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Statistics 5 Mark Each

LINEAR REGRASSION · Gujarat Board (GSEB) STD 12 Commerce (GSEB - English Medium). 31 questions.

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Question 15 Marks
A sample of seven students is taken from the students coming from abroad in the current year to study in university of the Gujarat State. The information regarding their $I.Q.$ and the marks obtained in an examination of $75$ marks is given below.
Obtain the regression line of $Y$ on $X$ and estimate the marks of a student whose $LQ.$ is $120.$ Also find the error in estimation when $1.0.$ is $100.$
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Question 25 Marks
In order to determine the relationship between monthly income $($in thousand $₹)$ and monthly expenditure $($in thousand $₹)$ of people of a group, a sample of seven persons is taken from that group and the following information is obtained.
Obtain the regression line of monthly expenditure on monthly income. If a person of the group has monthly income of $₹ 75$ thousand, estimate his monthly expenditure.
Answer
Since the regression line of monthly expenditure on monthly income is to be obtained, we shall take ‘monthly expenditure’ as variable Y and ‘monthly income’ as variable $X.$
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Question 35 Marks
In order to study the relationship between the repairing time of accident damaged cars and the cost of repair, the following information is collected.
Obtain the regression line of $Y ($repairing cost$)$ on $X ($repairing time$).$ If the time taken to repair a car is $50$ hours, find an estimate of the repairing cost.
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Question 45 Marks
The monthly sale of different types of laptops $($in hundred units$)$ and its profit $($in lakh $₹)$ for the last six months for a company is given below.
Obtain the regression line of $Y$ on $X.$ Also find the error in estimating $Y$ for $X = 7.$
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Question 55 Marks
In order to study the relationship between the density of population and the number of persons suffering from skin diseases, the following information is obtained for six cities regarding their density of population $($per sq. km$)$ and persons suffering from skin diseases $($per thousand$).$
Obtain the regression line of $Y$ on $X$ Estimate the number of patients suffering from skin diseases if density of population ofa city is $15000 ($per sq.km$).$ Examine the reliability of this regression model.
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Question 65 Marks
The following observations are obtained for life $($years of usage$)$ of cars and their average annual maintenance costs of a specific model of car of a particular company.

Obtain the regression line of maintenance cost on the life of cars. Also, estimate the maintenance cost if the life of a car is $10$ years.
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Question 75 Marks
The following results are obtained for a data. $n = 12, x = 30, y = 5, x^2 = 670, xy = 344$ Later on, it was known that one pair $(10, 14)$ was wrongly taken as $(11, 4)$. By correcting the above measures, obtain the regression line of Yon $X.$ Estimate $Y$ fro $X = 5$.
Answer
Here, $n=12 ; \Sigma x=30 ; \Sigma y=5 ; \Sigma x^2=670 ; \Sigma x y=344$ are given.
True pair of observation is $(10,14)$, i.e., $x=10, y=14$
False pair of observation is $(11,4)$, i.e., $x=11, y=5$
$\therefore \text { Correct } \Sigma x=30+10-11=29$
$\text { Correct } \Sigma y=5+14-4=15$
$\text { Correct } \Sigma x^2=670+(10)^2-(11)^2$
$=670+100-121$
$=649$
$\text { Correct } \Sigma x y=344+(10 \times 14)-(11 \times 4)$
$=344+140-44=440$
Now, $\overline{\mathrm{x}}=\frac{\text { Correct } \Sigma x}{n}=\frac{29}{12}=2.42$
$\bar{y}=\frac{\text { Correct } \Sigma y}{n}=\frac{15}{12}=1.5$
Regression line $\hat{y}=a+b x$ of $Y$ on $X$ is to obtained.
Now, b $=\frac{n \Sigma x y-(\Sigma x)(\Sigma y)}{n \Sigma x^2-(\Sigma x)^2}$
Putting the corrected values in the formula,
$b=\frac{12(440)-(29)(15)}{12(649)-(29)^2}$
$=\frac{5280-435}{7788-841}$
$=\frac{4845}{6947}$
$=0.70$
$a=\bar{y}-b \bar{x}$
$\therefore a=1.25-0.7(2.42)$
$=1.25-1.69=-0.44$
Regression line of $\mathrm{Y}$ on $\mathrm{X}$ :
Putting $a=-0.44$ and $b-0.7$ in $\hat{y}=a+b x$, we get
$\hat{y}=-0.44+0.7 x$
Estimated value of $\mathrm{Y}$ when $\mathrm{X}=5$ :
Putting $x=5$ in $\hat{y}=-0.44+0.7 x$, we get
$\hat{y}=-0.44+0.7(5)$
$=-0.44+3.50=3.06$
Hence, the estimated value of $Y$ obtained is $\hat{y}=3.06$.
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Question 85 Marks
Obtain the regression line of $Y$ on $X$ from the following data and estimate $Y$ for $X = 30. n = 10, x = 250, y = 300, xy = 7900, x^2 = 6500.$
Answer
Here, $n=10 ; \Sigma x=250 ; \Sigma y=300 ; \Sigma x y=7900 ; \Sigma x^2=6500$ are given,
Now, $\overline{\mathrm{x}}=\frac{\Sigma x}{n}=\frac{250}{10}=25 ; \bar{y}=\frac{\Sigma y}{n}=\frac{300}{10}=30$
We obtain the regression line $\hat{y}=a+b x$ of $Y$ on $X$.
