Gujarat Board Statistics Class 12 GSEB Solutions Part 1 Chapter 3 Linear Regression Ex 3.2 Textbook Exercise Questions and Answers.

## Gujarat Board Textbook Solutions Class 12 Statistics Part 1 Chapter 3 Linear Regression Ex 3.2

Question 1.

The following information is obtained from a study to know the effect of use of fertilizer on the yield of cotton:

Obtain the regression line of Y on X and estimate the yield of cotton per hectare if 300 kg fertilizer is used.

Answer:

Here, n = 7; X = Consumption of fertilizer and Y = Yield of cotton.

Now, xÌ„ = \(\frac{\Sigma x}{n}\) = \(\frac{177}{7}\) = 25.29; yÌ„ = \(\frac{\Sigma y}{n}\) = \(\frac{830}{7}\) = 118.57

We obtain the regression line of yield of cotton (Y) on the consumption of fertilizer (X) yÌ‚ = a + bx.

xÌ„ and yÌ„ are not integers. So we calculate ‘a’ and ‘b’ by shortcut method taking new variables

u = x – A, A = 25 and v = y – B, B = 120.

The table for calculation is prepared as follows:

b = \(\frac{n \Sigma u v-(\Sigma u)(\Sigma v)}{n \Sigma u^{2}-(\Sigma u)^{2}}\)

Putting n = 7; Î£uv = 399; Î£u = 2; Î£v = – 10 and Î£u^{2} =160 in the formula,

b = \(\frac{7(399)-(2)(-10)}{7(160)-(2)^{2}}\)

= \(\frac{2793+20}{1120-4}\)

= \(\frac{2813}{1116}\)

= 2.52

a = yÌ„ – bxÌ„

Putting yÌ„ = 118.57; xÌ„ = 25.29 and b = 2.52, we get

a = 118.57 – 2.52 (25.29)

= 118.57 – 63.73

= 54.84

Regression line of Y on X:

Putting a = 54.84 and b = 2.52 in yÌ‚ = a + bx, we get

yÌ‚ = 54.84 + 2.52

Estimate of Y when X = 30 :

Putting x = 30 in yÌ‚ = 54.84 + 2.52 x, we get

yÌ‚ = 54.84 + 2.52 (30)

= 54.84 + 75.6

= 130.44

Hence, the estimate of Y obtained is yÌ‚ 130.44 Quintals.

Question 2.

To know the relationship between the heights of father and sons, obtain the regression line of height of son on the height of father from the following information of eight pairs of fathers and adult sons.

Estimate the height of a son whose fatherâ€™s height is 170 cm.

Answer:

Here, n = 8; X = Height of father and Y = Height of son

Now, xÌ„ = \(\frac{\Sigma x}{n}\) = \(\frac{1346}{8}\) = 168.25 cm; yÌ„ = \(\frac{\Sigma y}{n}\) = \(\frac{1338}{8}\) = 167.25 cm

We obtain the regression line of height of son (Y) on height of father (Y) yÌ‚ = a + bx.

xÌ„ and yÌ„ are not integers. So we calculate the values of a and b by shortcut method taking new variables

u = x – A, A = 168 and v = y – B, B = 167.

The table for calculation is prepared as follows:

b = \(\frac{n \Sigma u v-(\Sigma u)(\Sigma v)}{n \Sigma u^{2}-(\Sigma u)^{2}}\)

Putting n = 8; Î£uv = 34; Î£u = 2; Î£v = 2 and Î£u^{2} = 50 in the formula,

b = \(\frac{8(34)-(2)(2)}{8(50)-(2)^{2}}\)

= \(\frac{272-4}{400-4}\)

= \(\frac{268}{396}\)

= 0.68

a = yÌ„ – bxÌ„

Putting yÌ„ = 167.25; xÌ„ = 168.25 and b = 0.68, we get

a = 167.25 – 0.68 (168.25)

= 167.25 – 114.41

= 52.84

Regression line of Y on X:

Putting a = 52.84 and b = 0.68 in yÌ‚ = a + bx, we get

yÌ‚ = 52.84 + 0.68x

Estimate of Y when X = 170 cm:

Putting x = 170 in yÌ‚ = 52.84 + 0.68 x, we get

yÌ‚ = 52.84 + 0.68 (170)

= 52.84 + 115.6

= 168.44

Hence, the estimate of son height obtained is yÌ‚ = 168.44 cm.

Question 3.

