.200} \)
\( \implies r = \frac{842}{847.16} \)
\( \implies r \approx 0.9940 \)
The correlation coefficient is approximately 0.9940. This value is very close to +1, which means there is a very strong positive and almost perfect linear correlation between X and Y.
In simple words: We choose estimated averages for X and Y, then subtract these from each data point to get smaller numbers (u and v). We sum these new numbers and their squares, then use a special formula. The result, about 0.994, tells us X and Y are nearly perfectly linked in a straight line, and they both increase together.

🎯 Exam Tip: A correlation coefficient close to 1 indicates an almost perfect positive linear relationship. Always show all steps, especially when using assumed means, to avoid errors in complex calculations.

 

Question 14. Calculate Karl Pearson's correlation coefficient between the marks in English and Hindi obtained by 10 students.

 

Question 14. Calculate Karl Pearson's correlation coefficient between the marks in English and Hindi obtained by 10 students.
Answer: To find Karl Pearson's correlation coefficient, we first organize the given marks for 10 students into a table. We assume a mean A = 20 for English marks (X) and B = 17 for Hindi marks (Y). Then we calculate the deviations u = X - 20 and v = Y - 17, along with uv, u², and v².

We use the formula for Pearson's correlation coefficient (r): \( r = \frac{n \Sigma uv - (\Sigma u)(\Sigma v)}{\sqrt{[n \Sigma u^2 - (\Sigma u)^2][n \Sigma v^2 - (\Sigma v)^2]}} \) Given \( n = 10 \), \( \Sigma u = -16 \), \( \Sigma v = 12 \), \( \Sigma uv = 102 \), \( \Sigma u^2 = 374 \), \( \Sigma v^2 = 142 \). Substituting these values into the formula: \( r = \frac{10 \times 102 - (-16)(12)}{\sqrt{[10 \times 374 - (-16)^2][10 \times 142 - (12)^2]}} \)
\( \implies r = \frac{1020 - (-192)}{\sqrt{[3740 - 256][1420 - 144]}} \)
\( \implies r = \frac{1020 + 192}{\sqrt{[3484][1276]}} \)
\( \implies r = \frac{1212}{\sqrt{4444544}} \)
\( \implies r = \frac{1212}{2108.2049} \)
\( \implies r \approx 0.575 \) So, the correlation coefficient is approximately 0.575. This indicates a moderate positive relationship between marks in English and Hindi.In simple words: We calculated a number called 'r' which tells us how much English and Hindi marks are connected. We found r to be about 0.575. This means if a student scores well in English, they are likely to score somewhat well in Hindi too, but it's not a very strong connection.

🎯 Exam Tip: When using assumed mean methods, double-check all calculations for \(u\), \(v\), \(uv\), \(u^2\), and \(v^2\) sums. A small error in summing can lead to a completely different correlation coefficient.

 

Question 15. Show that the coefficient of correlation \( \rho \) between two variables x and y is given by \( \rho=\frac{\sigma_x^2+\sigma_y^2-\sigma_{x-y}^2}{2 \sigma_y \sigma_x} \) where \( \sigma_x^2, \sigma_y^2 \) and \( \sigma_{x-y}^2 \) are the variances of x, y and x-y respectively.
Answer: We need to show how the correlation coefficient \( \rho \) is related to the variances of x, y, and (x-y). We know that the variance of the difference of two variables \( (x-y) \) is given by: \( \sigma_{x-y}^2 = \frac{1}{n} \Sigma[(x - \bar{x}) - (y - \bar{y})]^2 \) Let \( d_x = x - \bar{x} \) and \( d_y = y - \bar{y} \). So, \( \sigma_{x-y}^2 = \frac{1}{n} \Sigma (d_x - d_y)^2 \)
\( \implies \sigma_{x-y}^2 = \frac{1}{n} \Sigma (d_x^2 - 2 d_x d_y + d_y^2) \)
\( \implies \sigma_{x-y}^2 = \frac{1}{n} (\Sigma d_x^2 - 2 \Sigma d_x d_y + \Sigma d_y^2) \) We know that \( \sigma_x^2 = \frac{\Sigma d_x^2}{n} \), \( \sigma_y^2 = \frac{\Sigma d_y^2}{n} \), and the covariance \( \text{Cov}(x, y) = \frac{\Sigma d_x d_y}{n} \). And the correlation coefficient \( \rho = \frac{\text{Cov}(x, y)}{\sigma_x \sigma_y} \), which means \( \Sigma d_x d_y = n \rho \sigma_x \sigma_y \). Substitute these back into the equation for \( \sigma_{x-y}^2 \): \( \sigma_{x-y}^2 = \frac{\Sigma d_x^2}{n} - \frac{2 \Sigma d_x d_y}{n} + \frac{\Sigma d_y^2}{n} \)
\( \implies \sigma_{x-y}^2 = \sigma_x^2 - 2 \rho \sigma_x \sigma_y + \sigma_y^2 \) Now, rearrange the equation to solve for \( \rho \): \( 2 \rho \sigma_x \sigma_y = \sigma_x^2 + \sigma_y^2 - \sigma_{x-y}^2 \)
\( \implies \rho = \frac{\sigma_x^2 + \sigma_y^2 - \sigma_{x-y}^2}{2 \sigma_x \sigma_y} \) This proves the desired formula. This relationship is very useful for understanding how the spread of the difference between two variables is related to their individual spreads and their correlation.In simple words: This problem asks us to show a special way to find the correlation number (\( \rho \)). It says we can find \( \rho \) if we know how spread out the first set of numbers is (\( \sigma_x^2 \)), how spread out the second set of numbers is (\( \sigma_y^2 \)), and how spread out the differences between the numbers are (\( \sigma_{x-y}^2 \)). The formula links these spreads directly to the correlation.

