4 Marks Each

Question 14 Marks

From the data and calculation of illustration $12$ of the chapter of linear correlation, obtain the regression line of profit on the sales. Estimate the proﬁt when sales is $₹ 3$ crore.

Answer

From the illustration, we know that$ u=\frac{x-A}{c_{x}}=\frac{x-2}{0.1} \text { and } v=\frac{y-B}{c_{y}}=\frac{y-5600}{100}$
$ \therefore c_{x}=0.1 \text { and } c_{y}=100$
Note that $c_{x}$ is the divisor of $(x-A)$.
So, though we have multiplied $(x-A)$ by $10$ for simplicity of calculation, $c_{x}$ is $\frac{1}{10}=0.1$.$( \because$ To multiply by $10$ is same as to divide by $\frac{1}{10}=0.1)$
$\text { Now } b =\frac{n \Sigma u v-(\Sigma u)(\Sigma v)}{n \Sigma u^{2}-(\Sigma u)^{2}} \times \frac{c_{y}}{c_{x}}$
$ =\frac{9(121)-(0)(1)}{9(60)-(0)^{2}} \times \frac{100}{0.1}$
$ =\frac{1089}{540} \times \frac{100}{0.1}$
$ =\frac{108900}{54}$
$ =2016.6667$
$\therefore b \simeq 2016.67$
Now, $a=\bar{y}-b \bar{x}$$ =5611.11-2016.67(2)$
$ =5611.11-4033.34$
$\therefore a =1577.77$
So, the regression line of profit $(Y)$ on the sales $(X)$ is$\hat{y} =a+b x$
$\therefore \hat{y} =1577.77+2016.67 x$
Putting $X=3$,$\hat{y} =1577.77+2016.67(3)$
$ =1577.77+6050.01$
$\therefore \hat{y} =7627.78$
So, when sales is $₹ 3$ crore then the estimated profit is $7627.78 ($thousand $₹ ).$

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Question 24 Marks

For the data given in illustration $I,$ obtain the regression line of maintenance cost $(Y)$ on the life of cars $(X)$ by using short-cut method.

Answer

$ b =b_{v u} \cdot \frac{c_{y}}{c_{x}}=\frac{n \Sigma u v-(\Sigma u)(\Sigma v)}{n \Sigma u^{2}-(\Sigma u)^{2}} \times \frac{c_{y}}{c_{x}}$
$ =\frac{4(7)-2(1)}{4(6)-(2)^{2}} \times \frac{5}{2}$
$ =\frac{28-2}{24-4} \times \frac{5}{2}$
$ =\frac{26}{20} \times \frac{5}{2}$
$b =3.25 $
Now, $a=\bar{y}-b \bar{x}=21.25-3.25(5)=21.25-16.25=5$
$\therefore$ The regression line of $Y$ on $X$ is$\hat{y} =a+b x$
$\therefore \hat{y} =5+3.25 x$
Note : We can see that, $b_{v u}=\frac{26}{20}=1.3$
but when it is multiplied by $\frac{c_{y}}{c_{x}}=\frac{5}{2}$ then we get $b=1.3 \times \frac{5}{2}=$
$3.25 ($as obtained in illustration $1).$
So, we can understand that when the scale of variable $X$ and/or $Y$ are changed,
it is necessary to multiply $b_{v u}$ by $\frac{c_{y}}{c_{x}}$ to obtain $b$.

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Question 34 Marks

$(i)$ If the regression line is $\hat{y}=\frac{x}{2}+5$ and $s_{y}: s_{x}=5: 8$, find the coefficient of determination.
$(ii)$ If the regression line of $Y$ on $X$ is $4 x+5 y-65=0$, find the value of regression coefficient $b$.

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Question 44 Marks

$(i)$ If the regression line of $Y$ on $X$ is $\hat{y}=25+3 x$ and $\operatorname{Cov}(x, y)=48$, find the standard deviation of $X$. Also find coefficient of determination if the standard deviation of $Y$ is $15.$
$(ii)$ For the regression line given in the above question, how many units should be increased in the value of $X$ to increase approximately $15$ units in $Y ?$

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Question 54 Marks

Six pairs of father-son are selected in a sample of an experiment to know the relation between the heights of fathers in cm $(X)$ and the heights of their adult sons in em $(Y).$ The following results are obtained from it.$\Sigma x=1020, \Sigma y=990, \Sigma(x-170)^{2}=60, \Sigma(y-165)^{2}=105$ $\Sigma(x-170)(y-165)=45$ Obtain the regression line of the heights of sons $(Y)$ on the heights of fathers $(X).$ Also verify the reliability of the regression model.

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Question 64 Marks

The following table shows the experience of technicians $($in years$)$ employed at various companies and their monthly salary $($in thousand $₹).$

Calculate the coefﬁcient of determination and check the validity of the linearity assumption of regression between the years of experience and the monthly salary.

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Question 74 Marks

The electricity is generated by windmill manufactured by a company. The following information is obtained by recording five observations regarding the velocity of wind (km per hour) and generation of electricity $($in Watts$)$ by a unit of the company.Velocity of Wind $= X \ km$ per hour Electricity Generation $= Y$ Watts$\bar{x}=20, \bar{y}=186, \Sigma x y=23200, s_{x}^{2}=50$ Obtain the regression line of electricity generation $(Y)$ on velocity of wind $(X).$ Estimate the electricity generation if the velocity of wind is $25 \ km$ per hour.

