GSEB Class 12 Statistics Solutions Chapter 3 Linear Regression Exercise 3.1

Question 1. From the following data of price (in Rs.) and demand (in hundred units) of a commodity, obtain the regression line of demand on price. Also estimate the demand when price is Rs. 20.

Price (Rs.)	12	14	15	16	18	21
Demand (hundred units)	18	12	10	8	7	5

Answer:Here, the number of data points, \(n = 6\). Let \(X\) represent the price and \(Y\) represent the demand.
First, we calculate the mean of \(X\) and \(Y\).
The mean of price, \( \bar{x} = \frac{\Sigma x}{n} = \frac{96}{6} = 16 \).
The mean of demand, \( \bar{y} = \frac{\Sigma y}{n} = \frac{60}{6} = 10 \) hundred units.
To find the equation of the regression line of \(Y\) on \(X\), which is in the form \( \hat{y} = a + bx \), we need to calculate the values of \(a\) and \(b\). We prepare a calculation table:

X	y	\( (x-\bar{x}) \) \( x=16 \)	\( (y-\bar{y}) \) \( y=10 \)	\( (x-\bar{x})(y-\bar{y}) \)	\( (x-\bar{x})^2 \)
12	18	-4	8	-32	16
14	12	-2	2	-4	4
15	10	-1	0	0	1
16	8	0	-2	0	0
18	7	2	-3	-6	4
21	5	5	-5	-25	25
\( \Sigma x = 96 \)	\( \Sigma y = 60 \)	\( \Sigma (x-\bar{x}) = 0 \)	\( \Sigma (y-\bar{y}) = 0 \)	\( \Sigma (x-\bar{x})(y-\bar{y}) = -67 \)	\( \Sigma (x-\bar{x})^2 = 50 \)

The regression coefficient \(b\) is calculated as: \[ b = \frac{\Sigma(x-\bar{x})(y-\bar{y})}{\Sigma(x-\bar{x})^{2}} \]
Substituting the values: \[ b = \frac{-67}{50} \] \[ = -1.34 \]
Now, we calculate \(a\) using the formula: \( a = \bar{y} - b\bar{x} \)
Putting \( \bar{y} = 10 \), \( \bar{x} = 16 \), and \( b = -1.34 \): \[ a = 10 - (-1.34)(16) \] \[ = 10 + 21.44 \] \[ = 31.44 \]
So, the regression line for demand on price is: \[ \hat{y} = 31.44 - 1.34x \]
Next, we estimate the demand when the price \(X\) is Rs. 20.
Substitute \(x = 20\) into the regression equation: \[ \hat{y} = 31.44 - 1.34(20) \] \[ = 31.44 - 26.8 \] \[ = 4.64 \]
Therefore, when the price is Rs. 20, the estimated demand is \( \hat{y} = 4.64 \) hundred units.
In simple words: We found a mathematical rule to predict demand from price. For every increase in price, demand tends to decrease. If the price is 20 rupees, the estimated demand will be 4.64 hundred units.

🎯 Exam Tip: Remember to clearly show the calculation of means, the construction of the deviation table, and the step-by-step calculation of 'b' and 'a' for full marks. Ensure the final estimate is presented with correct units.

 

Question 2. To study the relationship between the time of usage of cars and its average annual maintenance cost, the following information is obtained: Obtain the regression line of Y on X. Find an estimate of average annual maintenance cost when the usage time of a car is 5 years. Also find its error.

Car	1	2	3	4	5	6
Time of usage of a car (years) x	3	1	2	2	5	3
Average annual maintenance cost (thousand Rs.) y	10	5	8	7	13	8

Answer:Here, the number of observations is \(n = 6\). Let \(X\) be the time of usage of a car and \(Y\) be the annual average maintenance cost.
First, we calculate the means:
Mean of usage time, \( \bar{x} = \frac{\Sigma x}{n} = \frac{16}{6} = 2.67 \).
Mean of maintenance cost, \( \bar{y} = \frac{\Sigma y}{n} = \frac{51}{6} = 8.5 \).
We need to find the regression line of average annual maintenance cost (Y) on the time of usage (X), which is \( \hat{y} = a + bx \). Since the values of \(X\) and \(Y\) are not very large, we can prepare a table to find the necessary sums for \(a\) and \(b\):

