Step 1: Calculate the sums of the variables.
Number of observations, $n=7$.
$\sum X=20+25+30+35+40+45+50=245$
$\sum Y=60+72+75+80+85+90+93=555$
$\sum X^2={20}^2+{25}^2+{30}^2+{35}^2+{40}^2+{45}^2+{50}^2=9275$
$\sum Y^2={60}^2+{72}^2+{75}^2+{80}^2+{85}^2+{90}^2+{93}^2=44783$
$\sum XY=(20\times 60)+(25\times 72)+(30\times 75)+(35\times 80)+(40\times 85)+(45\times 90)+(50\times 93)=20150$

Step 2: Calculate the regression coefficient Y on X ($b_{yx}$).
$$b_{yx}=\frac{n\sum XY-(\sum X)(\sum Y)}{n\sum X^2-(\sum X{)}^2}$$ $$b_{yx}=\frac{7\left(20150\right)-\left(245\right)\left(555\right)}{7\left(9275\right)-(245{)}^2}$$ $$b_{yx}=\frac{141050-135975}{64925-60025}=\frac{5075}{4900}\approx 1.036$$

Step 3: Calculate the regression coefficient X on Y ($b_{xy}$).
$$b_{xy}=\frac{n\sum XY-(\sum X)(\sum Y)}{n\sum Y^2-(\sum Y{)}^2}$$ $$b_{xy}=\frac{7\left(20150\right)-\left(245\right)\left(555\right)}{7\left(44783\right)-(555{)}^2}$$

(i) Find the two regression coefficients.
Using the relation between correlation coefficient and standard deviations: $$b_{yx}=r\times \frac{{\sigma }_y}{{\sigma }_x}=0.9\times \frac{8}{5}=0.9\times 1.6=1.44$$ $$b_{xy}=r\times \frac{{\sigma }_x}{{\sigma }_y}=0.9\times \frac{5}{8}=0.9\times 0.625=0.5625$$

(ii) Average increase in profit:
The average change in profit (Y) for a unit increase in capital employed (X) is given by the regression coefficient of Y on X ($b_{yx}$).
Therefore, the average increase is 1.44 crores.

(iii) Which is more variable?
To compare variability, we calculate the Coefficient of Variation (CV) for both variables: $$CV=\frac{SD}{Mean}\times 100$$ $$C{V}_x=\frac{5}{72}\times 100\approx 6.94\%$$ $$C{V}_y=\frac{8}{3}\times 100\approx 266.67\%$$ Since $C{V}_y>C{V}_x$, Profit earned (Y) is substantially more variable.

Step 1: Use the standard deviation method formula.
When deviations are given, the regression coefficients are calculated as: $$b_{yx}=\frac{n\sum dxdy-(\sum dx)(\sum dy)}{n\sum d{x}^2-(\sum dx{)}^2}$$ $$b_{xy}=\frac{n\sum dxdy-(\sum dx)(\sum dy)}{n\sum d{y}^2-(\sum dy{)}^2}$$

Step 2: Substitute the provided values.
Since $\sum dx=0$ and $\sum dy=0$, the formulas simplify directly to: $$b_{yx}=\frac{10\left(82\right)-0}{10\left(112\right)-0}=\frac{820}{1120}=\frac{82}{112}\approx 0.732$$ $$b_{xy}=\frac{10\left(82\right)-0}{10\left(97\right)-0}=\frac{820}{970}=\frac{82}{97}\approx 0.845$$

Step 1: Identify the correct and incorrect pairs.
Correct pairs to ADD: (8, 12) and (6, 8)
Incorrect pairs to SUBTRACT: (6, 14) and (8, 6)

Step 2: Calculate the corrected statistical sums.
$\sum X_{\text{new}}=125-6-8+8+6=125$
$\sum Y_{\text{new}}=100-14-6+12+8=100$
$\sum X_{\text{new}}^2=650-6^2-8^2+8^2+6^2=650$
$\sum X{Y}_{\text{new}}=508-(6\times 14)-(8\times 6)+(8\times 12)+(6\times 8)$
$\sum X{Y}_{\text{new}}=508-84-48+96+48=508-132+144=520$

Step 3: Calculate the corrected regression coefficient $b_{yx}$.
$$b_{yx}=\frac{n\sum X{Y}_{\text{new}}-(\sum X_{\text{new}})(\sum Y_{\text{new}})}{n\sum X_{\text{new}}^2-(\sum X_{\text{new}}{)}^2}$$ $$b_{yx}=\frac{25\left(520\right)-\left(125\right)\left(100\right)}{25\left(650\right)-(125{)}^2}$$ $$b_{yx}=\frac{13000-12500}{16250-15625}=\frac{500}{625}=0.8$$

Step 1: Simplify the data using substitutions.
Let $x=X-60$ and $y=Y-79$. The given data translates to:
$\sum x^2=36$, $\sum y^2=44$, $\sum xy=24$, and $\sum x=\sum y=0$.