Now, $\mathrm{b}=\frac{n \Sigma x y-(\Sigma x)(\Sigma y)}{n \Sigma x^2-(\Sigma x)^2}$
$\therefore b=\frac{10(7900)-(250)(300)}{10(6500)-(250)^2}$
$=\frac{79000-75000}{65000-62500}$
$=\frac{4000}{2500}$
$=1.6$
$a=\bar{y}-b \bar{x}$
$\therefore a=30-1.6(25)$
$=30-40$
$=-10$
Regression line of $\mathrm{Y}$ on $\mathrm{X}$ :
Putting $a=-10$ and $b=1.6$ in $\hat{y}=a+b x$, we get
$\hat{y}=-10+1.6 x$
Estimate of the value of $Y$ when $X=30$ :
Putting $x=30$ in $\hat{y}=-10+1.6 x$, we get
$\hat{y}=-10+1.6(30)$
$ =-10+48=38$
Hence, the estimate of $Y$ obtained is $\hat{y}=38$.
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Question 95 Marks
The information of eight construction companies regarding the number of contracts received in a year and the annual profit is as follows.
No. of contracts $2$ $5$ $9$ $12$ $6$ $4$ $8$ $10$
Annual profit $($lakh $Rs.)$ $100$ $300$ $700$ $1000$ $350$ $250$ $700$ $750$
Obtain the regression line of the annual profit on the number of contracts. Verify the reliability of the regression model.
Answer
Here, $n=8 ; X=$ No. of contracts and $Y=$ Annual profit.
Now, $\bar{x}=\frac{\Sigma x}{n}=\frac{56}{8}=7 ; \bar{y}=\frac{\Sigma y}{n}=\frac{4150}{8}=518.75$
We obtain the regression line $\hat{y}=a+b x$ of annual profit $(Y)$ on number of contracts $(X)$.
To make the calculation of values of $a$ and $b$ simple, we obtain new variables $u=x-A, A=8$ and $v$ $=\left(\frac{y-\mathrm{B}}{\mathrm{C}_y}\right), \mathrm{B}=700, \mathrm{C}_{\mathrm{y}}=50$
The table for calculation is prepared as follows :
Image
$\mathrm{b}=\frac{n \Sigma u v-(\Sigma u)(\Sigma v)}{n \Sigma u^2-(\Sigma u)^2} \times \frac{\mathrm{C}_y}{\mathrm{C}_x}$
Putting $\mathrm{n}=8 ; \Sigma \mathrm{uv}=172 ; \Sigma \mathrm{u}=-8 ; \Sigma \mathrm{v}=-29 ; \Sigma \mathrm{u}^2=86 ; C_{\mathrm{y}}=50$ and $C_{\mathrm{x}}=1$ in the formula,
$b=\frac{8(172)-(-8)(-29)}{8(86)-(-8)^2} \times \frac{50}{1}$
$=\frac{1376-232}{688-64} \times \frac{50}{1}$
$=\frac{1144 \times 50}{624}$
$=\frac{57200}{624}$
$=91.67$
$a=\bar{y}-b \bar{x}$
Putting $\bar{y}=518.75 ; \bar{x}=7$ and $b=91.67$, we get
$a=518.75-91.67(7)$
$=518.75-641.69$
$=-122.94$
Regression line of annual profit $(Y)$ on number of contracts $(X)$ :
Putting $a=-122.94$ and $b=91.67$ in
$\hat{y}=a+b x, \text { we get }$
$ \hat{y}=-122.94+91.67$
Coefficient of determination :
Now, $\mathrm{R}^2=\mathrm{r}^2=\left[\frac{n \Sigma u v-(\Sigma u)(\Sigma v)}{\sqrt{n \Sigma u^2-(\Sigma u)^2} \cdot \sqrt{n \Sigma v^2-(\Sigma v)^2}}\right]^2$
Putting $n=8 ; \Sigma \mathrm{u}=-8 ; \Sigma \mathrm{v}=-29 ; \Sigma \mathrm{uv}=172 ; \Sigma \mathrm{u}^2=86$ and $\Sigma \mathrm{v}^2=375$ in the formula,
$R^2 =\frac{[8(172)-(-8)(-29)]^2}{\left[8(86)-(-8)^2\right] \cdot\left[8(375)-(-29)^2\right]}$
$=\frac{[1376-232]^2}{[688-64] \cdot[3000-841]}$
$=\frac{[1144]^2}{624 \times 2159}=\frac{1308736}{1347216}=0.97$
$\mathrm{R}^2=0.97$, hence the regression model is reliable.
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Question 105 Marks
The following information is collected by a firm to know the effect of an advertisement compaign.
Year $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$
Advertisement cost $($ten thousand $Rs.)$ $12$ $15$ $15$ $23$ $24$ $38$ $42$ $48$
Sales $($crore $Rs.)$ $5$ $5.6$ $5.8$ $7$ $7.2$ $8.8$ $9.2$ $9.5$
Obtain the regression line of sales on the advertisement cost. Estimate the sales when the advertisement cost is $Rs. 5,00,000.$
Answer
Here, $n=8 ; X=$ Advertisement cost and $Y=$ Sales
Now, $\overline{\mathrm{x}}=\frac{\Sigma x}{n}=\frac{217}{8}=27.13 ; \overline{\mathrm{y}}=\frac{\Sigma y}{n}=\frac{58.1}{8}=7.26$
We obtain the regression line $\hat{y}=a+b x$ of sales $(Y)$ on the advertisement $\operatorname{cost}(X)$.