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Question 84 Marks

A person in a state of South India produces spoons from eatable materials. It can be eaten after using it. He launched such spoons for the purpose of selling in a state on an experimental level. The following results are obtained for the average price $($in $₹)$ and its demand $($in hundred units$)$ for the last six months.$n=6, \Sigma x=45, \Sigma y=122, \Sigma x^{2}=439, \Sigma x y=605$ Obtain the regression line of the demand $(Y)$ of spoons on the price $(X)$ and estimate the demand of spoons when the price of a spoon is $₹ 10.$

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Question 94 Marks

The information of price $($in $₹)$ of a ballpen and the supply of ballpen $($in units$)$ at the end of each month of a year for a company making ball pen is given below. Estimate the supply of ballpen when its price is $₹ 40.$

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Question 104 Marks

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 $₹ ) \bar{x}=40, \bar{y}=1,2, \Sigma x y= 306, S \times 2=20$ 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, $n=6 ; \bar{x}=40 ; \bar{y}=1.2 ; \Sigma x y=306$ and $S_x^2=20$ are given.
The regression line of sale of ice cream $(Y)$ on the maximum temperature $(X)$ :
$\hat{y}=a+b x$
$b=\frac{\Sigma x y-n \bar{x} \bar{y}}{n \cdot S _x^2}$
$=\frac{306-6(40)(1.2)}{6 \times 20}$
$=\frac{306-288}{120}$
$=\frac{18}{120}$
$=0.15$
$a = ȳ – bx̄$
$= 1.2 – 0.15 (40)$
$= 1.2 – 6$
$= – 4.8$
Putting $a=-4.8$ and $b=0.15$ in $\hat{y}=a+b x$,
we get $\hat{y}=-4.8+0.15 x$
Estimate of sale of ice cream $(Y)$ when temperature $X=42( $celsius$):$
Putting $x=42$ in $\hat{y}=-4.8+0.15 x$, we get
$ŷ = – 4.8 + 0.15 (42)$
$= – 4.8 + 6.3$
$= 1.5$
Hence, the estimate of sale of ice cream obtained is $\hat{y}=₹ 1.5$ lakh.

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Question 114 Marks

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 $(Y) :$
$n = 10, \sum x = 130, \sum y = 220, \sum x^2 = 2288$ and $\sum xy = 3467$
Obtain the regression line of $Y$ on $X$ and estimate the supply when price is ₹ $16.$

Answer

Here, $n = 10; \sum x = 130; \sum y = 220; \sum x^2 = 2288$ and $\sum xy = 3467$
Now, $x̄ = \frac{\Sigma x}{n}=\frac{130}{10} = 13; ȳ =\frac{\Sigma y}{n}= \frac{220}{10} = 22$
We obtain the regression line of $Y$ on $X.$
$ŷ = a + bx$
$b =\frac{n \Sigma x y-(\Sigma x)(\Sigma y)}{n \Sigma x^{2}-(\Sigma x)^{2}}$
$= \frac{10(3467)-(130)(220)}{10(2288)-(130)^{2}}$
$= \frac{34670-28600}{22880-16900}$
$= \frac{6070}{5980}$
$= 1.02$
$a = ȳ – bx̄$
$= 22 – 1.02 (13)$
$= 22 – 13.26$
$= 8.74$
Regression line of $Y$ on $X:$
Putting $a = 8.74$ and $b = 1.02$ in $ŷ = a + bx,$
we get $ŷ = 8.74 + 1.02x$
Estimate of supply $Y$ when price $X = ₹ 16 :$
Putting $x = 16$ in $ŷ = 8.74 + 1.02x,$ we get
$ŷ = 8.74 + 1.02 (16)$
$= 8.74 + 16.32$
$= 25.06$
Hence, the estimate of supply obtained is $ŷ = 25.06$ hundred units.

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Question 124 Marks

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, $x̄ = 82; ȳ = 180; r = 0.82; {S_x}^2 = 64$
$\therefore S_x = 8$ and ${S_y}^2 = 225$
$\therefore S_y = 15$ are given.
To estimate the yield of maize $(Y)$ when the rainfall $(X)$ is $60 \ cm$ we obtain the regression line of $Y$ on $X, ŷ = a + bx.$
Now, $b = r ∙ \frac{\mathrm{S}_{y}}{\mathrm{~S}_{x}};$
Putting $r=0.82 ; S_y=15$ and $S_x=8,$ we get
$b = 0.82 ∙ \frac{15}{8}$
$= \frac{12.3}{8}$
$= 1.54a = ȳ – bx̄$
Putting $ȳ = 180; b = 1.54$ and $x̄ = 82,$ we get
$a = 180 – 1.54 (82)$
$= 180 – 126.28 = 53.72$
Putting $a = 53.72$ and $b = 1.54$ in $ŷ = a + bx,$ the regression line of yield of maize $(Y)$ on the rainfall $(X)$ obtained is $ŷ = 53.72 + 1.54x$
Estimate of yield of maize $Y$ when rainfall $x = 60 \ cm :$
Putting $x = 60$ in $ŷ = 53.72 + 1.54x,$ we get
$ŷ = 53.72 + 1.54 (60)$
$= 53.72 + 92.40$
$= 146.12$
Hence, the estimate of yield of maize obtained is $ŷ = 146.12$ quintal per hectare.

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Statistics 4 Marks Each

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