GSEB Class 12 Statistics Solutions Chapter 2 Linear Correlation Exercise 2.2

GSEB Solutions Class 12 Statistics Part 1 Chapter 2 Linear Correlation Ex 2.2

Gujarat Board Textbook Solutions Class 12 Statistics Part 1 Chapter 2 Linear Correlation Ex 2.2

Question 1.From the following information obtained from a sample of 7 families of a society regarding height of father (in cm) and height of his adult son (in cm), calculate the correlation coefficient:

Height of father (cm)	170	169	168	167	166	165	164
Height of son (cm)	172	168	170	168	165	167	166

Answer:Here, we have a sample size of \( n = 7 \). Let \( x \) represent the height of the father and \( y \) represent the height of the son. First, we find the mean for \( x \) and \( y \):
\( \bar{x} = \frac{\Sigma x}{n} = \frac{1169}{7} = 167 \text{ cm} \)
\( \bar{y} = \frac{\Sigma y}{n} = \frac{1176}{7} = 168 \text{ cm} \) Since both \( x \) and \( y \) values are integers, we prepare a table to calculate the correlation coefficient \( r \) as follows:

Height of father (cm) x	Height of son (cm) y	\( (x-\bar{x}) \) \( x = 167 \)	\( (y-\bar{y}) \) \( y = 168 \)	\( (x-\bar{x})(y-\bar{y}) \)	\( (x-\bar{x})^2 \)	\( (y-\bar{y})^2 \)
170	172	3	4	12	9	16
169	168	2	0	0	4	0
168	170	1	2	2	1	4
167	168	0	0	0	0	0
166	165	-1	-3	3	1	9
165	167	-2	-1	2	4	1
164	166	-3	-2	6	9	4
\( \Sigma x = 1169 \)	\( \Sigma y = 1176 \)	\( \Sigma (x-\bar{x}) = 0 \)	\( \Sigma (y-\bar{y}) = 0 \)	\( \Sigma (x-\bar{x})(y-\bar{y}) = 25 \)	\( \Sigma (x-\bar{x})^2 = 28 \)	\( \Sigma (y-\bar{y})^2 = 34 \)

The formula for the correlation coefficient is: \[ r = \frac{\Sigma(x-\bar{x})(y-\bar{y})}{\sqrt{\Sigma(x-\bar{x})^{2}} \cdot \sqrt{\Sigma(y-\bar{y})^{2}}} \] Substitute the calculated values into the formula: \( \Sigma(x-\bar{x})(y-\bar{y}) = 25 \), \( \Sigma(x-\bar{x})^2 = 28 \), and \( \Sigma(y-\bar{y})^2 = 34 \).
\( r = \frac{25}{\sqrt{28} \cdot \sqrt{34}} \)
\( r = \frac{25}{\sqrt{952}} \)
\( r = \frac{25}{30.85} \)
\( r = 0.81 \) Thus, the correlation coefficient is \( 0.81 \).In simple words: We calculated how strongly the father's height and son's height are related. A value of 0.81 shows a strong positive connection, meaning taller fathers tend to have taller sons.

🎯 Exam Tip: Ensure accurate calculation of means (\( \bar{x} \) and \( \bar{y} \)) and careful computation of deviations and their squares/products to avoid errors in the final correlation coefficient.

Question 2.A local cottage industry making various snacks sells each snack in a packet of 100 gm. From a study for price determination regarding a new kind of wafer, the following information is obtained for the price and the demand:

Price (Rs.)	24	26	32	33	35	30
Demand (thousand units)	27	24	22	20	15	24

Find the correlation coefficient between price of wafer and its demand.
Answer:In this problem, we have \( n = 6 \) observations. Let \( x \) denote the price of the wafer and \( y \) represent its demand. First, we compute the means for \( x \) and \( y \):
\( \bar{x} = \frac{\Sigma x}{n} = \frac{180}{6} = 30 \)
\( \bar{y} = \frac{\Sigma y}{n} = \frac{132}{6} = 22 \) (in '000 units) Since both \( x \) and \( y \) values are integers, we construct a table to find the correlation coefficient \( r \):

Price (Rs.) x	Demand (thousand units) y	\( (x-\bar{x}) \) \( x = 30 \)	\( (y-\bar{y}) \) \( y = 22 \)	\( (x-\bar{x})(y-\bar{y}) \)	\( (x-\bar{x})^2 \)	\( (y-\bar{y})^2 \)
24	27	-6	5	-30	36	25
26	24	-4	2	-8	16	4
32	22	2	0	0	4	0
33	20	3	-2	-6	9	4
35	15	5	-7	-35	25	49
30	24	0	2	0	0	4
\( \Sigma x = 180 \)	\( \Sigma y = 132 \)	\( \Sigma (x-\bar{x}) = 0 \)	\( \Sigma (y-\bar{y}) = 0 \)	\( \Sigma (x-\bar{x})(y-\bar{y}) = -79 \)	\( \Sigma (x-\bar{x})^2 = 90 \)	\( \Sigma (y-\bar{y})^2 = 86 \)

The formula for the correlation coefficient is: \[ r = \frac{\Sigma(x-\bar{x})(y-\bar{y})}{\sqrt{\Sigma(x-\bar{x})^{2}} \cdot \sqrt{\Sigma(y-\bar{y})^{2}}} \] Now, substitute the values into the formula: \( \Sigma(x-\bar{x})(y-\bar{y}) = -79 \), \( \Sigma(x-\bar{x})^2 = 90 \), and \( \Sigma(y-\bar{y})^2 = 86 \).
\( r = \frac{-79}{\sqrt{90} \cdot \sqrt{86}} \)
\( r = \frac{-79}{\sqrt{7740}} \)
\( r = \frac{-79}{87.98} \)
\( r = -0.90 \) Thus, the correlation coefficient between the price of wafers and their demand is \( -0.90 \).In simple words: We calculated how price and demand are connected. A value of -0.90 means there is a strong negative link, indicating that as the price of wafers goes up, people buy fewer of them.