To make the calculation of values $a$ and $b$ simple, we obtain the new variable $u=(x-A), A=24$ and $v=\frac{y-\mathrm{B}}{\mathrm{C}_y}, \mathrm{~B}=7, \mathrm{C}_{\mathrm{y}}=0.1$
The table for calculation is prepared as follows :
Image
$\mathrm{b}=\frac{n \Sigma u v-(\Sigma u)(\Sigma v)}{n \Sigma u^2-(\Sigma u)^2} \times \frac{\mathrm{C}_y}{\mathrm{C}_x}$
Putting $n=8 ; \Sigma u v=1722 ; \Sigma u=25 ; \Sigma v=21 ; \Sigma u^2=1403 ; C_y=0.1$ and $C_x=1$ in the formula,
$b=\frac{8(1722)-(25)(21)}{8(1403)-(25)^2} \times \frac{0.1}{1}$
$=\frac{13776-525}{11224-625} \times \frac{0.1}{1}$
$=\frac{13251 \times 0.1}{10599}$
$=\frac{1325.1}{10599}$
$=0.13$
$a=\bar{y}-b \bar{x}$
Putting $\bar{y}=7.26 ; \bar{x}=27.13$ and $b=0.13$, we get
$a=7.26-0.13(27.13)$
$=7.26-3.53=3.73$
Regression line of sales $(Y)$ on advertisement $\operatorname{cost}(X)$ :
Putting $a=3.73$ and $b=0.13$ in $\hat{y}=a+b x$, we get
$\hat{y}=3.73+0.13 x$
Estimate of sales $(Y)$ when, $X=₹ 5,00,000 ($i.e., $50$ ten thousand$) :$
Putting $x=50$ in $\hat{y}=3.73+0.13 x$, we get
$\hat{y}=3.73+0.13(50)$
$ =3.73+6.5$
$ =10.23$
Hence, the estimate of sales $Y$ obtained is $\hat{y}=₹ 10.23$ crore.
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Question 115 Marks
The information regarding the daily income $($in $Rs.)$ and expenditure $($in $Rs.)$ of five labour families earning by daily work :
Daily income $(Rs.)$ $200$ $300$ $400$ $600$ $900$
Expenditure $(Rs.)$ $180$ $270$ $320$ $480$ $700$
Obtain the regression line of the expenditure on the daily income. Estimate the expenditure of a family having daily income of $Rs. 500.$
Answer
Here, $n=5 ; X=$ Daily income and $Y=$ Expenditure.
Now, $\bar{x}=\frac{\Sigma x}{n}=\frac{2400}{5}=480 ; \bar{y}=\frac{\Sigma y}{n}=\frac{1950}{5}=390$;
We obtain the regression line $\hat{y}=a+b x$ of expenditure $(Y)$ on daily income $(X)$.
To make the calculation of values a and $\mathrm{b}$ simple, we take new variables $u=\frac{x-A}{C_x}, A=400, C_x=$ 100 and $v=\frac{y-B}{C_y}, B=320, C_Y=10$
The table for calculation is prepared as follows :
Image
$b=\frac{n \Sigma u v-(\Sigma u)(\Sigma v)}{n \Sigma u^2-(\Sigma u)^2} \times \frac{\mathrm{C}_y}{\mathrm{C}_x}$
Putting $n=5 ; \Sigma \mathrm{uv}=255 ; \Sigma \mathrm{u}=4 ; \Sigma \mathrm{v}=35 ; \Sigma \mathrm{u}^2=34 ; C_{\mathrm{x}}=100$ and $C_{\mathrm{y}}=10$ in the formula,
$b=\frac{5(255)-(4)(35)}{5(34)-(4)^2} \times \frac{10}{100}$
$=\frac{1275-140}{170-16} \times \frac{1}{10}$
$=\frac{1275-140}{170-16} \times \frac{1}{10}$
$=\frac{1135}{1540}$
$=0.74$
$a=\bar{y}-b \bar{x}$
Putting $\bar{y}=390 ; \bar{x}=480$ and $b=0.74$, we get
$a=390-0.74(480)$
$=390-355.2$
$=34.8$
Regression line of expenditure $(Y)$ on daily income $(X)$ :
Putting $a=34.8 ; b=0.74$ in $y=a+b x$, we get
$\hat{y}=34.8+0.74 x$
Estimate of expenditure $Y$ when $X=₹ 500$ :
Putting $x=500$ in $\hat{y}=34.8+0.74 x$, we get
$\hat{y}=34.8+0.74(500)$
$ =34.8+370=404.8$
Hence, the estimate of expenditure obtained is $y=₹ 404.8$.
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Question 125 Marks
The information regarding the experience $($In years$)$ of eight workers on a machine and their performance ratings based on the non-defective units they manufactured in every $100$ units is as follows :
Experience of worker $($years$)$ $5$ $12$ $15$ $8$ $20$ $18$ $22$ $25$
Performance rating $80$ $82$ $85$ $81$ $90$ $90$ $95$ $97$
Obtain the regression line of the performance rating on the experience and estimate the performance rating if a worker has an experience of $17$ years.
Answer
Here, $n=8 ; X=$ Experience of worker and $Y=$ Performance rating
Now, $\bar{x}=\frac{\Sigma x}{n}=\frac{125}{8}=15.63 ; \bar{y}=\frac{\Sigma y}{n}=\frac{700}{8}=87.5$
We obtain the regressoin line $\hat{y}=a+b x$ of performance rating $(Y)$ on the experience of workers
$(X)$. To make the calculation of $a$ and $b$ simple, we take new variables $u=x-A, A=15$ and $v=y-$ $B, B=90$
The table for calculation is prepared as follows :
Image
$b=\frac{n \Sigma u v-(\Sigma u)(\Sigma v)}{n \Sigma u^2-(\Sigma u)^2}$
Putting $n=8 ; \Sigma \mathrm{uv}=292 ; \Sigma \mathrm{u}=5 ; \Sigma \mathrm{u}=-20$ and $\Sigma \mathrm{u}^2=341$ in the formula,
$b=\frac{8(292)-(5)(-20)}{8(341)-(5)^2}$
$ =\frac{2336+100}{2728-25}$
$ =\frac{2436}{2703}$
$ =0.9$
$a=\bar{y}-b \bar{x}$
Putting $\bar{y}=87.5 ; \bar{x}=15.63 ; b=0.9$, we get
$a=87.5-0.9(15.63)$
$=87.5-14.07$
$=73.43$
Regression line of performance rating $(Y)$ on experience of worker $(X)$ :
Putting $a=73.43 ; b=0.9$ in $\hat{y}=a+b x$, we get $\hat{y}=73.43+0.9 x$
Estimate of performance rating $Y$ when $X=17$ year :
Putting $x=17$ in $\hat{y}=73.43+0.9 x$, we get
$\hat{y}=73.43+0.9(17)$
$ =73.43+15.3$
$ =88.73$
Hence, the estimate of performance rating $(Y)$ obtained is $\hat{y}=88.73$.