From the following information of altitude and the amount of effective oxygen in air at the place, obtain the regression line of amount of effective oxygen (Y) on the altitude (X): (305 metre = 1000 feet)

If the altitude of a place is 7 units (1 unit = 305 metre), estimate the percentage of effective oxygen in air at that place.

Answer:

Here, n = 7; X = Altitude and Y = Effective oxygen

Now, xÌ„ = \(\frac{\Sigma x}{n}\) = \(\frac{21}{7}\) = 3; yÌ„ = \(\frac{\Sigma y}{n}\) = \(\frac{130.1}{7}\) = 18.59

We obtain the regression line of effective oxygen (Y) on altitude (X) yÌ‚ = a + bx.

yÌ‚ is not integer and its values are multiple of 0.1. We calculate the values of a and b by short-cut method taking new variables

u = \(\frac{x-\mathrm{A}}{\mathrm{C}_{x}}\); A = 3, C_{x} = 1 and v = \(\frac{y-B}{C_{y}}\); B = 17.9, C_{y} = 0.1.

The table for calculation is prepared as follows:

b = b_{uv} = \(\frac{n \Sigma u v-(\Sigma u)(\Sigma v)}{n \Sigma u^{2}-(\Sigma u)^{2}} \times \frac{\mathrm{C}_{y}}{\mathrm{C}_{x}}\)

Putting n = 7; Î£uv = – 200; Î£u = 0; Î£u = 48; Î£u^{2} = 28; C_{y} = 0.1 and C_{x} = 1 in the formula,

b = \(\frac{7(-200)-(0)(48)}{7(28)-(0)^{2}} \times \frac{0.1}{1}\)

= \(\frac{-1400-0}{196-0} \times \frac{0.1}{1}\)

= \(\frac{-1400 \times 0.1}{196}\)

= \(\frac{-140}{196}\)

= – 0.71

a = yÌ„ – bxÌ„

Putting yÌ„ = 18.59; xÌ„ = 3 and b = – 0.71, we get

a = 18.59 – (- 0.71) (3)

= 18.59 + 2.13

= 20.72

The regression line og Y on X:

Putting a = 20.72 and b = – 0.71, we get yÌ‚ = 20.72 – 0.71x

The estimate of percentage of oxygen (Y) when X = 7:

Putting x = 7 in y = 20.72 – 0.7 lx, we get

yÌ‚ = 20.72 – 0.71 (7)

= 20.72 – 4.97

= 15.75

Hence, the estimate of percentage of oxygen obtained is yÌ‚ = 15.75%.

Question 4.

The following information is obtained to study the relation between the carpet area in a house and its monthly rent in a city:

Obtain the regression line of Y on X. Estimate the monthly rent of a house having carpet area of 110 square metre.

Answer:

Here, n = 7; X = Carpet area and Y = Monthly rent.

Now, xÌ„ = \(\frac{\Sigma x}{n}\) = \(\frac{630}{7}\) = 90; yÌ„ = \(\frac{\Sigma y}{n}\) = \(\frac{182000}{7}\) = 26000

To obtain the regression line yÌ‚ = a + bx of monthly rent (Y) on the carpet area (X), we calculate

the values of a and b by shortcut method, taking new variables u = \(\frac{x-\mathrm{A}}{\mathrm{C}_{x}}\), A = 80, C_{x} = 5 and v = \(\frac{y-\mathrm{B}}{\mathrm{C}_{y}}\), B = 25000, C_{y} = 1000

The table for calculation is prepared as follows:

b = b_{vu} = \(\frac{n \Sigma u v-(\Sigma u)(\Sigma v)}{n \Sigma u^{2}-(\Sigma u)^{2}} \times \frac{\mathrm{C}_{y}}{\mathrm{C}_{x}}\)

Putting n = 7; Î£uv = 404; Î£u = 14; Î£v = 7; Î£u^{2} = 266; C_{x} = 5 and C_{y} = 1000 in the formula,

b = \(\frac{7(404)-(14)(7)}{7(266)-(14)^{2}} \times \frac{1000}{5}\)

= \(\frac{2828-98}{1862-196}\) Ã— 200

= \(\frac{2730 \times 200}{1666}\)

= \(\frac{546000}{1666}\)

= 327.73

a = yÌ„ – bxÌ„

Putting yÌ„ = 26000; b = 327.73 and xÌ„ = 90, we get

a = 26000 – 327.73 (90)

= 26000 – 29495.7

= – 3495.7

Regression line of Y on X:

Putting a = – 3495.7 and b = 327.73 in yÌ‚ = a + bx, we get

yÌ‚ = – 3495.7 + 327.73x

Estimate of monthly rent (Y) when X=110:

Putting x = 110 in yÌ‚ = – 3495.7 + 327.73x, we get

yÌ‚ = – 3495.7 + 327.73 (110)

= – 3495.7 + 36050.3

= 32554.6

Hence, the estimate of monthly rate obtained is yÌ‚ = â‚¹ 32554.6

Question 5.