🎯 Exam Tip: For problems involving price and demand, expect a negative correlation coefficient, as these variables typically move in opposite directions. Double-check all subtractions and multiplications in the deviation table.

Question 3.From the following information of a sample of ten students of a school regarding their marks in two subjects Accountancy and Statistics, find the coefficient of correlation between the marks of two subjects:

Marks in Accountancy	60	80	50	80	95	40	70	40	35	90
Marks in Statistics	50	75	60	85	90	40	65	30	45	70

Answer:Here, the number of students \( n = 10 \). Let \( x \) be the marks in Accountancy and \( y \) be the marks in Statistics. First, we calculate the average marks for both subjects:
\( \bar{x} = \frac{\Sigma x}{n} = \frac{640}{10} = 64 \text{ marks} \)
\( \bar{y} = \frac{\Sigma y}{n} = \frac{610}{10} = 61 \text{ marks} \) Since both averages are integers, we create a table to calculate the correlation coefficient \( r \):

Marks in Accountancy (x)	Marks in Statistics (y)	\( (x-\bar{x}) \) \( \bar{x} = 64 \)	\( (y-\bar{y}) \) \( \bar{y} = 61 \)	\( (x-\bar{x})(y-\bar{y}) \)	\( (x-\bar{x})^2 \)	\( (y-\bar{y})^2 \)
60	50	-4	-11	44	16	121
80	75	16	14	224	256	196
50	60	-14	-1	14	196	1
80	85	16	24	384	256	576
95	90	31	29	899	961	841
40	40	-24	-21	504	576	441
70	65	6	4	24	36	16
40	30	-24	-31	744	576	961
35	45	-29	-16	464	841	256
90	70	26	9	234	676	81
\( \Sigma x = 640 \)	\( \Sigma y = 610 \)	\( \Sigma (x-\bar{x}) = 0 \)	\( \Sigma (y-\bar{y}) = 0 \)	\( \Sigma (x-\bar{x})(y-\bar{y}) = 3535 \)	\( \Sigma (x-\bar{x})^2 = 4390 \)	\( \Sigma (y-\bar{y})^2 = 3490 \)

The formula for the correlation coefficient is: \[ r = \frac{\Sigma(x-\bar{x})(y-\bar{y})}{\sqrt{\Sigma(x-\bar{x})^{2}} \cdot \sqrt{\Sigma(y-\bar{y})^{2}}} \] Now, we input the calculated values: \( \Sigma(x-\bar{x})(y-\bar{y}) = 3535 \), \( \Sigma(x-\bar{x})^2 = 4390 \), and \( \Sigma(y-\bar{y})^2 = 3490 \).
\( r = \frac{3535}{\sqrt{4390} \cdot \sqrt{3490}} \)
\( r = \frac{3535}{\sqrt{15321100}} \)
\( r = \frac{3535}{3914.22} \)
\( r = 0.90 \) Therefore, the correlation coefficient between marks in Accountancy and Statistics is \( 0.90 \).In simple words: We found how much a student's marks in Accountancy and Statistics are related. A value of 0.90 shows a very strong positive connection, meaning students who score well in one subject are likely to score well in the other.

🎯 Exam Tip: When both means are integers, the deviation method is generally simpler. Be sure to correctly sum up all the products and squares for an accurate result.

Question 4.To check the ability of Mathematics and Logic of the students of a city a private educational institute gives twenty puzzles based on these two subjects to six children selected from various schools. The number of puzzles solved by them is given below:

No. of puzzles solved based on Mathematics	12	8	9	10	8	11
No. of puzzles solved based on Logic	11	10	4	7	13	16

Compute the correlation coefficient between performances of children in two types of puzzles using the given data.
Answer:Here, the number of children \( n = 6 \). Let \( x \) represent the number of puzzles solved based on Mathematics and \( y \) represent the number of puzzles solved based on Logic. First, we calculate the means:
\( \bar{x} = \frac{\Sigma x}{n} = \frac{58}{6} = 9.67 \)
\( \bar{y} = \frac{\Sigma y}{n} = \frac{61}{6} = 10.17 \) Since \( \bar{x} \) and \( \bar{y} \) are fractions and the values of \( x \) and \( y \) are not very large, we can use the direct method (product-moment formula) to calculate \( r \). We prepare the following table:

x	y	xy	\( x^2 \)	\( y^2 \)
12	11	132	144	121
8	10	80	64	100
9	4	36	81	16
10	7	70	100	49
8	13	104	64	169
11	16	176	121	256
\( \Sigma x = 58 \)	\( \Sigma y = 61 \)	\( \Sigma xy = 598 \)	\( \Sigma x^2 = 574 \)	\( \Sigma y^2 = 711 \)