Car	x	y	xy	\( x^2 \)
1	3	10	30	9
2	1	5	5	1
3	2	8	16	4
4	2	7	14	4
5	5	13	65	25
6	3	8	24	9
n = 6	\( \Sigma x = 16 \)	\( \Sigma y = 51 \)	\( \Sigma xy = 154 \)	\( \Sigma x^2 = 52 \)

The regression coefficient \(b\) is calculated using the formula: \[ b = \frac{n \Sigma xy - (\Sigma x)(\Sigma y)}{n \Sigma x^{2} - (\Sigma x)^{2}} \]
Substituting the values from the table: \[ b = \frac{6(154) - (16)(51)}{6(52) - (16)^{2}} \] \[ = \frac{924 - 816}{312 - 256} \] \[ = \frac{108}{56} \] \[ = 1.93 \]
Now, we calculate \(a\): \( a = \bar{y} - b\bar{x} \)
Substituting \( \bar{y} = 8.5 \), \( \bar{x} = 2.67 \), and \( b = 1.93 \): \[ a = 8.5 - 1.93(2.67) \] \[ = 8.5 - 5.15 \] \[ = 3.35 \]
So, the regression line for average annual maintenance cost (Y) on car usage time (X) is: \[ \hat{y} = 3.35 + 1.93x \]
To estimate the average annual maintenance cost \(Y\) when the car usage time \(X\) is 5 years:
Substitute \(x = 5\) into the regression equation: \[ \hat{y} = 3.35 + 1.93(5) \] \[ = 3.35 + 9.65 \] \[ = 13 \]
Therefore, the estimated average annual maintenance cost is Rs. 13 thousand when the car is used for 5 years.
Finally, we calculate the error. The actual maintenance cost for a car used for 5 years is \(y = 13\) (from the first table).
Error = \( \hat{y} - y = 13 - 13 = 0 \)
This means that the point (5, 13) lies exactly on the fitted regression line \( \hat{y} = 3.35 + 1.93x \).
In simple words: We found a rule to guess car maintenance cost based on how long a car is used. A car used longer will likely have higher maintenance costs. For a car used for 5 years, the estimated maintenance cost is 13 thousand rupees, and in this case, our guess was perfectly accurate.

🎯 Exam Tip: When calculating 'a' and 'b', be careful with decimal places for intermediate values. The "error" calculation requires comparing the estimated value with an actual value given in the original data, if available for that specific X. If not, the question would usually omit the error part or provide additional data.

 

Question 3. The information for a year regarding the average rainfall (in cm) and total production of crop (in tons) of five districts is given below: Find the regression line of production of crop on rainfall and estimate the crop if average rainfall is 35 cm.

Average rainfall (cm)	25	32	38	29	31
Crop (tons)	84	90	95	88	93

Answer:Here, the number of districts is \(n = 5\). Let \(X\) represent the average rainfall and \(Y\) represent the crop production.
First, we calculate the means:
Mean of rainfall, \( \bar{x} = \frac{\Sigma x}{n} = \frac{155}{5} = 31 \) cm.
Mean of crop production, \( \bar{y} = \frac{\Sigma y}{n} = \frac{450}{5} = 90 \) tons.
We need to find the regression line of crop production (Y) on average rainfall (X), which is \( \hat{y} = a + bx \). Since \(x\) and \(y\) are integers, we prepare a calculation table to find the necessary sums:

X	y	\( (x-\bar{x}) \) \( x = 31 \)	\( (y-\bar{y}) \) \( y = 90 \)	\( (x-\bar{x})(y-\bar{y}) \)	\( (x-\bar{x})^2 \)
25	84	-6	-6	36	36
32	90	1	0	0	1
38	95	7	5	35	49
29	88	-2	-2	4	4
31	93	0	3	0	0
\( \Sigma x = 155 \)	\( \Sigma y = 450 \)	\( \Sigma (x-\bar{x}) = 0 \)	\( \Sigma (y-\bar{y}) = 0 \)	\( \Sigma (x-\bar{x})(y-\bar{y}) = 75 \)	\( \Sigma (x-\bar{x})^2 = 90 \)