The following sample data is obtained to study the relation between the number of customers visiting a mall per day and the sales (ten thousand â‚¹) :

Obtain the regression line of Y on X. Estimate the sales of a mall if 80 customers have visited the mall on a particular day.

Answer:

Here, n = 6; X = No. of customers and Y = sales.

Now, xÌ„ = \(\frac{\Sigma x}{n}\) = \(\frac{560}{6}\) = 93.33; yÌ„ = \(\frac{\Sigma y}{n}\) = \(\frac{14.4}{6}\) = 2.4

To obtain tire regression line yÌ‚ = a + bx of sales (Y) on the number of customers (X), we calculate the values of a and b by shortcut method, taking new variables u = \(\frac{x-\mathrm{A}}{\mathrm{C}_{x}}\), A = 100, C_{x} = 10 and v = \(\frac{y-\mathrm{B}}{\mathrm{C}_{y}}\), B = 2.5, C_{y} = 0.1.

The table for calculation is prepared as follows:

b = \(\frac{n \Sigma u v-(\Sigma u)(\Sigma v)}{n \Sigma u^{2}-(\Sigma u)^{2}} \times \frac{\mathrm{C}_{y}}{\mathrm{C}_{x}}\)

Putting n = 6; Î£uv = 148; Î£u = – 4; Î£v = – 6;

Î£u^{2} = 72; C_{x} = 10 and C_{y} = 0.1 in the formula.

b = \(\frac{6(148)-(-4)(-6)}{6(72)-(-4)^{2}} \times \frac{0.1}{10}\)

= \(\frac{888-24}{432-16} \times \frac{0.1}{10}\)

= \(\frac{864 \times 0.1}{416 \times 10}\)

= \(\frac{86.4}{4160}\)

= 0.02

a = yÌ„ – bxÌ„

Putting yÌ„ = 2.4; xÌ„ = 93.33 and b = 0.02, we get

a = 2.4 – 0.02 (93.33)

= 2.4- 1.87

= 0.53

Regression line of Y on X:

Putting a = 0.53 and b = 0.02 in yÌ‚ = a + bx,

we get yÌ‚ = 0.53 + 0.02x

Estimate of sales Y when X = 80 :

Putting x = 80 in yÌ‚ = 0.53 + 0.02x, we get

yÌ‚ = 0.53 + 0.02 (80)

= 0.53 + 1.6 = 2.13

Hence, the estimate of sales obtained is yÌ‚ = â‚¹ 2.13 (ten thousand).

Question 6.

The following information is given for ten firms running business of clothes in a city regarding their average annual profit (in lakh â‚¹) and average annual administrative cost (in lakh â‚¹) :

Answer:

Here, xÌ„ = 60; yÌ„ = 25; S_{x} = 6; S_{y} = 3 and Cov (x, y) = 10.4

Regression line of Y on X:

yÌ‚ = a + bx

b = \(\frac{{Cov}(x, y)}{{S}_{x}^{2}}\)

= \(\frac{10.4}{(6)^{2}}\)

= \(\frac{10.4}{36}\)

= 0.29

a = yÌ„ – bxÌ„

= 25 – 0.29(60)

= 25 – 17.4

= 7.6

Putting a = 7.6 and b = 0.29 in yÌ‚ = a + bx, we get yÌ‚ = 7.6 + 0.29 x

Question 7.

The following information is obtained to study the relationship between average rainfall (in cm) and the yield of maize (in quintal per hectare) in different talukaa of Gujarat:

Estimate the yield of maize when the rainfall is 60 cm.

Answer:

Here, xÌ„ = 82; yÌ„ = 180; r = 0.82; S_{x}^{2} = 64

âˆ´ S_{x} = 8 and S_{y}^{2} = 225

âˆ´ S_{y} = 15 are given.

To estimate the yield of maize (Y) when the rainfall (X) is 60 cm we obtain the regression line of Y on X, yÌ‚ = a + bx.