The formula for the correlation coefficient using the direct method is: \[ r = \frac{n \Sigma xy - (\Sigma x)(\Sigma y)}{\sqrt{n \Sigma x^{2} - (\Sigma x)^{2}} \cdot \sqrt{n \Sigma y^{2} - (\Sigma y)^{2}}} \] Substitute the sums into the formula: \( n = 6 \), \( \Sigma xy = 598 \), \( \Sigma x = 58 \), \( \Sigma y = 61 \), \( \Sigma x^2 = 574 \), and \( \Sigma y^2 = 711 \).
\( r = \frac{6 \times 598 - (58)(61)}{\sqrt{6 \times 574 - (58)^{2}} \cdot \sqrt{6 \times 711 - (61)^{2}}} \)
\( r = \frac{3588 - 3538}{\sqrt{3444 - 3364} \cdot \sqrt{4266 - 3721}} \)
\( r = \frac{50}{\sqrt{80} \cdot \sqrt{545}} \)
\( r = \frac{50}{\sqrt{43600}} \)
\( r = \frac{50}{208.81} \)
\( r = 0.24 \) Hence, the correlation coefficient between the children's performance in two types of puzzles is \( 0.24 \).In simple words: We checked how good children are at solving math puzzles compared to logic puzzles. A correlation of 0.24 means there is a weak positive link; good performance in one type of puzzle does not strongly predict good performance in the other.

🎯 Exam Tip: When means are not integers, the direct method (using \( \Sigma xy, \Sigma x^2, \Sigma y^2 \)) is often more practical than the deviation method. Be careful with large numbers in calculations and use a calculator accurately.

Question 5.Find the correlation coefficient between capital (in crore Rs.) invested and the profit (in crore Rs.) from the following data:

Company	A	B	C	D	E	F	G
Capital investment (crore Rs.)	15	22	12	10	17	20	14
Profit (crore Rs.)	9	12	8	6	10	9	10

Answer:For this problem, \( n = 7 \) companies. Let \( x \) be the capital investment and \( y \) be the profit. First, we compute the means for \( x \) and \( y \):
\( \bar{x} = \frac{\Sigma x}{n} = \frac{110}{7} = 15.71 \) crore Rs.
\( \bar{y} = \frac{\Sigma y}{n} = \frac{64}{7} = 9.14 \) crore Rs. Since \( \bar{x} \) and \( \bar{y} \) are not integers and the values are not extremely large, we will use the direct method (product-moment formula) to find the correlation coefficient \( r \). We construct the following table:

Capital investment (crore Rs.) x	Profit (crore Rs.) y	xy	\( x^2 \)	\( y^2 \)
15	9	135	225	81
22	12	264	484	144
12	8	96	144	64
10	6	60	100	36
17	10	170	289	100
20	9	180	400	81
14	10	140	196	100
\( \Sigma x = 110 \)	\( \Sigma y = 64 \)	\( \Sigma xy = 1045 \)	\( \Sigma x^2 = 1838 \)	\( \Sigma y^2 = 606 \)

The formula for the correlation coefficient is: \[ r = \frac{n \Sigma xy - (\Sigma x)(\Sigma y)}{\sqrt{n \Sigma x^{2} - (\Sigma x)^{2}} \cdot \sqrt{n \Sigma y^{2} - (\Sigma y)^{2}}} \] Substitute the totals from the table into the formula: \( n = 7 \), \( \Sigma xy = 1045 \), \( \Sigma x = 110 \), \( \Sigma y = 64 \), \( \Sigma x^2 = 1838 \), and \( \Sigma y^2 = 606 \).
\( r = \frac{7 \times 1045 - (110)(64)}{\sqrt{7 \times 1838 - (110)^{2}} \cdot \sqrt{7 \times 606 - (64)^{2}}} \)
\( r = \frac{7315 - 7040}{\sqrt{12866 - 12100} \cdot \sqrt{4242 - 4096}} \)
\( r = \frac{275}{\sqrt{766} \cdot \sqrt{146}} \)
\( r = \frac{275}{\sqrt{111836}} \)
\( r = \frac{275}{334.42} \)
\( r = 0.82 \) Therefore, the correlation coefficient between capital investment and profit is \( 0.82 \).In simple words: We looked at how much money a company invests and how much profit it makes. A correlation of 0.82 shows a strong positive link, meaning more investment usually leads to more profit.

🎯 Exam Tip: When dealing with fractional means or larger numbers, using the direct formula involving raw sums (\( \Sigma x, \Sigma y, \Sigma xy, \Sigma x^2, \Sigma y^2 \)) can simplify calculations and reduce rounding errors compared to the deviation method.