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Question 135 Marks
Obtain the regression line of the demand on the price using the following information collected for the demand and the price of a commodity. Estimate the demand of the commodity of price is $Rs. 40.$
Cost $(Rs.)$ $38$ $36$ $37$ $37$ $36$ $38$ $39$ $36$ $38$
Demand $($hundred units$)$ $12$ $18$ $15$ $12$ $17$ $13$ $13$ $15$ $12$
Answer
Here, $n=9 ; X=$ Price and $Y=$ Demand
Now, $\overline{\mathrm{x}}=\frac{\Sigma x}{n}=\frac{335}{9}=37.22 ; \overline{\mathrm{y}}=\frac{\Sigma y}{n}=\frac{127}{9}=14.11$
We have to obtain the regression line $\hat{y}=a+b x$ of $\operatorname{demand}(Y)$ on price $(X)$.
We calculate the values of $a$ and $b$ by shortcut method taking new variables $u=x-A, A=30, v=y$ $-B, B=15$
The table for calculation is prepared as follows :
Image
$\mathrm{b}=\frac{n \Sigma u v-(\Sigma u)(\Sigma v)}{n \Sigma u^2-(\Sigma u)^2}$
Putting $\mathrm{n}=9 ; \Sigma \mathrm{uv}=-25: \Sigma \mathrm{u}=11 ; \Sigma \mathrm{v}=-8$ and $\Sigma \mathrm{u}^2=23$ in the formula,
$b=\frac{9(-25)-(11)(-8)}{9(23)-(11)^2}$
$=\frac{-225+88}{207-121}$
$=\frac{-137}{86}$
$=-1.59$
$a=\bar{y}-b \bar{x}$
Putting $\bar{y}=14.11 ; \bar{x}=37.22$ and $b=-1.59$, we get
$a=14.11-(-1.59)(37.22)$
$ =14.11+59.18$
$ =73.29$
Regression line of demand $(Y)$ on price $(X)$ :
Putting $a=73.29$ and $b=-1.59$ in $\hat{y}=a+b x$, we get
$\hat{y}=73.29-1.59 x$
Estimate of demand $(Y)$ when price $(X)=₹ 40$ :
Putting $x=40$ in $\hat{y}=73.29-1.59 x$, we get
$\hat{y}=73.29-1.59(40)$
$=73.29-63.6$
$=9.69$
Hence, the estimate of demand obtained is $\hat{y}=9.69 ('00$ units$)$
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Question 145 Marks

The information of investment $($in lakh $₹)$ and Its market price $($in lakh $₹)$ after six months in share market in the last seven years for a Mutual Fund Company is obtained as follows :

Obtain the regression line of $Y$ on $X$ and estimate the market price in the share market after six months if there is an investment of $₹\ 45$ lakh in a year.
Answer
Here, $X =$ Investment and $Y =$ Market price after six months.
$\therefore x̄ = 40; ȳ = 50; S_x^2 = 100; S_y^2 = 256$
Cov $(x, y) = 80$ Now, $b = \frac{{Cov}(x, y)}{\mathrm{S}_{x}^{2}}$
$\therefore b = \frac{80}{100} = 0.8$
$a = ȳ – bx̄$
$\therefore a = 50 – 0.8 (40)$
$= 50 – 32 = 18$
The regression line of $Y$ on $X :$
Putting $a = 18$ and $b = 0.8,$ in $ŷ = a + bx,$ we get
$ŷ = 18 + 0.8x$
Estimate of market price $(Y)$ for investment $X = ₹\ 45$ lakh :
Putting $x = 45$ in $ŷ = 18 + 0.8x,$ we get
$ŷ = 18 + 0.8 (45)$
$= 18 + 36 = 54$
Hence, the estimate of market price in the share market after six months obtained is $₹\ 54$ lakh.
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Question 155 Marks
The following results are obtained from the information of average rain and yield of a crop per acre in the last ten years of an arid region :
Image
Estimate the yield of the crop if it rains $20 \ cms.$
Answer
Here, $X =$ Rainfall and $Y =$ Yield of crop
$\therefore x̄ = 18 \ cm;ȳ = 970 \ kg, S_x = 2, S_y = 38, r = 0.6.$
Now, $b=r \cdot \frac{S_y}{S_x}$
$\therefore b=0.6 \times \frac{38}{2}$
$=11.4$
$a=\bar{y}-b \bar{x}$
$\therefore a=970-11.4(18)$
$=970-205.2$
$=764.8$
The regression line of yield of crop $(Y)$ on rainfall $(X) :$
Putting $a = 764.8$ and $b = 11.4$ in $ŷ = a + bx,$ we get
$ŷ = 764.8 + 11.4x$
Estimate of yield of crop $(Y) $if rainfall $X = 20 \ cms $:
Putting $x = 20$ in $ŷ = 764.8 + 11.4x$, we get
$ŷ = 764.8 + 11.4 (20)$
$= 764.8 + 228$
$= 992.8$
Hence, the estimate of yield of crop obtained is $992.8 \ kg.$
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Question 165 Marks
The information collected regarding price (in ₹) of a commodity and its supply (in hundred units) is as follows:

Obtain the regression line of the supply on the price.