The regression coefficient \(b\) is calculated as: \[ b = \frac{\Sigma(x-\bar{x})(y-\bar{y})}{\Sigma(x-\bar{x})^{2}} \]
Substituting the values from the table: \[ b = \frac{75}{90} \] \[ = 0.83 \]
Now, we calculate \(a\): \( a = \bar{y} - b\bar{x} \)
Putting \( \bar{y} = 90 \), \( \bar{x} = 31 \), and \( b = 0.83 \): \[ a = 90 - 0.83(31) \] \[ = 90 - 25.73 \] \[ = 64.27 \]
So, the regression line for crop production (Y) on average rainfall (X) is: \[ \hat{y} = 64.27 + 0.83x \]
To estimate the crop production \(Y\) when the average rainfall \(X\) is 35 cm:
Substitute \(x = 35\) into the regression equation: \[ \hat{y} = 64.27 + 0.83(35) \] \[ = 64.27 + 29.05 \] \[ = 93.32 \]
Therefore, when the average rainfall is 35 cm, the estimated crop production is \( \hat{y} = 93.32 \) tons.
In simple words: We found a way to predict how much crop will grow based on the amount of rain. More rain usually means more crop. If there is 35 cm of rain, we expect about 93.32 tons of crop.

🎯 Exam Tip: Always state the final regression equation clearly. Ensure units are correctly included in your final estimated value, as this shows a complete understanding of the problem context.

 

Question 4. The following data gives the experience of machine operators and their performance ratings: Calculate the regression line of performance ratings on the experience and estimate the performance rating of an operator having 7 years of experience.

Operator	1	2	3	4	5	6	7	8
Experience (years) x	12	5	10	3	18	4	12	16
Performance rating y	83	75	80	78	89	68	88	87

Answer:Here, the number of operators is \(n = 8\). Let \(X\) represent the experience in years and \(Y\) represent the performance rating.
First, we calculate the means:
Mean experience, \( \bar{x} = \frac{\Sigma x}{n} = \frac{80}{8} = 10 \) years.
Mean performance rating, \( \bar{y} = \frac{\Sigma y}{n} = \frac{648}{8} = 81 \).
We need to find the regression line of performance ratings (Y) on experience (X), which is \( \hat{y} = a + bx \). Since \(x\) and \(y\) are integers, we prepare a calculation table:

Operator	X	y	\( (x-\bar{x}) \) \( x = 10 \)	\( (y-\bar{y}) \) \( y = 81 \)	\( (x-\bar{x})(y-\bar{y}) \)	\( (x-\bar{x})^2 \)
1	12	83	2	2	4	4
2	5	75	-5	-6	30	25
3	10	80	0	-1	0	0
4	3	78	-7	-3	21	49
5	18	89	8	8	64	64
6	4	68	-6	-13	78	36
7	12	88	2	7	14	4
8	16	87	6	6	36	36
n = 8	\( \Sigma x = 80 \)	\( \Sigma y = 648 \)	\( \Sigma (x-\bar{x}) = 0 \)	\( \Sigma (y-\bar{y}) = 0 \)	\( \Sigma (x-\bar{x})(y-\bar{y}) = 247 \)	\( \Sigma (x-\bar{x})^2 = 218 \)

The regression coefficient \(b\) is calculated as: \[ b = \frac{\Sigma(x-\bar{x})(y-\bar{y})}{\Sigma(x-\bar{x})^{2}} \]
Substituting the values from the table: \[ b = \frac{247}{218} \] \[ = 1.13 \]
Now, we calculate \(a\): \( a = \bar{y} - b\bar{x} \)
Putting \( \bar{y} = 81 \), \( \bar{x} = 10 \), and \( b = 1.13 \): \[ a = 81 - 1.13(10) \] \[ = 81 - 11.3 \] \[ = 69.7 \]
So, the regression line for performance ratings (Y) on experience (X) is: \[ \hat{y} = 69.7 + 1.13x \]
To estimate the performance rating \(Y\) for an operator with 7 years of experience:
Substitute \(x = 7\) into the regression equation: \[ \hat{y} = 69.7 + 1.13(7) \] \[ = 69.7 + 7.91 \] \[ = 77.61 \]
Therefore, the estimated performance rating for an operator with 7 years of experience is \( \hat{y} = 77.61 \).
In simple words: We created a formula to predict an operator's work performance based on how many years they've worked. More experience generally means better performance. An operator with 7 years of experience is expected to have a performance rating of 77.61.

🎯 Exam Tip: Pay close attention to the positive or negative sign of the regression coefficient 'b', as it indicates the direction of the relationship between the variables. Clearly label all sums from your calculation table to avoid errors in the formulas for 'b' and 'a'.