Question 6.The following information is available for five students selected from a school regarding the average number of study hours per day and the average number of sleeping hours:

No. of study hours	10	5	7	5	3
No. of sleeping hours	6	9	7	8	10

Calculate the correlation coefficient between the study hours and sleeping hours.
Answer:Here, the sample size is \( n = 5 \) students. Let \( x \) be the number of study hours and \( y \) be the number of sleeping hours. First, we find the average study hours and sleeping hours:
\( \bar{x} = \frac{\Sigma x}{n} = \frac{30}{5} = 6 \text{ hours} \)
\( \bar{y} = \frac{\Sigma y}{n} = \frac{40}{5} = 8 \text{ hours} \) Since both \( \bar{x} \) and \( \bar{y} \) are integers, we will use the deviation method. The table for calculating \( r \) is as follows:

No. of study hours (x)	No. of sleeping hours (y)	\( (x-\bar{x}) \) \( \bar{x} = 6 \)	\( (y-\bar{y}) \) \( \bar{y} = 8 \)	\( (x-\bar{x})(y-\bar{y}) \)	\( (x-\bar{x})^2 \)	\( (y-\bar{y})^2 \)
10	6	4	-2	-8	16	4
5	9	-1	1	-1	1	1
7	7	1	-1	-1	1	1
5	8	-1	0	0	1	0
3	10	-3	2	-6	9	4
\( \Sigma x = 30 \)	\( \Sigma y = 40 \)	\( \Sigma (x-\bar{x}) = 0 \)	\( \Sigma (y-\bar{y}) = 0 \)	\( \Sigma (x-\bar{x})(y-\bar{y}) = -16 \)	\( \Sigma (x-\bar{x})^2 = 28 \)	\( \Sigma (y-\bar{y})^2 = 10 \)

The formula for the correlation coefficient is: \[ r = \frac{\Sigma(x-\bar{x})(y-\bar{y})}{\sqrt{\Sigma(x-\bar{x})^{2}} \cdot \sqrt{\Sigma(y-\bar{y})^{2}}} \] Now, substitute the values: \( \Sigma(x-\bar{x})(y-\bar{y}) = -16 \), \( \Sigma(x-\bar{x})^2 = 28 \), and \( \Sigma(y-\bar{y})^2 = 10 \).
\( r = \frac{-16}{\sqrt{28} \cdot \sqrt{10}} \)
\( r = \frac{-16}{\sqrt{280}} \)
\( r = \frac{-16}{16.73} \)
\( r = -0.96 \) Therefore, the correlation coefficient between study hours and sleeping hours is \( -0.96 \).In simple words: We checked if more study hours lead to less sleep. A correlation of -0.96 means there is a very strong negative connection, suggesting that students who study more tend to sleep less, and vice-versa.

🎯 Exam Tip: A negative correlation coefficient is expected between study hours and sleeping hours, as these tend to be inversely related. Ensure your deviation calculations are precise, especially with negative numbers, to avoid sign errors.

Question 7.From the following information of the age (in years) and blood pressure (in mm), find the correlation coefficient between age and blood pressure:

Age (years)	58	55	65	52	48	68	62	56
Systolic blood pressure (mm)	130	150	150	130	140	158	155	140

Answer:Here, the number of observations \( n = 8 \). Let \( x \) denote age in years and \( y \) denote systolic blood pressure in mm. First, we calculate the means:
\( \bar{x} = \frac{\Sigma x}{n} = \frac{464}{8} = 58 \text{ years} \)
\( \bar{y} = \frac{\Sigma y}{n} = \frac{1153}{8} = 144.125 \text{ mm} \) Since \( \bar{x} \) is an integer but \( \bar{y} \) is a fraction, and the values of \( x \) and \( y \) are relatively large, we will use the shortcut method (change of origin). Let \( A = 58 \) for \( x \) and \( B = 150 \) for \( y \). We define new variables: \( u = x - A = x - 58 \) and \( v = y - B = y - 150 \). The table for calculating \( r \) is prepared as follows:

Age (years) x	Blood pressure (mm) y	\( u = (x-58) \)	\( v = (y-150) \)	\( u^2 \)	\( v^2 \)	uv
58	130	0	-20	0	400	0
55	150	-3	0	9	0	0
65	150	7	0	49	0	0
52	130	-6	-20	36	400	120
48	140	-10	-10	100	100	100
68	158	10	8	100	64	80
62	155	4	5	16	25	20
56	140	-2	-10	4	100	20
\( \Sigma x = 464 \)	\( \Sigma y = 1153 \)	\( \Sigma u = 0 \)	\( \Sigma v = -47 \)	\( \Sigma u^2 = 314 \)	\( \Sigma v^2 = 1089 \)	\( \Sigma uv = 340 \)

The formula for the correlation coefficient using the shortcut method is: \[ r = \frac{n \Sigma uv - (\Sigma u)(\Sigma v)}{\sqrt{n \Sigma u^{2} - (\Sigma u)^{2}} \cdot \sqrt{n \Sigma v^{2} - (\Sigma v)^{2}}} \] Substitute the sums into the formula: \( n = 8 \), \( \Sigma uv = 340 \), \( \Sigma u = 0 \), \( \Sigma v = -47 \), \( \Sigma u^2 = 314 \), and \( \Sigma v^2 = 1089 \).
\( r = \frac{8 \times 340 - (0)(-47)}{\sqrt{8 \times 314 - (0)^{2}} \cdot \sqrt{8 \times 1089 - (-47)^{2}}} \)
\( r = \frac{2720}{\sqrt{2512 - 0} \cdot \sqrt{8712 - 2209}} \)
\( r = \frac{2720}{\sqrt{2512} \cdot \sqrt{6503}} \)
\( r = \frac{2720}{\sqrt{16335536}} \)
\( r = \frac{2720}{4041.72} \)
\( r = 0.67 \) Hence, the correlation coefficient between age and blood pressure is \( 0.67 \).In simple words: We checked how age and blood pressure are linked. A correlation of 0.67 indicates a moderate positive relationship, suggesting that as people get older, their blood pressure tends to increase.