Answer

Here, $n = 8; X =$ Price and $Y =$ Supply
Now, $x̄ = \frac{\Sigma x}{n} = \frac{480}{8} = 60; ȳ = \frac{\Sigma y}{n} = \frac{632}{8} =79$
We obtain the regression line of supply $(Y)$ on the price $(X),$
$ŷ = a + bx$
$x̄$ and $ȳ$ are integers. So the table for calculating the values of $a$ and $b $ is prepared as follows :Image
$b = \frac{\Sigma(x-\bar{x})(y-\bar{y})}{\Sigma(x-\bar{x})^{2}}$
Putting $\sum (x – x̄) (y – ȳ) = 24 $ and $\sum (x – x̄)^2 = 36$ in the formula,
$b = \frac{24}{36}$
$= 0.67$
$a = ȳ – bx̄$
Putting $ȳ = 79; x̄ = 60$ and $b = 0.67$, we get
$a = 79 – 0.67 (60)$
$= 79 – 40.2 = 38.8$
Regression line of $Y$ on $X:$
Putting $a = 38.8$ and $b = 0.67$ in $ŷ = a + bx,$ we get
$ŷ = 38.8 + 0.67x$
Hence, the regression line of supply $(Y) $ on price $(X)$ obtained is $ŷ = 38.8 + 0.67x$
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Question 175 Marks
A manager of an $I.T.$ company has collected the following information regarding the years of job and monthly income of seven marketing executives;

Obtain the regression line of the monthly income on the years of job of the marketing executives.
Answer
Here, $n = 7; X =$ Years of job and $Y =$ Monthly income
Now, $x̄ = \frac{\Sigma x}{n} = \frac{56}{7} = 8$ and $ȳ = \frac{\Sigma y}{n} = \frac{56}{7} = 8$
We obtain the regression line of monthly income $(Y)$ on the years of job $(X), ŷ = a + bx.$
The table for calculating the values of $a$ and $b$ is prepared as follows :
Image
$b = \frac{n \Sigma x y-(\Sigma x)(\Sigma y)}{n \Sigma x^{2}-(\Sigma x)^{2}}$
Putting $n = 7; \sum xy = 469; \sum x^2 = 476; \sum x = 56$ and $\sum y = 56$ in the formula,
$b = \frac{7(469)-(56)(56)}{7(476)-(56)^{2}}$
$= \frac{3283-3136}{3332-3136}$
$= \frac{147}{196}$
$= 0.75$
$a = ȳ – bx̄$
Putting $ȳ = 8; x̄ = 8$ and $b = 0.75,$ we get
$a = 8 – 0.75 (8)$
$=8 – 6 = 2$
Regression line of monthly income $(Y)$ on the years of job $(X) :$
Putting $a = 2$ and $b = 0.75$ in $ŷ = a + bx,$ we get $ŷ = 2 + 0.75x.$
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Question 185 Marks
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 $₹) :$
Image
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, $\bar{x}=\frac{\Sigma x}{n}=\frac{560}{6}=93.33 ; \bar{y}=\frac{\Sigma y}{n}=\frac{14.4}{6}=2.4$
To obtain tire regression line $ŷ = a + bx$ of sales $(Y)$ on the number of customers $(X),$ we calculate the values of $a$ and $b$ by shortcut method, taking new variables
$u =\frac{x- A }{ C _x}, A =100, C _{ x }=10$ and $v =$ $\frac{y- B }{ C _y}, B =2.5, C _{ y }=0.1$
The table for calculation is prepared as follows :
Image
$ b =\frac{n \Sigma u v-(\Sigma u)(\Sigma v)}{n \Sigma u^2-(\Sigma u)^2} \times \frac{ C _y}{ C _x}$
$\text { Putting } n =6 ; \Sigma uv =148 ; \Sigma u =-4 ; \Sigma v =-6 ;$
$\Sigma u ^2=72 ; C_{ x }=10 \text { and } C _{ y }=0.1 \text { in the formula. }$
$b =\frac{6(148)-(-4)(-6)}{6(72)-(-4)^2} \times \frac{0.1}{10}$
$=\frac{888-24}{432-16} \times \frac{0.1}{10}$
$=\frac{864 \times 0.1}{416 \times 10}$
$=\frac{86.4}{4160}$
$=0.02$
$a=\bar{y}-b \bar{x}$
Putting $\bar{y}=2.4 ; \bar{x}=93.33$ 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 $ŷ = a + bx,$
we get $ŷ = 0.53 + 0.02x$
Estimate of sales $Y$ when $X = 80 :$
Putting $x = 80$ in $ŷ = 0.53 + 0.02x,$ we get
$ŷ = 0.53 + 0.02 (80)$
$= 0.53 + 1.6 = 2.13$
Hence, the estimate of sales obtained is $ŷ = ₹ 2.13 ($ten thousand$).$
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Question 195 Marks
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, $x̄ = \frac{\Sigma x}{n} = \frac{630}{7} = 90; ȳ = \frac{\Sigma y}{n} = \frac{182000}{7} = 26000$
To obtain the regression line $ŷ = a + bx$ 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 :
Image
$b = b_{vu} = \frac{n \Sigma u v-(\Sigma u)(\Sigma v)}{n \Sigma u^{2}-(\Sigma u)^{2}} \times \frac{\mathrm{C}_{y}}{\mathrm{C}_{x}}$
Putting $n = 7; \sum uv = 404; \sum u = 14; \sum v = 7; \sum u^2 = 266; C_x = 5$ and $C_y = 1000$ in the formula,
$b = \frac{7(404)-(14)(7)}{7(266)-(14)^{2}} \times \frac{1000}{5}$
$= \frac{2828-98}{1862-196} \times 200$
$= \frac{2730 \times 200}{1666}$
$= \frac{546000}{1666}$
$= 327.73$
$a = ȳ – bx̄$