Because $\sum x=0$ and $\sum y=0$, we know that 60 and 79 are the actual arithmetic means $\overline{X}$ and $\overline{Y}$.

Step 2: Calculate the regression coefficients.
Using the direct formula for deviations from the mean: $$b_{yx}=\frac{\sum xy}{\sum x^2}=\frac{24}{36}=\frac{2}{3}\approx 0.667$$ $$b_{xy}=\frac{\sum xy}{\sum y^2}=\frac{24}{44}=\frac{6}{11}\approx 0.545$$

Step 3: Compare to determine which variable changes more.
The coefficient $b_{yx}$ indicates the change in Y for a unit change in X. The coefficient $b_{xy}$ indicates the change in X for a unit change in Y.
Since $\left|b_{yx}\right|>\left|b_{xy}\right|$ (0.667 > 0.545), Y changes more for a unit change in X than X does for a unit change in Y.

Step 1: Find the number of observations (n).
We know that $\overline{X}=\frac{\sum X}{n}$. Given $\sum X=50$ and $\overline{X}=5$, we have: $$5=\frac{50}{n}\Rightarrow n=10$$

Step 2: Calculate the regression coefficients using the raw score formula.
For Y on X: $$b_{yx}=\frac{n\sum XY-(\sum X)(\sum Y)}{n\sum X^2-(\sum X{)}^2}$$ $$b_{yx}=\frac{10\left(350\right)-\left(50\right)\left(60\right)}{10\left(300\right)-(50{)}^2}=\frac{3500-3000}{3000-2500}=\frac{500}{500}=1$$
For X on Y: $$b_{xy}=\frac{n\sum XY-(\sum X)(\sum Y)}{n\sum Y^2-(\sum Y{)}^2}$$ $$b_{xy}=\frac{500}{10\left(420\right)-(60{)}^2}=\frac{500}{4200-3600}=\frac{500}{600}=\frac{5}{6}\approx 0.833$$

The regression line passes through the point $(\overline{x},\overline{y})$. Substituting $\overline{y}=32$: $$8\overline{x}+5\left(32\right)=240\Rightarrow 8\overline{x}+160=240\Rightarrow 8\overline{x}=80\Rightarrow \overline{x}=10$$

Observing the changes: $\frac{\Delta x}{\Delta p}=\frac{3}{2}=1.5$ universally across all data points. This denotes a perfect linear relationship ($x=1.5p+9$) with a positive slope. Thus, the correlation coefficient $r$ is precisely 1.

Assume $x+4y=34$ is $y$ on $x$. Then $y=-0.25x+8.5$ ($b_{yx}=-0.25$).
Assume $9x+5y=74$ is $x$ on $y$. Then $x=-\frac{5}{9}y+\frac{74}{9}$ ($b_{xy}=-\frac{5}{9}$).
Product $r^2=(-0.25)\times (-\frac{5}{9})=\frac{5}{36}\le 1$. This is correct. Using $y$ on $x$ to predict $y$ at $x=6$:
$6+4y=34\Rightarrow 4y=28\Rightarrow y=7$.

Regression coefficients scale proportionally: $b_{vu}=\frac{\text{scale of}v}{\text{scale of}u}\times b_{yx}$.
The scale of $u$ is 2, the scale of $v$ is -2. Given $b_{yx}=1.5$.
$$b_{vu}=\frac{-2}{2}\times 1.5=-1\times 1.5=-1.5$$

$r=\pm \sqrt{b_{yx}\times b_{xy}}$. The correlation coefficient shares the sign of the regression coefficients (negative).
$$r=-\sqrt{(-2)\times (-\frac{1}{8})}=-\sqrt{\frac{2}{8}}=-\sqrt{0.25}=-0.5$$