🎯 Exam Tip: When raw data values are large, using a change of origin (shortcut method) simplifies calculations. Choose convenient assumed means (A and B) close to the actual means or data values to keep \( u \) and \( v \) small.

Question 8.An Engineer Association wants to know the relation between the production (thousand units) and the unit production cost of different factories. The information collected from six factories regarding their production and unit production cost is given below:

Production (thousand units)	15	20	35	24	18	31
Cost per unit of production (Rs.)	95	90	75	80	87	70

Find the correlation coefficient between production and cost per unit of production.
Answer:Here, the number of factories \( n = 6 \). Let \( x \) be the production (in thousand units) and \( y \) be the cost per unit of production (in Rs.). First, we find the means:
\( \bar{x} = \frac{\Sigma x}{n} = \frac{143}{6} = 23.83 \) thousand units
\( \bar{y} = \frac{\Sigma y}{n} = \frac{497}{6} = 82.83 \) Rs. Since \( \bar{x} \) and \( \bar{y} \) are not integers and the data values are relatively large, we will use the shortcut method (change of origin). Let \( A = 23 \) for \( x \) and \( B = 82 \) for \( y \). We define new variables: \( u = x - A = x - 23 \) and \( v = y - B = y - 82 \). The table for calculating \( r \) is prepared as follows:

Production (thousand units) x	Cost per unit of production (Rs.) y	\( u = (x-23) \)	\( v = (y-82) \)	\( u^2 \)	\( v^2 \)	uv
15	95	-8	13	64	169	-104
20	90	-3	8	9	64	-24
35	75	12	-7	144	49	-84
24	80	1	-2	1	4	-2
18	87	-5	5	25	25	-25
31	70	8	-12	64	144	-96
\( \Sigma x = 143 \)	\( \Sigma y = 497 \)	\( \Sigma u = 5 \)	\( \Sigma v = 5 \)	\( \Sigma u^2 = 307 \)	\( \Sigma v^2 = 455 \)	\( \Sigma uv = -335 \)

The formula for the correlation coefficient using the shortcut method is: \[ r = \frac{n \Sigma uv - (\Sigma u)(\Sigma v)}{\sqrt{n \Sigma u^{2} - (\Sigma u)^{2}} \cdot \sqrt{n \Sigma v^{2} - (\Sigma v)^{2}}} \] Substitute the sums into the formula: \( n = 6 \), \( \Sigma uv = -335 \), \( \Sigma u = 5 \), \( \Sigma v = 5 \), \( \Sigma u^2 = 307 \), and \( \Sigma v^2 = 455 \).
\( r = \frac{6 \times (-335) - (5)(5)}{\sqrt{6 \times 307 - (5)^{2}} \cdot \sqrt{6 \times 455 - (5)^{2}}} \)
\( r = \frac{-2010 - 25}{\sqrt{1842 - 25} \cdot \sqrt{2730 - 25}} \)
\( r = \frac{-2035}{\sqrt{1817} \cdot \sqrt{2705}} \)
\( r = \frac{-2035}{\sqrt{4914985}} \)
\( r = \frac{-2035}{2216.98} \)
\( r = -0.92 \) Hence, the correlation coefficient between production and cost per unit of production is \( -0.92 \).In simple words: We examined the link between how much a factory produces and how much each unit costs. A correlation of -0.92 shows a very strong negative link, meaning that when factories produce more units, the cost per unit usually goes down.

🎯 Exam Tip: When using the shortcut method, choose arbitrary origins (A and B) that make the transformed values (u and v) small and easy to work with. Pay close attention to negative signs, especially in the \( \Sigma uv \) term and when squaring negative numbers.

Question 9.Find the correlation coefficient between the yearly per capita income (in Rs.) and the price index of the people of six different cities from the following data:

City	A	B	C	D	E	F
Yearly per capita income (Rs.)	32,000	29,000	40,000	36,000	30,000	39,000
Price index	120	100	250	180	110	220

Answer:Here, the number of cities \( n = 6 \). Let \( x \) be the yearly per capita income and \( y \) be the price index. To simplify calculations, we will use a change of origin and scale. The values of \( x \) are multiples of 1000, and \( y \) are multiples of 10. Let \( A = 30000 \) and \( C_x = 1000 \) for \( x \). Let \( B = 180 \) and \( C_y = 10 \) for \( y \). We define new variables:
\( u = \frac{x - A}{C_x} = \frac{x - 30000}{1000} \)
\( v = \frac{y - B}{C_y} = \frac{y - 180}{10} \) The table for calculating \( r \) is prepared as follows:

Yearly per capita income (Rs.) x	Price index y	\( u = \frac{x-30000}{1000} \)	\( v = \frac{y-180}{10} \)	\( u^2 \)	\( v^2 \)	uv
32000	120	2	-6	4	36	-12
29000	100	-1	-8	1	64	8
40000	250	10	7	100	49	70
36000	180	6	0	36	0	0
30000	110	0	-7	0	49	0
39000	220	9	4	81	16	36
Total	\( n = 6 \)	\( \Sigma u = 26 \)	\( \Sigma v = -10 \)	\( \Sigma u^2 = 222 \)	\( \Sigma v^2 = 214 \)	\( \Sigma uv = 102 \)