🎯 Exam Tip: Remember the basic definitions of variance and covariance in terms of deviations from the mean. This problem relies on expanding \( (A-B)^2 \) and then recognizing these definitions.

 

Question 16. A computer expert while calculating correlation coefficient between X and Y from 25 pairs of observations obtained the following results : n = 25, ΣX = 125, ΣX² = 650, ΣY = 100, ΣY² = 460, ΣXY = 508. It was, however, later discovered at the time of checking that he had copied down two pairs as X=6, Y=14 and X=8, Y=6 while the correct values were X=8, Y=12 and X=6, Y=8. Obtain the correct value of correlation coefficient.
Answer: We need to correct the sums for \( \Sigma X \), \( \Sigma Y \), \( \Sigma X^2 \), \( \Sigma Y^2 \), and \( \Sigma XY \) by removing the incorrect pairs and adding the correct ones. Then we will calculate the correlation coefficient using the corrected sums. Given: \( n = 25, \Sigma X = 125, \Sigma X^2 = 650, \Sigma Y = 100, \Sigma Y^2 = 460, \Sigma XY = 508 \). Incorrect pairs: \( (X_1=6, Y_1=14) \) and \( (X_2=8, Y_2=6) \). Correct pairs: \( (X_3=8, Y_3=12) \) and \( (X_4=6, Y_4=8) \). 1. **Corrected \( \Sigma X \)**: \( \text{Corrected } \Sigma X = \Sigma X_{\text{old}} - (X_1 + X_2) + (X_3 + X_4) \) \( = 125 - (6 + 8) + (8 + 6) = 125 - 14 + 14 = 125 \) 2. **Corrected \( \Sigma Y \)**: \( \text{Corrected } \Sigma Y = \Sigma Y_{\text{old}} - (Y_1 + Y_2) + (Y_3 + Y_4) \) \( = 100 - (14 + 6) + (12 + 8) = 100 - 20 + 20 = 100 \) 3. **Corrected \( \Sigma X^2 \)**: \( \text{Corrected } \Sigma X^2 = \Sigma X^2_{\text{old}} - (X_1^2 + X_2^2) + (X_3^2 + X_4^2) \) \( = 650 - (6^2 + 8^2) + (8^2 + 6^2) = 650 - (36 + 64) + (64 + 36) = 650 - 100 + 100 = 650 \) 4. **Corrected \( \Sigma Y^2 \)**: \( \text{Corrected } \Sigma Y^2 = \Sigma Y^2_{\text{old}} - (Y_1^2 + Y_2^2) + (Y_3^2 + Y_4^2) \) \( = 460 - (14^2 + 6^2) + (12^2 + 8^2) = 460 - (196 + 36) + (144 + 64) = 460 - 232 + 208 = 436 \) 5. **Corrected \( \Sigma XY \)**: \( \text{Corrected } \Sigma XY = \Sigma XY_{\text{old}} - (X_1Y_1 + X_2Y_2) + (X_3Y_3 + X_4Y_4) \) \( = 508 - (6 \times 14 + 8 \times 6) + (8 \times 12 + 6 \times 8) = 508 - (84 + 48) + (96 + 48) = 508 - 132 + 144 = 520 \) Now, use the corrected values to calculate Pearson's correlation coefficient \( r \): \( n = 25 \), Corrected \( \Sigma X = 125 \), Corrected \( \Sigma Y = 100 \), Corrected \( \Sigma X^2 = 650 \), Corrected \( \Sigma Y^2 = 436 \), Corrected \( \Sigma XY = 520 \). The formula for \( r \) is: \( r = \frac{n \Sigma XY - (\Sigma X)(\Sigma Y)}{\sqrt{[n \Sigma X^2 - (\Sigma X)^2][n \Sigma Y^2 - (\Sigma Y)^2]}} \) Substitute the corrected values: \( r = \frac{25 \times 520 - (125)(100)}{\sqrt{[25 \times 650 - (125)^2][25 \times 436 - (100)^2]}} \)
\( \implies r = \frac{13000 - 12500}{\sqrt{[16250 - 15625][10900 - 10000]}} \)
\( \implies r = \frac{500}{\sqrt{[625][900]}} \)
\( \implies r = \frac{500}{\sqrt{562500}} \)
\( \implies r = \frac{500}{750} \)
\( \implies r = \frac{2}{3} \approx 0.667 \) The corrected correlation coefficient is approximately 0.667. This shows a strong positive correlation between X and Y.In simple words: First, we updated all the total numbers (sums) by removing the wrong data and putting in the right data. Then, we used these new, correct totals in the main formula for the correlation coefficient. This gave us a new 'r' value of about 0.667. This number tells us that the two sets of observations are quite strongly linked, meaning if one increases, the other generally increases too.