Now, b = r âˆ™ \(\frac{\mathrm{S}_{y}}{\mathrm{~S}_{x}}\);

Putting r = 0.82; S_{y} = 15 and S_{x} = 8, we get

b = 0.82 âˆ™ \(\frac{15}{8}\)

= \(\frac{12.3}{8}\)

= 1.54

a = yÌ„ – bxÌ„

Putting yÌ„ = 180; b = 1.54 and xÌ„ = 82, we get

a = 180 – 1.54 (82)

= 180 – 126.28 = 53.72

Putting a = 53.72 and b = 1.54 in yÌ‚ = a + bx, the regression line of yield of maize (Y) on the rainfall (X) obtained is yÌ‚ = 53.72 + 1.54x

Estimate of yield of maize Y when rainfall x = 60 cm:

Putting x = 60 in yÌ‚ = 53.72 + 1.54x, we get

yÌ‚ = 53.72 + 1.54 (60)

= 53.72 + 92.40

= 146.12

Hence, the estimate of yield of maize obtained is yÌ‚ = 146.12 quintal per hectare.

Question 8.

The following results are obtained to study the relation between the price of battery (cell) of wrist watch in rupees (X) and its supply in hundred units (Y) :

n = 10, Î£x = 130, Î£y = 220, Î£x^{2} = 2288 and Î£xy = 3467

Obtain the regression line of Y on X and estimate the supply when price is â‚¹ 16.

Answer:

Here, n = 10; Î£x = 130; Î£y = 220; Î£x^{2} = 2288 and Î£xy = 3467

Now, xÌ„ = \(\frac{\Sigma x}{n}\) = \(\frac{130}{10}\) = 13; yÌ„ = \(\frac{\Sigma y}{n}\) = \(\frac{220}{10}\) = 22

We obtain the regression line of Y on X.

yÌ‚ = a + bx

b = \(\frac{n \Sigma x y-(\Sigma x)(\Sigma y)}{n \Sigma x^{2}-(\Sigma x)^{2}}\)

= \(\frac{10(3467)-(130)(220)}{10(2288)-(130)^{2}}\)

= \(\frac{34670-28600}{22880-16900}\)

= \(\frac{6070}{5980}\)

= 1.02

a = yÌ„ – bxÌ„

= 22 – 1.02 (13)

= 22 – 13.26

= 8.74

Regression line of Y on X:

Putting a = 8.74 and b = 1.02 in yÌ‚ = a + bx,

we get yÌ‚ = 8.74 + 1.02x

Estimate of supply Y when price X = â‚¹ 16 :

Putting x = 16 in yÌ‚ = 8.74 + 1.02x, we get

yÌ‚ = 8.74 + 1.02 (16)

= 8.74 + 16.32

= 25.06

Hence, the estimate of supply obtained is yÌ‚ = 25.06 hundred units.

Question 9.

The information regarding maximum temperature (X) and sale of ice cream (Y) of six different days in summer for a city is given below:

Maximum temperature = X (in celsius).

Sale of ice cream = Y (in lakh â‚¹)

xÌ„ = 40, yÌ„ = 1.2, Î£xy = 306, S_{x}^{2} = 20

Obtain the regression line of sale of ice cream on maximum temperature. Estimate the sale of ice cream if the maximum temperature on a day is 42 Celsius.

Answer:

Here, n = 6; xÌ„ = 40; yÌ„ = 1.2; Î£xy = 306 and S_{x}^{2} = 20 are given.

The regression line of sale of ice cream (Y) on the maximum temperature (X) :

yÌ‚ = a + bx

b = \(\frac{\Sigma x y-n \bar{x} \bar{y}}{n \cdot \mathrm{S}_{x}{ }^{2}}\)

= \(\frac{306-6(40)(1.2)}{6 \times 20}\)

= \(\frac{306-288}{120}\)

= \(\frac{18}{120}\)

= 0.15

a = yÌ„ – bxÌ„

= 1.2 – 0.15 (40)

= 1.2 – 6

= – 4.8

Putting a = – 4.8 and b = 0.15 in yÌ‚ = a + bx,

we get yÌ‚ = – 4.8 + 0.15 x

Estimate of sale of ice cream (Y) when temperature X = 42 (celsius):

Putting x = 42 in yÌ‚ = – 4.8 + 0.15x, we get

yÌ‚ = – 4.8 + 0.15 (42)

= – 4.8 + 6.3

= 1.5

Hence, the estimate of sale of ice cream obtained is yÌ‚ = â‚¹ 1.5 lakh.