The formula for the correlation coefficient using the change of origin and scale method is: \[ r = \frac{n \Sigma uv - (\Sigma u)(\Sigma v)}{\sqrt{n \Sigma u^{2} - (\Sigma u)^{2}} \cdot \sqrt{n \Sigma v^{2} - (\Sigma v)^{2}}} \] Substitute the sums into the formula: \( n = 6 \), \( \Sigma uv = 102 \), \( \Sigma u = 26 \), \( \Sigma v = -10 \), \( \Sigma u^2 = 222 \), and \( \Sigma v^2 = 214 \).
\( r = \frac{6 \times 102 - (26)(-10)}{\sqrt{6 \times 222 - (26)^{2}} \cdot \sqrt{6 \times 214 - (-10)^{2}}} \)
\( r = \frac{612 + 260}{\sqrt{1332 - 676} \cdot \sqrt{1284 - 100}} \)
\( r = \frac{872}{\sqrt{656} \cdot \sqrt{1184}} \)
\( r = \frac{872}{\sqrt{776704}} \)
\( r = \frac{872}{881.31} \)
\( r = 0.99 \) Hence, the correlation coefficient between yearly per capita income and the price index is \( 0.99 \).In simple words: We studied the link between how much people earn in a year and how high prices are in their city. A correlation of 0.99 shows an almost perfect positive link, meaning cities with higher average incomes also tend to have higher price indexes.

🎯 Exam Tip: When data values are very large or have consistent multiples (like 1000s or 10s), applying both a change of origin (subtracting a constant) and a change of scale (dividing by a constant) can significantly simplify calculations without affecting the correlation coefficient.

Question 10.The following data are given to study the relation between the number of persons in a family who drive vehicle and usage of petrol (in litre) per week:

No. of members per family who drive vehicle	3	5	2	4	3	6	1
Weekly usage of petrol (litre)	11.5	21	14.5	15.5	7	22.5	10

Find the correlation coefficient between the number of members in a family and usage of petrol.
Answer:Here, the number of families \( n = 7 \). Let \( x \) be the number of members in a family who drive a vehicle, and \( y \) be the weekly usage of petrol in litres. To simplify calculations, we apply a change of origin and scale. Let \( A = 4 \) and \( C_x = 1 \) for \( x \). Let \( B = 15.5 \) and \( C_y = 0.5 \) for \( y \). We define new variables:
\( u = \frac{x - A}{C_x} = \frac{x - 4}{1} \)
\( v = \frac{y - B}{C_y} = \frac{y - 15.5}{0.5} \) The table for calculating \( r \) is prepared as follows:

No. of members in a family x	Weekly usage of petrol y	\( u = \frac{x-4}{1} \)	\( v = \frac{y-15.5}{0.5} \)	\( u^2 \)	\( v^2 \)	uv
3	11.5	-1	-8	1	64	8
5	21	1	11	1	121	11
2	14.5	-2	-2	4	4	4
4	15.5	0	0	0	0	0
3	7	-1	-17	1	289	17
6	22.5	2	14	4	196	28
1	10	-3	-11	9	121	33
Total \( n = 7 \)		\( \Sigma u = -4 \)	\( \Sigma v = -13 \)	\( \Sigma u^2 = 20 \)	\( \Sigma v^2 = 795 \)	\( \Sigma uv = 101 \)

The formula for the correlation coefficient using the change of origin and scale method is: \[ r = \frac{n \Sigma uv - (\Sigma u)(\Sigma v)}{\sqrt{n \Sigma u^{2} - (\Sigma u)^{2}} \cdot \sqrt{n \Sigma v^{2} - (\Sigma v)^{2}}} \] Substitute the sums into the formula: \( n = 7 \), \( \Sigma uv = 101 \), \( \Sigma u = -4 \), \( \Sigma v = -13 \), \( \Sigma u^2 = 20 \), and \( \Sigma v^2 = 795 \).
\( r = \frac{7 \times 101 - (-4)(-13)}{\sqrt{7 \times 20 - (-4)^{2}} \cdot \sqrt{7 \times 795 - (-13)^{2}}} \)
\( r = \frac{707 - 52}{\sqrt{140 - 16} \cdot \sqrt{5565 - 169}} \)
\( r = \frac{655}{\sqrt{124} \cdot \sqrt{5396}} \)
\( r = \frac{655}{\sqrt{669104}} \)
\( r = \frac{655}{817.99} \)
\( r = 0.80 \) Therefore, the correlation coefficient between the number of members in a family and weekly petrol usage is \( 0.80 \).In simple words: We checked how many people in a family drive and how much petrol they use each week. A correlation of 0.80 shows a strong positive link, meaning that more drivers in a family generally lead to higher weekly petrol consumption.

🎯 Exam Tip: When dealing with decimal values in the raw data, using a change of scale (dividing by a suitable constant, like 0.5 for y in this case) can convert decimals to integers, making calculations much simpler and less prone to errors.