Putting $ȳ = 26000; b = 327.73$ and $x̄ = 90,$ 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 $ŷ = a + bx,$ we get
$ŷ = – 3495.7 + 327.73x$
Estimate of monthly rent $(Y)$ when $X=110 :$
Putting $x = 110$ in $ŷ = – 3495.7 + 327.73x,$ we get
$ŷ = – 3495.7 + 327.73 (110)$
$= – 3495.7 + 36050.3$
$= 32554.6$
Hence, the estimate of monthly rate obtained is $ŷ = ₹ 32554.6$
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Question 205 Marks
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, $x̄ = \frac{\Sigma x}{n} = \frac{21}{7} = 3; ȳ = \frac{\Sigma y}{n} = \frac{130.1}{7} = 18.59$
We obtain the regression line of effective oxygen $(Y)$ on altitude $(X) ŷ = a + bx.$
$ŷ$ 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 :
Image
$b = b_{uv} = \frac{n \Sigma u v-(\Sigma u)(\Sigma v)}{n \Sigma u^{2}-(\Sigma u)^{2}} \times \frac{\mathrm{C}_{y}}{\mathrm{C}_{x}}$
Putting $n = 7; \sum uv = – 200; \sum u = 0; \sum u = 48; \sum u^2 = 28; C_y = 0.1$ and $C_x = 1$ in the formula,
$b = \frac{7(-200)-(0)(48)}{7(28)-(0)^{2}} \times \frac{0.1}{1}$
$= \frac{-1400-0}{196-0} \times \frac{0.1}{1}$
$= \frac{-1400 \times 0.1}{196}$
$= \frac{-140}{196}$
$= – 0.71$
$a = ȳ – bx̄$
Putting$ ȳ = 18.59; x̄ = 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 $ŷ = 20.72 – 0.71x$
The estimate of percentage of oxygen $(Y)$ when $X = 7 :$
Putting $x = 7$ in $y = 20.72 – 0.7 lx,$ we get
$ŷ = 20.72 – 0.71 (7)$
$= 20.72 – 4.97$
$= 15.75$
Hence, the estimate of percentage of oxygen obtained is $ŷ = 15.75\%.$
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Question 215 Marks
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 father’s height is $170\ cm$.
Answer

Here, $n = 8; X =$ Height of father and $Y =$ Height of son
Now, $x̄ =$ $\frac{\Sigma x}{n}$ = $\frac{1346}{8}$$ = 168.25\ cm$; ȳ = $\frac{\Sigma y}{n}$ = $\frac{1338}{8}$ $= 167.25\ cm$
We obtain the regression line of height of son $(Y)$ on height of father $(Y) ŷ = a + bx$.
$x̄$ and $ȳ$ 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 :
Image
b = $\frac{n \Sigma u v-(\Sigma u)(\Sigma v)}{n \Sigma u^{2}-(\Sigma u)^{2}}$
Putting $n = 8; Σuv = 34; Σu = 2; Σv = 2$ and $\sum u^2 = 50$ in the formula,
b = $\frac{8(34)-(2)(2)}{8(50)-(2)^{2}}$
= $\frac{272-4}{400-4}$
= $\frac{268}{396}$
$= 0.68$
$a = ȳ – bx̄$
Putting $ȳ = 167.25; x̄ = 168.25$ 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 $ŷ = a + bx$, we get
$ŷ = 52.84 + 0.68x$
Estimate of $Y$ when X = 170 cm :
Putting x = 170 in ŷ = 52.84 + 0.68 x, we get
$ŷ = 52.84 + 0.68 (170)$
$= 52.84 + 115.6$
$= 168.44$
Hence, the estimate of son height obtained is $ŷ = 168.44\ cm.$
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Question 225 Marks
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.
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Question 235 Marks
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.
Answer
Here, $n = 8; X =$ Experience and $Y =$ Performance rating.
Now,$x̄$ = $\frac{\Sigma x}{n}$ = $\frac{80}{8}$ $= 10$ year; $ȳ$ = $\frac{\Sigma y}{n}$ = $\frac{648}{8}$ = 81
We obtain the regression line of Performance rating $(Y)$ on experience $(X)\ ŷ = a + bx$.
$x̄$ and $ȳ$ are integers. So to calculate a and b we prepare the following table : Image
b = $\frac{\Sigma(x-\bar{x})(y-\bar{y})}{\Sigma(x-\bar{x})^{2}}$
Putting, $Σ(x – x̄) (y – ȳ) = 247$ and $Σ(x – x̄)^2 = 218$ in the formula,
$b =$ $\frac{247}{218}$
$= 1.13$
$a = ȳ – bx̄$
Putting $ȳ = 81; x̄ = 10$ and $b = 1.13,$ we get
$a = 81 – 1.13 (31)$
$= 81 – 11.3$
$= 69.7$
Regression line of $Y$ on $X$:
Putting $a = 69.7$ and $b = 1.13$ in $ŷ = a + bx$, we get
$ŷ = 69.7 + 1.13x$
Estimate of $Y$ when $X = 7$ years :
Putting $x = 7$ in $ŷ = 69.7 + 1.13x$, we get
$ŷ = 69.7 + 1.13 (7)$
$= 69.7 + 7.91$
$= 77.61$
Hence, the estimate of performance rating obtained is $ŷ = 77.61$.
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Question 245 Marks
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 $35\ cm$.
Answer
Here, $n = 5; X =$ Average rainfall and $Y =$ Crop
Now, $x̄ = \frac{\Sigma x}{n} = \frac{155}{5} = 31\ cm; ȳ = \frac{\Sigma y}{n} = \frac{450}{5} = 90\ tons$
We obtain the regression line of crop $(Y)$ on the average rainfall $(X)$
$ŷ = a + bx.$
$x̄$ and $ȳ$ are integers.
