1. **Problem (a): Draw Histogram and Estimate Mode** Given class intervals and frequencies: | C.I | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | | Frequency | 12 | 20 | 26 | 18 | 10 | 6 | The histogram consists of 6 bars with heights equal to the frequencies. To estimate the mode, use the formula for grouped data: $$\text{Mode} = L + \frac{(f_1 - f_0)}{(2f_1 - f_0 - f_2)} \times h$$ where: - $L$ = lower boundary of modal class - $f_1$ = frequency of modal class - $f_0$ = frequency of class before modal class - $f_2$ = frequency of class after modal class - $h$ = class width Here, modal class is 20-30 with $f_1=26$, $f_0=20$ (10-20), $f_2=18$ (30-40), $L=20$, $h=10$. Calculate: $$\text{Mode} = 20 + \frac{(26 - 20)}{(2 \times 26 - 20 - 18)} \times 10 = 20 + \frac{6}{(52 - 38)} \times 10 = 20 + \frac{6}{14} \times 10 = 20 + 4.29 = 24.29$$ 2. **Problem (b): Find Mean Using Step Deviation Method** Given marks and frequencies: | Marks | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 | 80-90 | 90-100 | | Students | 8 | 12 | 24 | 16 | 9 | 7 | 4 | Step 1: Find midpoints ($x_i$) of each class: $$35, 45, 55, 65, 75, 85, 95$$ Step 2: Choose assumed mean $A = 65$ (midpoint of 60-70 class). Step 3: Calculate deviations $d_i = \frac{x_i - A}{h}$ where $h=10$ (class width): $$d_i = \frac{35-65}{10} = -3, \frac{45-65}{10} = -2, \frac{55-65}{10} = -1, \frac{65-65}{10} = 0, \frac{75-65}{10} = 1, \frac{85-65}{10} = 2, \frac{95-65}{10} = 3$$ Step 4: Multiply $f_i$ by $d_i$: $$8 \times (-3) = -24, 12 \times (-2) = -24, 24 \times (-1) = -24, 16 \times 0 = 0, 9 \times 1 = 9, 7 \times 2 = 14, 4 \times 3 = 12$$ Step 5: Sum frequencies and $f_i d_i$: $$\sum f_i = 8 + 12 + 24 + 16 + 9 + 7 + 4 = 80$$ $$\sum f_i d_i = -24 - 24 - 24 + 0 + 9 + 14 + 12 = -37$$ Step 6: Calculate mean: $$\text{Mean} = A + h \times \frac{\sum f_i d_i}{\sum f_i} = 65 + 10 \times \frac{-37}{80} = 65 - 4.625 = 60.375$$ 3. **Problem (c): Matrix X** Given: $$\begin{bmatrix}6 & -2 \\ 2 & 3\end{bmatrix} X = \begin{bmatrix}10 \\ -4\end{bmatrix}$$ (i) The matrix on the left is $2 \times 2$, and the result is $2 \times 1$, so $X$ must be $2 \times 1$. (ii) To find $X$, solve: $$\begin{bmatrix}6 & -2 \\ 2 & 3\end{bmatrix} \begin{bmatrix}x_1 \\ x_2\end{bmatrix} = \begin{bmatrix}10 \\ -4\end{bmatrix}$$ This gives the system: $$6x_1 - 2x_2 = 10$$ $$2x_1 + 3x_2 = -4$$ Multiply second equation by 3: $$6x_1 + 9x_2 = -12$$ Subtract first equation: $$(6x_1 + 9x_2) - (6x_1 - 2x_2) = -12 - 10$$ $$11x_2 = -22$$ $$x_2 = -2$$ Substitute $x_2$ into first equation: $$6x_1 - 2(-2) = 10$$ $$6x_1 + 4 = 10$$ $$6x_1 = 6$$ $$x_1 = 1$$ So, $$X = \begin{bmatrix}1 \\ -2\end{bmatrix}$$