Question 11.The following information is obtained to study the effect of the use of fertilizer on yield of corn in a rural area:

Use of fertilizer (quintal)	1.5	2.1	0.9	1.8	1.1	1.2
Yield of corn per hectare (quintal)	60	95	50	75	45	75

Find the correlation coefficient between use of fertilizer and yield of corn.
Answer:Here, the number of observations \( n = 6 \). Let \( x \) be the use of fertilizer (quintal) and \( y \) be the yield of corn per hectare (quintal). To make calculations easy, we will use a change of origin and scale. Let \( A = 1.5 \) and \( C_x = 0.1 \) for \( x \). Let \( B = 75 \) and \( C_y = 5 \) for \( y \). We define new variables:
\( u = \frac{x - A}{C_x} = \frac{x - 1.5}{0.1} \)
\( v = \frac{y - B}{C_y} = \frac{y - 75}{5} \) The table for calculating \( r \) is prepared as follows:

Use of fertilizer (quintal) x	Yield of corn (quintal) y	\( u = \frac{x-1.5}{0.1} \)	\( v = \frac{y-75}{5} \)	\( u^2 \)	\( v^2 \)	uv
1.5	60	0	-3	0	9	0
2.1	95	6	4	36	16	24
0.9	50	-6	-5	36	25	30
1.8	75	3	0	9	0	0
1.1	45	-4	-6	16	36	24
1.2	75	-3	0	9	0	0
Total	\( n = 6 \)	\( \Sigma u = -4 \)	\( \Sigma v = -10 \)	\( \Sigma u^2 = 106 \)	\( \Sigma v^2 = 86 \)	\( \Sigma uv = 78 \)

The formula for the correlation coefficient using the change of origin and scale method is: \[ r = \frac{n \Sigma uv - (\Sigma u)(\Sigma v)}{\sqrt{n \Sigma u^{2} - (\Sigma u)^{2}} \cdot \sqrt{n \Sigma v^{2} - (\Sigma v)^{2}}} \] Substitute the sums into the formula: \( n = 6 \), \( \Sigma uv = 78 \), \( \Sigma u = -4 \), \( \Sigma v = -10 \), \( \Sigma u^2 = 106 \), and \( \Sigma v^2 = 86 \).
\( r = \frac{6 \times 78 - (-4)(-10)}{\sqrt{6 \times 106 - (-4)^{2}} \cdot \sqrt{6 \times 86 - (-10)^{2}}} \)
\( r = \frac{468 - 40}{\sqrt{636 - 16} \cdot \sqrt{516 - 100}} \)
\( r = \frac{428}{\sqrt{620} \cdot \sqrt{416}} \)
\( r = \frac{428}{\sqrt{257920}} \)
\( r = \frac{428}{507.86} \)
\( r = 0.84 \) Hence, the correlation coefficient between the use of fertilizer and the yield of corn is \( 0.84 \).In simple words: We looked at the connection between how much fertilizer is used and how much corn is harvested. A correlation of 0.84 shows a strong positive link, meaning that using more fertilizer generally leads to a higher corn yield.

🎯 Exam Tip: When dealing with decimal values in both variables, select appropriate constants for change of origin and scale (A, \( C_x \), B, \( C_y \)) to convert them into smaller, whole numbers, simplifying calculations while maintaining accuracy.

Question 12.Find the correlation coefficient from the following information of rainfall (X) (in cm) and yield (Y) (tons per hectare) for the last 10 years of a district: n = 10, Cov (x, y) = 30, S.D. of X = 5 and variance of Y = 144
Answer:Here, we are given: \( n = 10 \) Covariance of \( x \) and \( y \), \( \text{Cov}(x, y) = 30 \) Standard Deviation of \( x \), \( S_x = 5 \) Variance of \( y \), \( S_y^2 = 144 \) From the variance of \( y \), we can find its standard deviation: \( S_y = \sqrt{S_y^2} = \sqrt{144} = 12 \) The formula for the correlation coefficient \( r \) using covariance and standard deviations is: \[ r = \frac{\text{Cov}(x, y)}{S_x \cdot S_y} \] Substitute the given values into the formula:
\( r = \frac{30}{5 \times 12} \)
\( r = \frac{30}{60} \)
\( r = 0.5 \) Hence, the correlation coefficient between rainfall and crop yield is \( 0.5 \).In simple words: We calculated the relationship between rainfall and how much crop grows. A value of 0.5 means there is a moderate positive link; more rainfall tends to lead to more crop, but not always very strongly.

🎯 Exam Tip: Remember the relationship between variance and standard deviation (\( \text{S.D.} = \sqrt{\text{Variance}} \)). This direct formula for correlation coefficient is useful when covariance and standard deviations are provided, saving time on lengthy table calculations.

Question 13.The following information is obtained regarding the height (X) and weight (Y) from a sample of ten students of a school: \( \bar{x} = 160 \), \( \bar{y} = 55 \), \( \Sigma xy = 90000 \), \( S_x = 25 \), \( S_y = 10 \) Find the correlation coefficient between the height and weight from it.
Answer:Here, we are given information for \( n = 10 \) students: Mean height \( \bar{x} = 160 \) Mean weight \( \bar{y} = 55 \) Sum of products of \( x \) and \( y \), \( \Sigma xy = 90000 \) Standard deviation of height \( S_x = 25 \) Standard deviation of weight \( S_y = 10 \) The formula for the correlation coefficient \( r \) when means, sum of products, and standard deviations are given is: \[ r = \frac{\Sigma xy - n \bar{x} \bar{y}}{n \cdot S_x \cdot S_y} \] Substitute the given values into the formula:
\( r = \frac{90000 - 10 \times 160 \times 55}{10 \times 25 \times 10} \)
\( r = \frac{90000 - 88000}{2500} \)
\( r = \frac{2000}{2500} \)
\( r = 0.8 \) Hence, the correlation coefficient between height and weight is \( 0.8 \).In simple words: We calculated how much a student's height and weight are related. A value of 0.8 shows a strong positive link, meaning taller students generally tend to be heavier.