From the equation $4y=5x+15$, we have $b_{yx}=\frac{5}{4}$.
Using $r^2=b_{yx}\times b_{xy}$ with $r=0.5$:
$$(0.5{)}^2=\frac{1}{4}=\frac{5}{4}\times b_{xy}\Rightarrow b_{xy}=\frac{1}{4}\times \frac{4}{5}=\frac{1}{5}$$

Using the definition of the regression coefficient: $b_{yx}=r\frac{{\sigma }_y}{{\sigma }_x}$
$$0.64=0.8\times \frac{{\sigma }_y}{5}$$ $${\sigma }_y=\frac{0.64\times 5}{0.8}=\frac{3.2}{0.8}=4$$

The intersection of the two regression lines gives $(\overline{X},\overline{Y})$.
From $2Y-X=50$, we have $X=2Y-50$.
Substitute into $3Y-2X=10$: $$3Y-2(2Y-50)=10\Rightarrow 3Y-4Y+100=10\Rightarrow Y=90$$ Now find $X$: $X=2\left(90\right)-50=180-50=130$.

In a regression equation $Y=a+bX$, the coefficient $b$ is the slope ($b_{yx}$). This explicitly represents the expected or average change in $Y$ corresponding to a one-unit change in $X$.

The regression line of $X$ on $Y$ must pass through the point $(\overline{X},\overline{Y})$. Substitute $\overline{Y}=20$ to find $\overline{X}$:
$$\overline{X}=0.5\left(20\right)+10=10+10=20$$

The slope in the regression equation of Y on X (which is 0.65 here) mathematically describes the expected change in the dependent variable (Y) for every single unit increase in the independent variable (X).

Step 1: Analyze Assertion (A).
The regression equation of X on Y is used to predict X given Y and yields the regression coefficient $b_{xy}$. The regression coefficient $b_{yx}$ corresponds to the regression equation of Y on X. Therefore, you indeed cannot determine $b_{yx}$ directly from the equation of X on Y alone. Assertion (A) is TRUE.

Step 2: Analyze Reason (R).
In the regression equation X on Y, Y acts as the independent variable and X as the dependent variable. In the equation Y on X, X is independent and Y is dependent. The roles interchange. Reason (R) is TRUE.

Step 3: Check Explanation.
Although both statements are true, the real reason we cannot determine $b_{yx}$ from X on Y is that the two coefficients mathematically represent different slopes derived from minimizing different errors (horizontal vs vertical distances), not merely because the labels swap. Hence, (R) is true but not the direct, complete mathematical explanation for (A).

Step 1: Analyze Reason (R).
The mathematical formula for the angle $\theta$ between two lines of regression is indeed given by: $$\tan \theta =\frac{1-r^2}{\left|r\right|}\left(\frac{{\sigma }_x{\sigma }_y}{{\sigma }_x^2+{\sigma }_y^2}\right)$$ This statement is TRUE.

Step 2: Analyze Assertion (A).
If the lines are perpendicular (angle = ${90}^{\circ }$), then $\tan \theta =\tan {90}^{\circ }=\infty$.
Looking at the formula in (R), the expression approaches infinity when the denominator is 0. That happens when $\left|r\right|=0$, which means $r=0$.
Thus, if the angle is ${90}^{\circ }$, there is no correlation ($r=0$). Assertion (A) is TRUE.

Step 1: Analyze the Reason (R).
We know from the properties of regression analysis that the correlation coefficient ($r$) is exactly the geometric mean of the two regression coefficients ($b_{yx}$ and $b_{xy}$). $$r=\pm \sqrt{b_{yx}\times b_{xy}}$$ This statement is factually TRUE.

Step 2: Analyze the Assertion (A).
Because the correlation coefficient $r$ always lies in the range $[-1,1]$, its square $r^2$ must be less than or equal to 1 ($r^2\le 1$).
Therefore, the product of the regression coefficients must be: $$b_{yx}\times b_{xy}\le 1$$ If one regression coefficient (e.g., $b_{yx}$) is strictly greater than 1, then for their product to remain less than or equal to 1, the other regression coefficient ($b_{xy}$) must be strictly less than 1.
It is mathematically impossible for both regression coefficients to be greater than 1 simultaneously. Hence, the Assertion (A) is FALSE.

Step 1: Analyze Assertion (A).
The signs of $b_{yx}$, $b_{xy}$, and $r$ are identically linked to the covariance between X and Y. Thus, they must all be positive or all be negative. Assertion (A) is TRUE.