🎯 Exam Tip: When correcting data, be very systematic: subtract the old values for each sum (\( \Sigma X, \Sigma Y, \Sigma X^2, \Sigma Y^2, \Sigma XY \)) and then add the new values. This ensures all parts of the correlation formula are based on accurate data.

 

Question 17. A computer obtained the data: n = 30, Σx = 120, Σy = 90, Σx² = 600, Σy² = 250 and Σxy = 356. Later on, it was found that pairs (X=8, Y=10) and (X=12, Y=7) are wrong while the correct values are (X=8, Y=12) and (X=10, Y=8). Find the correct values of \( \rho(X, Y) \).
Answer: We must first correct the given sums based on the identified errors and then compute Pearson's correlation coefficient. Given: \( n = 30, \Sigma x = 120, \Sigma y = 90, \Sigma x^2 = 600, \Sigma y^2 = 250, \Sigma xy = 356 \). Incorrect pairs: \( (X_1=8, Y_1=10) \) and \( (X_2=12, Y_2=7) \). Correct pairs: \( (X_3=8, Y_3=12) \) and \( (X_4=10, Y_4=8) \). 1. **Corrected \( \Sigma x \)**: \( \text{Corrected } \Sigma x = 120 - (8 + 12) + (8 + 10) = 120 - 20 + 18 = 118 \) 2. **Corrected \( \Sigma y \)**: \( \text{Corrected } \Sigma y = 90 - (10 + 7) + (12 + 8) = 90 - 17 + 20 = 93 \) 3. **Corrected \( \Sigma x^2 \)**: \( \text{Corrected } \Sigma x^2 = 600 - (8^2 + 12^2) + (8^2 + 10^2) = 600 - (64 + 144) + (64 + 100) = 600 - 208 + 164 = 556 \) 4. **Corrected \( \Sigma y^2 \)**: \( \text{Corrected } \Sigma y^2 = 250 - (10^2 + 7^2) + (12^2 + 8^2) = 250 - (100 + 49) + (144 + 64) = 250 - 149 + 208 = 309 \) 5. **Corrected \( \Sigma xy \)**: \( \text{Corrected } \Sigma xy = 356 - (8 \times 10 + 12 \times 7) + (8 \times 12 + 10 \times 8) = 356 - (80 + 84) + (96 + 80) = 356 - 164 + 176 = 368 \) Now we calculate the correlation coefficient \( \rho \) using the corrected values: \( n = 30 \), Corrected \( \Sigma x = 118 \), Corrected \( \Sigma y = 93 \), Corrected \( \Sigma x^2 = 556 \), Corrected \( \Sigma y^2 = 309 \), Corrected \( \Sigma xy = 368 \). Using the formula: \( \rho = \frac{n \Sigma xy - (\Sigma x)(\Sigma y)}{\sqrt{[n \Sigma x^2 - (\Sigma x)^2][n \Sigma y^2 - (\Sigma y)^2]}} \) Substitute the corrected values: \( \rho = \frac{30 \times 368 - (118)(93)}{\sqrt{[30 \times 556 - (118)^2][30 \times 309 - (93)^2]}} \)
\( \implies \rho = \frac{11040 - 10974}{\sqrt{[16680 - 13924][9270 - 8649]}} \)
\( \implies \rho = \frac{66}{\sqrt{[2756][621]}} \)
\( \implies \rho = \frac{66}{\sqrt{1711276}} \)
\( \implies \rho = \frac{66}{1308.15} \)
\( \implies \rho \approx 0.0504 \) The corrected correlation coefficient is approximately 0.0504. This shows a very weak positive correlation between X and Y, almost negligible.In simple words: We first corrected all the total numbers by removing the wrong data points and adding the right ones. For example, we took out the sum for the wrong X values and added the sum for the correct X values. After updating all the sums, we put them into the correlation formula. The result, about 0.0504, means there's a tiny, almost non-existent, positive link between the two sets of data.

