Draw a cumulative frequency curve for The median mark (ii) the inter quartile range and semi interquartile range (iii) If only 5 % of the candidates attained the distinction level, estimate the lowest mark for this level.

For a $2 \times 2$ matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$, the determinant is $ad - bc$.

$$\det(A) = (4)(2) - (-3)(1) = 8 - (-3) = 8 + 3 = 11$$

The inverse of $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ is $A^{-1} = \frac{1}{\det(A)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$.

$$A^{-1} = \frac{1}{11} \begin{pmatrix} 2 & -(-3) \\ -1 & 4 \end{pmatrix} = \frac{1}{11} \begin{pmatrix} 2 & 3 \\ -1 & 4 \end{pmatrix}$$

$$A^{-1} = \begin{pmatrix} \frac{2}{11} & \frac{3}{11} \\ -\frac{1}{11} & \frac{4}{11} \end{pmatrix}$$

The inverse matrix is $\boxed{\begin{pmatrix} \frac{2}{11} & \frac{3}{11} \\ -\frac{1}{11} & \frac{4}{11} \end{pmatrix}}$.

Part 7b:
Under the linear transformation $P$ of a plane, the image of $(2,1)$ is $(4,2)$ and that of $(-7,1)$ is $(-1,7)$. Find the matrix $P$ of the transformation.

Let the transformation matrix be $P = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$.
The transformation can be written as $P \mathbf{x} = \mathbf{x}'$.
Given $P \begin{pmatrix} 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 4 \\ 2 \end{pmatrix}$ and $P \begin{pmatrix} -7 \\ 1 \end{pmatrix} = \begin{pmatrix} -1 \\ 7 \end{pmatrix}$.
These two equations can be combined into a single matrix equation:

$$P \begin{pmatrix} 2 & -7 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} 4 & -1 \\ 2 & 7 \end{pmatrix}$$

Let $X = \begin{pmatrix} 2 & -7 \\ 1 & 1 \end{pmatrix}$ and $Y = \begin{pmatrix} 4 & -1 \\ 2 & 7 \end{pmatrix}$. So, $PX = Y$.

First, calculate the determinant of $X$:

$$\det(X) = (2)(1) - (-7)(1) = 2 - (-7) = 2 + 7 = 9$$

Now, find $X^{-1}$:

$$X^{-1} = \frac{1}{9} \begin{pmatrix} 1 & -(-7) \\ -1 & 2 \end{pmatrix} = \frac{1}{9} \begin{pmatrix} 1 & 7 \\ -1 & 2 \end{pmatrix}$$

Since $PX = Y$, then $P = YX^{-1}$.

$$P = \begin{pmatrix} 4 & -1 \\ 2 & 7 \end{pmatrix} \frac{1}{9} \begin{pmatrix} 1 & 7 \\ -1 & 2 \end{pmatrix}$$ $$P = \frac{1}{9} \begin{pmatrix} 4 & -1 \\ 2 & 7 \end{pmatrix} \begin{pmatrix} 1 & 7 \\ -1 & 2 \end{pmatrix}$$

$$P = \frac{1}{9} \begin{pmatrix} (4)(1) + (-1)(-1) & (4)(7) + (-1)(2) \\ (2)(1) + (7)(-1) & (2)(7) + (7)(2) \end{pmatrix}$$ $$P = \frac{1}{9} \begin{pmatrix} 4 + 1 & 28 - 2 \\ 2 - 7 & 14 + 14 \end{pmatrix}$$ $$P = \frac{1}{9} \begin{pmatrix} 5 & 26 \\ -5 & 28 \end{pmatrix}$$

$$P = \begin{pmatrix} \frac{5}{9} & \frac{26}{9} \\ -\frac{5}{9} & \frac{28}{9} \end{pmatrix}$$

The matrix $P$ of the transformation is $\boxed{\begin{pmatrix} \frac{5}{9} & \frac{26}{9} \\ -\frac{5}{9} & \frac{28}{9} \end{pmatrix}}$.