Step 2: Analyze Reason (R).
We know that the product of the two regression coefficients is bounded by 1: $b_{yx}\times b_{xy}\le 1$.
If one is greater than 1, the other must be less than 1. However, if one is less than 1 (say 0.5), the other does NOT have to be greater than 1 (it could also be 0.5). Therefore, the statement "is always greater than one" makes Reason (R) FALSE.

Let $X$ represent marks in Statistics and $Y$ represent marks in Mathematics. Number of observations, $N=8$.

Step 1: Calculate the arithmetic means.
$\sum X=25+35+32+31+36+34+42+45=280$
$\overline{X}=\frac{280}{8}=35$

$\sum Y=43+44+49+41+35+30+32+46=320$
$\overline{Y}=\frac{320}{8}=40$

Step 2: Construct the calculation table.
Let $x=X-\overline{X}$ and $y=Y-\overline{Y}$.

(a) Calculating the coefficient of correlation ($r$):
Using the formula for direct deviation method: $$r=\frac{\sum xy}{\sqrt{\sum x^2\times \sum y^2}}$$ $$r=\frac{-52}{\sqrt{276\times 332}}=\frac{-52}{\sqrt{91632}}\approx \frac{-52}{302.71}\approx -0.1718$$

(b) Calculating the most likely marks in Statistics ($X$) when Mathematics ($Y$) is 48:
We need to formulate the regression equation of $X$ on $Y$. First, find the regression coefficient $b_{xy}$: $$b_{xy}=\frac{\sum xy}{\sum y^2}=\frac{-52}{332}\approx -0.1566$$ The regression equation of $X$ on $Y$ is given by: $$X-\overline{X}=b_{xy}(Y-\overline{Y})$$ Substitute the known values: $$X-35=-0.1566(Y-40)$$ Now, to estimate $X$ when $Y=48$: $$X-35=-0.1566(48-40)$$ $$X-35=-0.1566\left(8\right)$$ $$X-35=-1.2528$$ $$X=35-1.2528=33.7472$$

Let $X$ represent height of mothers and $Y$ represent height of daughters. Number of observations, $N=8$.

Step 1: Calculate the arithmetic means.
$\sum X=148+149+150+152+153+155+157+160=1224$
$\overline{X}=\frac{1224}{8}=153$

$\sum Y=120+125+122+128+130+140+133+150=1048$
$\overline{Y}=\frac{1048}{8}=131$

Step 2: Construct the calculation table to find deviations.
Let $x=X-\overline{X}$ and $y=Y-\overline{Y}$.

(a) Line of regression of mother ($X$) on daughter ($Y$):
First, find the regression coefficient $b_{xy}$: $$b_{xy}=\frac{\sum xy}{\sum y^2}=\frac{268}{694}\approx 0.38616$$ The equation for the line of regression of $X$ on $Y$ is: $$X-\overline{X}=b_{xy}(Y-\overline{Y})$$ $$X-153=0.386(Y-131)$$

(b) Estimating height of mother ($X$) when height of daughter ($Y$) is 145 cm:
Substitute $Y=145$ into the regression equation: $$X-153=0.386(145-131)$$ $$X-153=0.386\left(14\right)$$ $$X-153=5.404$$ $$X=153+5.404=158.404$$

Let $X$ represent height and $Y$ represent weight. Number of observations, $N=10$.

Step 1: Calculate the arithmetic means.
$\sum X=61+62+58+63+64+65+64+67+69+72=645$
$\overline{X}=\frac{645}{10}=64.5$

$\sum Y=112+115+108+118+120+125+119+125+130+140=1212$
$\overline{Y}=\frac{1212}{10}=121.2$

Step 2: Use Assumed Means to simplify calculations.
Let assumed mean for $X$ be $A=64$ and for $Y$ be $B=120$.
Define deviations: $dx=X-64$ and $dy=Y-120$.