🎯 Exam Tip: Be very careful with arithmetic when dealing with square terms (\( X^2, Y^2 \)) and product terms (\( XY \)). Small calculation errors during data correction can significantly alter the final correlation coefficient.

 

Question 18. Show that Pearson's coefficient of correlation lies between -1 and +1, i.e., \( -1 \leq r \leq 1 \) or \( |r| \leq 1 \).
Answer: We can prove that Pearson's correlation coefficient \( r \) always falls between -1 and +1 using the properties of sums of squares. Let \( x \) and \( y \) be the deviations of the variables from their respective means (i.e., \( x = X - \bar{X} \) and \( y = Y - \bar{Y} \)). The standard deviations are \( \sigma_x = \sqrt{\frac{\Sigma x^2}{n}} \) and \( \sigma_y = \sqrt{\frac{\Sigma y^2}{n}} \). The correlation coefficient is \( r = \frac{\Sigma xy}{n \sigma_x \sigma_y} \). Consider the expression \( \Sigma \left( \frac{x}{\sigma_x} + \frac{y}{\sigma_y} \right)^2 \). Since the square of any real number is non-negative, this sum must be greater than or equal to zero. \( \Sigma \left( \frac{x}{\sigma_x} + \frac{y}{\sigma_y} \right)^2 \geq 0 \)
\( \implies \Sigma \left( \frac{x^2}{\sigma_x^2} + \frac{2xy}{\sigma_x \sigma_y} + \frac{y^2}{\sigma_y^2} \right) \geq 0 \)
\( \implies \frac{\Sigma x^2}{\sigma_x^2} + \frac{2 \Sigma xy}{\sigma_x \sigma_y} + \frac{\Sigma y^2}{\sigma_y^2} \geq 0 \) We know \( \sigma_x^2 = \frac{\Sigma x^2}{n} \implies \frac{\Sigma x^2}{\sigma_x^2} = n \), and similarly \( \frac{\Sigma y^2}{\sigma_y^2} = n \). Also, \( r = \frac{\Sigma xy}{n \sigma_x \sigma_y} \implies \frac{\Sigma xy}{\sigma_x \sigma_y} = nr \). Substitute these into the inequality: \( n + 2(nr) + n \geq 0 \)
\( \implies 2n + 2nr \geq 0 \)
\( \implies 2n(1 + r) \geq 0 \) Since \( n \) (number of observations) is always positive, we can divide by \( 2n \): \( 1 + r \geq 0 \)
\( \implies r \geq -1 \) Next, consider the expression \( \Sigma \left( \frac{x}{\sigma_x} - \frac{y}{\sigma_y} \right)^2 \). This sum must also be non-negative: \( \Sigma \left( \frac{x}{\sigma_x} - \frac{y}{\sigma_y} \right)^2 \geq 0 \)
\( \implies \Sigma \left( \frac{x^2}{\sigma_x^2} - \frac{2xy}{\sigma_x \sigma_y} + \frac{y^2}{\sigma_y^2} \right) \geq 0 \)
\( \implies \frac{\Sigma x^2}{\sigma_x^2} - \frac{2 \Sigma xy}{\sigma_x \sigma_y} + \frac{\Sigma y^2}{\sigma_y^2} \geq 0 \) Substitute \( \frac{\Sigma x^2}{\sigma_x^2} = n \), \( \frac{\Sigma y^2}{\sigma_y^2} = n \), and \( \frac{\Sigma xy}{\sigma_x \sigma_y} = nr \): \( n - 2(nr) + n \geq 0 \)
\( \implies 2n - 2nr \geq 0 \)
\( \implies 2n(1 - r) \geq 0 \) Divide by \( 2n \) (which is positive): \( 1 - r \geq 0 \)
\( \implies r \leq 1 \) Combining both results, \( r \geq -1 \) and \( r \leq 1 \), we get \( -1 \leq r \leq 1 \). This means the correlation coefficient always falls within this range. The range of values for 'r' reflects the strength and direction of the linear relationship, from perfect negative (-1) to perfect positive (+1).In simple words: This question asks us to prove that the correlation coefficient, which tells us how two things are linked, can only be a number between -1 and +1. We show this by using a basic math rule: when you square any number, the answer is always zero or a positive number. By carefully building up equations using this rule, we can show that the correlation coefficient cannot be smaller than -1 or larger than +1.

🎯 Exam Tip: This proof relies on the fact that the sum of squares of real numbers is always non-negative. Clearly state this principle at the beginning of your derivation to establish the foundation of the proof.