So to calculate $a$ and $b,$ we prepare the following table :Image
$b = \frac{\Sigma(x-\bar{x})(y-\bar{y})}{\Sigma(x-\bar{x})^{2}}$
Putting, $\sum (x – x̄) (y – ȳ) = 75$ and $\sum (x – x̄)^2 = 90$ in the formula,
$b = \frac{75}{50}$
$= 0.83$
$a = ȳ – bx̄$
Putting $ȳ = 90; x̄ = 31$ and $b = 0.83,$ we get
$a = 90 – 0.83 (31)$
$= 90 – 25.73$
$= 64.27$
Regression line of crop $(Y)$ on average rainfall $(X) :$
Putting $b = 0.83$ and $a = 64.27$ in $ŷ = a + bx,$ we get
$ŷ = 64.27 + 0.83x.$
Estimate of $Y$ when $X = 35 \ cm :$
Putting $x = 35$ in $ŷ = 64.27 + 0.83x,$ we get
$ŷ = 64.27 + 0.83 (35) = 64.27 + 29.05 = 93.32$
Hence, the estimate of crop obtained is $ŷ = 93.32$ tons.
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Question 255 Marks
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.
Answer
Here, $n = 6; X =$ Time of usage of a car and $Y =$ Annual average maintenance cost
Now, $x̄ = \frac{\Sigma x}{n} = \frac{16}{6} = 2.67; ȳ = \frac{\Sigma y}{n} = \frac{51}{6} = 8.5$
We obtain the regression line of average annual maintenance cost $(Y)$ on the time of usage $(X) ŷ = a + bx$.
The values of $X$ and $Y$ are not large. So we calculate the values of a and b by preparing the following table :
Image
$b = \frac{n \Sigma x y-(\Sigma x)(\Sigma y)}{n \Sigma x^{2}-(\Sigma x)^{2}}$
Putting $n = 6, \sum xy = 154; \sum x = 16; \sum y = 15$ and $\sum x^2 = 52$ in the formula
$= \frac{6(154)-(16)(51)}{6(52)-(16)^{2}}$
$= \frac{924-816}{312-256}$
$= \frac{108}{56} = 1.93$
$a = ȳ – bx̄$
Putting $ȳ = 8.5; x̄ = 2.67$ and $b = 1.93, $ we get,
$a = 8.5 – 1.93 (2.67)$
$= 8.5 – 5.15$
$= 3.35$
Regression line of $Y$ on $X:$
Putting $a = 3.35$ and $b = 1.93$ in $ŷ = a + bx,$ we obtain $ŷ = 3.35 + 1.93x$
The estimate of average annual maintenance cost $Y$ when usage of car $X = 5$ years:
Putting $x = 5$ in $ŷ = 3.35 + 1.93x,$ we get
$ŷ = 3.35 + 1.93 (5)$
$= 3.35 + 9.65 = ₹ 13$ thousand
Hence, the estimate of average annual maintenance cost obtained is $y = ₹ 13$ thousand.
Error: $ŷ = 13, y = 13$
$\therefore $ Error $= ŷ – y = 13 – 13 = 0$
Hence, point $(5, 13)$ is on he fitted regression line $ŷ = 3.35 + 1.93 x.$
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Question 265 Marks
From the following data of price $($in $₹)$ and demand $($in hundred units$)$ of a commodity, obtain the regression line of demand on price. Also estimate the demand when price is $₹ 20.$
Answer

Here, $n = 6; X =$ Price and $Y =$ Demand
Now, $x̄ = \frac{\Sigma x}{n} = \frac{96}{6} = ₹ 16, ȳ = \frac{\Sigma y}{n} = \frac{60}{6} = 10 (’00)$ units.
To obtain the equation of regression line of $Y$ on $X,$
$ŷ = a + bx,$ we calculate a and $b.$
$x̄$ and $ȳ$ are integers. So the table of calculation is prepared as follows :Image
$b = \frac{\Sigma(x-\bar{x})(y-\bar{y})}{\Sigma(x-\bar{x})^{2}}$
Putting, $\sum (x – x̄) (y – ȳ) = – 67$ and $\sum (x – x̄)^2 = 50$ in the formula,
$b = \frac{-67}{50}$
$= – 1.34$
$a = ȳ – bx̄$
Putting $ȳ = 10; x̄ = 16$ and $b = – 1.34,$ we get
$a = 10 – (- 1.34) (16)$
$= 10 + 21.44$
$= 31.44$
Regression line of demand $(Y)$ on price $(X) :$
Putting $a = 31.44$ and $b = – 1.34$ in $ŷ = a + bx$, we get
$ŷ = 31.44 – 1.34x$
When price $X = ₹ 20,$ the estimate of demand $Y:$
Putting $x = 20$ in $ŷ = 31.44 – 1.34x$, we get
$ŷ = 31.44 – 1.34 (20)$
$= 31.44 – 26.8 = 4.64$
Hence, when price is $₹ 20,$ the estimated demand $Y$ obtained is $ŷ = 4.64 (’00)$ units.
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Question 275 Marks

The following results are obtained from the information of average rain and yield of a crop per acre in the last ten years of an arid region:

Estimate the yield of the crop if it rains $20$ cms.
Answer
Here, $X =$ Rainfall and $Y =$ Yield of crop
$\therefore x̄ = 18$ cm; $ȳ = 970$ kg, $S_x = 2, S_y = 38, r = 0.6$.Now, $b = r ∙$
$\frac{\mathrm{S}_{y}}{\mathrm{~S}_{x}}$
$\therefore b = 0.6 \times   \frac{38}{2} $
$= 11.4$
$a = ȳ – bx̄$
$\therefore a = 970 – 11.4 (18)$
$= 970 – 205.2$
$= 764.8$
The regression line of yield of crop $(Y)$ on rainfall $(X) :$
Putting $a = 764.8$ and $ b = 11.4$ in $ŷ = a + bx$, we get
$ŷ = 764.8 + 11.4x$
Estimate of yield of crop $(Y)$ if rainfall $X = 20$ cms :
Putting $x = 20$ in $ŷ = 764.8 + 11.4x$, we get
$ŷ = 764.8 + 11.4 (20)$
$= 764.8 + 228$
$= 992.8$
Hence, the estimate of yield of crop obtained is $992.8$ kg.