Step 3: Calculate the regression coefficient of Y on X ($b_{yx}$):
Since we need to predict weight ($Y$) given height ($X$), we compute $b_{yx}$ using the shortcut formula: $$b_{yx}=\frac{N\sum dxdy-(\sum dx)(\sum dy)}{N\sum d{x}^2-(\sum dx{)}^2}$$ $$b_{yx}=\frac{10\left(338\right)-\left(5\right)\left(12\right)}{10\left(149\right)-(5{)}^2}$$ $$b_{yx}=\frac{3380-60}{1490-25}=\frac{3320}{1465}\approx 2.2662$$

Step 4: Formulate regression line Y on X and predict Y.
The equation is: $$Y-\overline{Y}=b_{yx}(X-\overline{X})$$ $$Y-121.2=2.2662(X-64.5)$$ Substitute the requested height $X=70$ inches: $$Y=121.2+2.2662(70-64.5)$$ $$Y=121.2+2.2662\left(5.5\right)$$ $$Y=121.2+12.4641=133.6641$$

Step 1: Calculate the regression coefficients.
Using standard deviations, the regression coefficient of $Y$ on $X$ ($b_{yx}$) is: $$b_{yx}=r\frac{S_y}{S_x}=0.8\times \frac{3}{2}=0.8\times 1.5=1.2$$ The regression coefficient of $X$ on $Y$ ($b_{xy}$) is: $$b_{xy}=r\frac{S_x}{S_y}=0.8\times \frac{2}{3}=\frac{1.6}{3}\approx 0.533$$

Step 2: Form the regression equation of Y on X. $$Y-\overline{Y}=b_{yx}(X-\overline{X})$$ $$Y-40=1.2(X-50)$$ $$Y-40=1.2X-60\Rightarrow Y=1.2X-20$$

Step 3: Form the regression equation of X on Y. $$X-\overline{X}=b_{xy}(Y-\overline{Y})$$ $$X-50=0.533(Y-40)$$ $$X-50=0.533Y-21.32\Rightarrow X=0.533Y+28.68$$

Step 1: Identify the variables and knowns.
Let $X$ be Rainfall and $Y$ be Production Yield. We need to estimate $Y$ given $X=36$. We must find the regression equation of Y on X.
$\overline{X}=25$, ${\sigma }_x=4$
$\overline{Y}=40$, ${\sigma }_y=6$
$r=0.9$

Step 2: Calculate $b_{yx}$. $$b_{yx}=r\frac{{\sigma }_y}{{\sigma }_x}=0.9\times \frac{6}{4}=0.9\times 1.5=1.35$$

Step 3: Construct the regression line and estimate $Y$. $$Y-\overline{Y}=b_{yx}(X-\overline{X})$$ $$Y-40=1.35(X-25)$$ Substitute $X=36$: $$Y=40+1.35(36-25)$$ $$Y=40+1.35\left(11\right)$$ $$Y=40+14.85=54.85$$

Step 1: Understand the objective.
We are asked to estimate $X$ when $Y$ is given. This requires the regression line of X on Y.

Step 2: Calculate the means.
Given $N=5$.
$\overline{X}=\frac{\sum X}{N}=\frac{15}{5}=3$
$\overline{Y}=\frac{\sum Y}{N}=\frac{25}{5}=5$

Step 3: Calculate the regression coefficient $b_{xy}$.
Using raw data formula: $$b_{xy}=\frac{N\sum XY-(\sum X)(\sum Y)}{N\sum Y^2-(\sum Y{)}^2}$$ $$b_{xy}=\frac{5\left(83\right)-\left(15\right)\left(25\right)}{5\left(135\right)-(25{)}^2}$$ $$b_{xy}=\frac{415-375}{675-625}=\frac{40}{50}=0.8$$

Step 4: Formulate the regression line and estimate X.
Equation of X on Y: $$X-\overline{X}=b_{xy}(Y-\overline{Y})$$ $$X-3=0.8(Y-5)$$ Substitute $Y=12$: $$X=3+0.8(12-5)$$ $$X=3+0.8\left(7\right)=3+5.6=8.6$$

Step 1: Find the means by solving the equations simultaneously.
Since the regression lines intersect at $(\overline{X},\overline{Y})$, we solve:
(1) $X=2Y-50$
(2) $2X=3Y-10$
Substitute (1) into (2): $$2(2Y-50)=3Y-10$$ $$4Y-100=3Y-10\Rightarrow Y=90$$ Substitute $Y=90$ into (1): $$X=2\left(90\right)-50=180-50=130$$ So, $\overline{X}=130$ and $\overline{Y}=90$.