🎯 Exam Tip: Understand how to calculate the correlation coefficient using different formulas based on the given information. This formula is efficient when means and standard deviations are already known, avoiding the need for deviation tables.

Question 14.Determine the value of correlation coefficient from the following data:
(1) \( \Sigma (x-\bar{x})^2 = 72 \), \( \Sigma (y-\bar{y})^2 = 32 \), \( \Sigma (x-\bar{x})(y-\bar{y}) = 45 \)
Answer:(1) Here, we are given: Sum of squares of deviations for \( x \), \( \Sigma (x-\bar{x})^2 = 72 \) Sum of squares of deviations for \( y \), \( \Sigma (y-\bar{y})^2 = 32 \) Sum of products of deviations, \( \Sigma (x-\bar{x})(y-\bar{y}) = 45 \) The formula for the correlation coefficient \( r \) using deviations from the mean is: \[ r = \frac{\Sigma(x-\bar{x})(y-\bar{y})}{\sqrt{\Sigma(x-\bar{x})^{2}} \cdot \sqrt{\Sigma(y-\bar{y})^{2}}} \] Substitute the given values into the formula:
\( r = \frac{45}{\sqrt{72} \times \sqrt{32}} \)
\( r = \frac{45}{\sqrt{2304}} \)
\( r = \frac{45}{48} \)
\( r = 0.94 \) The correlation coefficient obtained is \( 0.94 \).In simple words: We found the connection between two sets of data by using how much each point differs from its average. A value of 0.94 shows a very strong positive link between them.

🎯 Exam Tip: This question directly provides the necessary sums of squares and products of deviations, making it a straightforward application of the formula. Ensure correct square root calculations and division for accuracy.

Question 14 (Continuation).
(2) \( n = 6 \), \( \Sigma x = 16 \), \( \Sigma y = 51 \), \( \Sigma xy = 154 \), \( \Sigma x^2 = 52 \), \( \Sigma y^2 = 471 \)
Answer:(2) Here, we are given: Number of observations \( n = 6 \) Sum of \( x \), \( \Sigma x = 16 \) Sum of \( y \), \( \Sigma y = 51 \) Sum of products of \( x \) and \( y \), \( \Sigma xy = 154 \) Sum of squares of \( x \), \( \Sigma x^2 = 52 \) Sum of squares of \( y \), \( \Sigma y^2 = 471 \) The formula for the correlation coefficient \( r \) using raw data sums is: \[ r = \frac{n \Sigma xy - (\Sigma x)(\Sigma y)}{\sqrt{n \Sigma x^{2} - (\Sigma x)^{2}} \cdot \sqrt{n \Sigma y^{2} - (\Sigma y)^{2}}} \] Substitute the given values into the formula:
\( r = \frac{6 \times 154 - (16)(51)}{\sqrt{6 \times 52 - (16)^{2}} \cdot \sqrt{6 \times 471 - (51)^{2}}} \)
\( r = \frac{924 - 816}{\sqrt{312 - 256} \cdot \sqrt{2826 - 2601}} \)
\( r = \frac{108}{\sqrt{56} \times \sqrt{225}} \)
\( r = \frac{108}{\sqrt{12600}} \)
\( r = \frac{108}{112.25} \)
\( r = 0.96 \) Hence, the correlation coefficient obtained is \( 0.96 \).In simple words: We used the sums of the data points to find their relationship. A value of 0.96 shows a very strong positive connection between the two sets of data.

🎯 Exam Tip: When given raw sums, this formula is highly efficient. Double-check all multiplications and subtractions, especially the squared terms in the denominator, to ensure an accurate final correlation coefficient.

Question 15.Find the value of r from the following data.
(1)

Particulars	x	y
Average	60	95
The sum of squares of deviations taken from their mean	920	1050
The sum of product of deviations taken from their mean	-545

Answer:Here, we are given: Mean of \( x \), \( \bar{x} = 60 \) Mean of \( y \), \( \bar{y} = 95 \) Sum of squares of deviations for \( x \), \( \Sigma (x-\bar{x})^2 = 920 \) Sum of squares of deviations for \( y \), \( \Sigma (y-\bar{y})^2 = 1050 \) Sum of products of deviations, \( \Sigma (x-\bar{x})(y-\bar{y}) = -545 \) The formula for the correlation coefficient \( r \) using deviations from the mean is: \[ r = \frac{\Sigma(x-\bar{x})(y-\bar{y})}{\sqrt{\Sigma(x-\bar{x})^{2}} \cdot \sqrt{\Sigma(y-\bar{y})^{2}}} \] Substitute the given values into the formula:
\( r = \frac{-545}{\sqrt{920} \cdot \sqrt{1050}} \)
\( r = \frac{-545}{\sqrt{966000}} \)
\( r = \frac{-545}{982.85} \)
\( r = -0.55 \) Hence, the correlation coefficient obtained is \( -0.55 \).In simple words: We used the average values and how far data points are from these averages to find their relationship. A value of -0.55 shows a moderate negative link between the two data sets.

🎯 Exam Tip: This question directly provides all the necessary components for the deviation method. Pay attention to the negative sign in the sum of products of deviations, as it directly impacts the sign of the correlation coefficient.

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