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Question 285 Marks
To study the relationship between the time of usage of cars and its average annual maintenance cost of a car manufacturing company, the following information is obtained :
Car123456
Time of usage of a car (Years) x312253
Average annual maintenance cost (thousand ₹) y10587138
Obtain the regression line of Y on X. Find an estimate of average annual maintenance cost when the usage of a car is 5 years. Also find its error.
Answer
Total $x\left(\sum x\right): 3+1+2+2+5+3=16$
Total $y\left(\sum y\right): 10+5+8+7+13+8=51$
Mean $\bar{x} : \frac{16}{6}=2.67$ years
Mean $\bar{y} : \frac{51}{6}=8.5$ (₹1000)
$b=$ $\frac{n \sum x y-\left(\sum x\right)\left(\sum y\right)}{n \sum x^2-\left(\sum x\right)^2}$
$b \approx 1.93$.
$a=\bar{y}-b \bar{x}$
$a \approx 3.36$
The regression line of $Y$ on $X$ is :
$\hat{y}=3.36+1.93 x$
$x=5$
$\hat{y}=3.36+1.93(5)$
= $3.36+9.65=13.01 \approx 13$
$(e=y-\hat{y})$
when $x=5$, the actual $y=13$.
$=13 -$ $13.01 \approx 0$.
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Question 295 Marks
In order to determine the relationship between monthly income (in thousand ₹) and monthly expenditure (in thousand ₹) of people of a group, a sample of seven persons is taken from that group and the following information is obtained :
Person1234567
Monthly Income (thousand ₹)60706468626572
Monthly Expenditure (thousand ₹)50595750535860
Obtain the regression line of monthly expenditure on monthly income. If a person of the group has monthly income of ₹ 75 thousand, estimate his monthly expenditure.
Answer
$A=65$ and $B=55$
$u=x-65$
$V=y-55$
PersonxyuV$u^2$uV
16050-5-52525
27059542520
36457-121-2
468503-59-15
56253-3-296
665580300
77260754935
Sum4613876211869
$\bar{x}=\frac{461}{7} \approx 65.86$
$\bar{y}=\frac{387}{7} \approx 55.29$
$b=\frac{n \sum u V-\left(\sum u\right)\left(\sum V\right)}{n \sum u^2-\left(\sum u\right)^2}$
$=\frac{7(69)-(6)(2)}{7(118)-(6)^2}=\frac{483-12}{826-36}=\frac{471}{790} \approx 0.596$
$a=\bar{y}-b \bar{x}$
$=55.29-(0.596 \times 65.86)=55.29-39.25=16.04$
To estimate the monthly expenditure when $x=75$ :
$\hat{y}=16.04+0.596(75)$
$=16.04+44.7$
$\hat{y}=60.74$
Monthly Expenditure: ₹ 60.74 thousand (or ₹ 60,740).
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Question 305 Marks
The following information is obtained from a study to know the effect of use of fertilizer in the yield of cotton :
Consumption of
Fertilizer ( 10 kg ) $x$		28352524202520
Yield of cotton per hectare (quintals) y128140115120105122100
Obtain the regression line of Y on X and estimate the yield of cotton per hectare if 300 kg fertilizers is used.
Answer
$x$ (Fertilizer)$y$ (Yield)$x^2$$x y$
281287843584
3514012254900
251156252875
241205762880
201054002100
251226253050
201004002000
$\sum x=177$$\sum y=830$$\sum x^2=4,635$$\sum x y=21,389$
$\bar{x}=\frac{\sum x}{n}=\frac{177}{7} \approx 25.286$
$\bar{y}=\frac{\sum y}{n}=\frac{830}{7} \approx 118.571$
$b=\frac{n \sum x y-\left(\sum x\right)\left(\sum y\right)}{n \sum x^2-\left(\sum x\right)^2}$
$=\frac{7(21,389)-(177)(830)}{7(4,635)-(177)^2}$
$=\frac{149,723-146,910}{32,445-31,329}=\frac{2,813}{1,116} \approx 2.52$
$a=\bar{y}-b \bar{x}$
$=118.571-(2.52 \times 25.286)$
$=118.571-63.721 \approx 54.85$
The regression line of $y$ on $x$ is :
$\hat{y}=54.85+2.52 x$
$\hat{y}=54.85+2.52(30)$
$\hat{y}=54.85+75.6$
$\hat{y}=130.45$ quintals
The estimated yield of cotton per hectare is $1 3 0 . 4 5$ quintals.
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Question 315 Marks
From the following data of price (in ₹) and demand (in hundred units) of a commodity, obtain the regression line of demand on price. Also estimate the demand when price is ₹ 60 :
Price (in ₹) x364245485463
Demand (hundred units) y543630242115
Answer
Price (x)Demand (y)$x-\bar{x}(u)$$y-\bar{y}(v)$$u^2$uv
3654-1224144-288
4236-6636-36
4530-3090
48240-600
54216-936-54
631515-15225-225
Sum: 288Sum: 18000450-603
$(\bar{x}) : 288 / 6=48$
$(\bar{y}) : 180 / 6=30$
$b_{y x}=\frac{\sum(x-\bar{x})(y-\bar{y})}{\sum(x-\bar{x})^2}=\frac{\sum u v}{\sum u^2}$
$=\frac{-603}{450}=-1.34$
$a=\bar{y}-b_{y x} \bar{x}$
$=30-(-1.34 \times 48)$
$=30+64.32=94.32$
$\hat{y}=94.32-1.34 x$
$=94.32-1.34(60)$
$=94.32-80.4$
$\hat{y}=13.92$
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