Step 2: Identify the regression equations to find correlation ($r$).
Assume $2Y-X-50=0$ is the line of Y on X. Rearranging for Y: $$2Y=X+50\Rightarrow Y=0.5X+25$$ Thus, $b_{yx}=0.5$.
Assume $3Y-2X-10=0$ is the line of X on Y. Rearranging for X: $$2X=3Y-10\Rightarrow X=1.5Y-5$$ Thus, $b_{xy}=1.5$.
Check validity: $r^2=b_{yx}\times b_{xy}=0.5\times 1.5=0.75$. Since $0.75\le 1$, our assumption is correct.

Step 3: Calculate $r$. $$r=\pm \sqrt{0.75}$$ Since both coefficients are positive, $r$ must be positive: $$r=\sqrt{0.75}\approx 0.866$$

(i) Arithmetic means:
Solve the system: (1) $2x+3y=8$ and (2) $x+2y=5\Rightarrow x=5-2y$.
Substitute $x$ into (1): $$2(5-2y)+3y=8$$ $$10-4y+3y=8\Rightarrow y=2$$ So, $x=5-2\left(2\right)=1$. Therefore, $\overline{x}=1$, $\overline{y}=2$.

(ii) Identify regression equation of y on x:
Assume $2y=5-x$ is y on x. Then $y=-0.5x+2.5$, so $b_{yx}=-0.5$.
Assume $2x=8-3y$ is x on y. Then $x=-1.5y+4$, so $b_{xy}=-1.5$.
$r^2=(-0.5)\times (-1.5)=0.75\le 1$. This is valid. The equation y on x is $2y=5-x$.

(iii) Compute correlation coefficient:
$$r=-\sqrt{0.75}\approx -0.866$$ (Negative because coefficients are negative).

(iv) Find standard deviation of y:
Given $S_x^2=16\Rightarrow S_x=4$.
We know $b_{yx}=r\frac{S_y}{S_x}$. $$-0.5=-0.866\left(\frac{S_y}{4}\right)$$ $$S_y=\frac{4\times 0.5}{0.866}=\frac{2}{0.866}\approx 2.31$$ (Alternatively, using $\frac{b_{yx}}{b_{xy}}=\frac{S_y^2}{S_x^2}$: $\frac{-0.5}{-1.5}=\frac{1}{3}\Rightarrow S_y^2=\frac{16}{3}\Rightarrow S_y=\frac{4}{\sqrt{3}}\approx 2.31$)

Step 1: Identify the correct regression lines.
Assume $8x+10y=25$ is the line of Y on X. Rearranging gives $10y=-8x+25\Rightarrow y=-0.8x+2.5$. So $b_{yx}=-0.8$.
Assume $16x+5y=12$ is the line of X on Y. Rearranging gives $16x=-5y+12\Rightarrow x=-\frac{5}{16}y+\frac{12}{16}$. So $b_{xy}=-0.3125$.
Product $r^2=(-0.8)\times (-0.3125)=0.25\le 1$. The assumption is correct.

Step 2: Use the formula linking variances and regression coefficients.
We know: $$\frac{b_{yx}}{b_{xy}}=\frac{r\frac{S_y}{S_x}}{r\frac{S_x}{S_y}}=\frac{S_y^2}{S_x^2}$$ Given $S_x^2=25$: $$\frac{-0.8}{-0.3125}=\frac{S_y^2}{25}$$ $$2.56=\frac{S_y^2}{25}$$ $$S_y^2=2.56\times 25=64$$ $$S_y=\sqrt{64}=8$$

(i) Direction and Degree:
If the scatter plot points are densely clustered along a straight line that slopes upwards from left to right, this indicates a strong, positive (or direct) correlation. As the distance travelled increases, the card charges also increase proportionally.

(ii) Regression Equation:
Using the provided standard answers from the text analysis, the regression equation of Y on X can be framed using the computed slope $b_{yx}=1.33$ and means $\overline{X}=3.16$, $\overline{Y}=4.28$: $$Y-\overline{Y}=b_{yx}(X-\overline{X})$$ $$Y-4.28=1.33(X-3.16)$$ $$Y=1.33X-4.2028+4.28\Rightarrow Y=1.33X+0.08$$

(iii) Prediction:
Substitute $X=6$ (since X is measured in thousands of miles) into the equation: $$Y=1.33\left(6\right)+0.08$$ $$Y=7.98+0.08=8.06$$ Since Y is measured in thousands of dollars, the predicted charge is $8.06\times 1000$.