Table 2.2 shows the scores of candidates in an entrance examination into a college. (a) Prepare a cumulative frequency table and draw the cumulative frequency curve for the distribution. (b) Use your curve to estimate the: Median mark, Upper and lower quartiles, Semi-inter quartile range.

The latitude ($\alpha$) is $38^\circ$. The Earth's radius (R) is 6400 km.

The Earth rotates $2\pi$ radians in 24 hours.

The speed of X due to the rotation of the earth is $\boxed{1320 \text{ km/h}}$.

10. (b) Distance between X and Y along the great circle

Longitude of X = $50^\circ E$. Longitude of Y = $130^\circ W$. Since they are on opposite sides, add the longitudes.

Using $\pi = \frac{22}{7}$:

The distance between X and Y along the great circle is $\boxed{20100 \text{ km}}$.

10. (c) Distance between X and Y along the parallel of latitude

From part (a), $r \approx 5043.26$ km.
From part (b), $\theta = \pi$ radians.

Using $\pi = \frac{22}{7}$:

The distance between X and Y along the parallel of latitude is $\boxed{15900 \text{ km}}$.

11. (a) Copy and complete Table 2.1 for the relation $y = 2x^2 - 5x - 6$, for $-2 \le x \le 5$.

For $x = -1$: $y = 2(-1)^2 - 5(-1) - 6 = 2(1) + 5 - 6 = 1$.
For $x = 0$: $y = 2(0)^2 - 5(0) - 6 = 0 - 0 - 6 = -6$.
For $x = 1$: $y = 2(1)^2 - 5(1) - 6 = 2 - 5 - 6 = -9$.
For $x = 4$: $y = 2(4)^2 - 5(4) - 6 = 2(16) - 20 - 6 = 32 - 20 - 6 = 6$.

11. (b) Using a scale of 2 cm to represent 1 unit on x-axis and 2 cm to represent 5 units on y-axis, draw the graph of $y = 2x^2 - 5x - 6$ for $-2 \le x \le 5$.

This step requires drawing a graph, which cannot be performed by an AI. You would plot the points from the table in part (a) on graph paper using the specified scale and draw a smooth curve through them.

11. (c) Use your graph to find the:

(i) Roots of the equation $2x^2 - 5x - 6 = 0$, correct to one decimal place.
These are the x-intercepts of the graph $y = 2x^2 - 5x - 6$. You would read these values from your graph where the curve crosses the x-axis ($y=0$).
Using the quadratic formula for $2x^2 - 5x - 6 = 0$:

The roots are approximately $\boxed{x = 3.4 \text{ and } x = -0.9}$.

(ii) Values of $x$ for which $2x^2 - 5x - 8 = 0$, correct to one decimal place.
This equation can be rewritten as $2x^2 - 5x - 6 = 2$. You would draw the line $y = 2$ on your graph and find the x-coordinates of the intersection points with the curve $y = 2x^2 - 5x - 6$.
Using the quadratic formula for $2x^2 - 5x - 8 = 0$:

The values of $x$ are approximately $\boxed{x = 3.6 \text{ and } x = -1.1}$.

(iii) Gradient of the tangent at the point where $x = 2$.
You would draw a tangent line to the curve at $x=2$ on your graph and calculate its slope.
Alternatively, using calculus:

The gradient of the tangent at the point where $x=2$ is $\boxed{3}$.

12. (a) Prepare a cumulative frequency table and draw the cumulative frequency curve for the distribution.

Use upper class boundaries for plotting the curve.

The points to plot are (30.5, 9), (40.5, 29), (50.5, 60), (60.5, 102), (70.5, 152), (80.5, 184), (90.5, 200). Start the curve from (20.5, 0).

12. (b) Use your curve to estimate the:
Total number of students (N) = 200.

(i) Median mark
The median is at the $\frac{N}{2}$-th position, which is $\frac{200}{2} = 100$-th student.

• Graphical estimation: Locate 100 on the cumulative frequency axis, draw a horizontal line to the curve, then a vertical line down to the marks axis. Read the value.
• Analytical estimation (linear interpolation): The 100th student falls in the 51-60 class. $$\text{Median} = L + \left( \frac{\frac{N}{2} - cf_b}{f_m} \right) c$$ $$\text{Median} = 50.5 + \left( \frac{100 - 60}{42} \right) 10$$ $$\text{Median} = 50.5 + \left( \frac{40}{42} \right) 10 \approx 50.5 + 9.52 \approx 60.02$$

The median mark is approximately $\boxed{60.0}$.

(ii) Upper and lower quartiles

• Lower Quartile ($Q_1$): Position = $\frac{N}{4} = \frac{200}{4} = 50$-th student.

  • Graphical estimation: Locate 50 on the cumulative frequency axis, draw a horizontal line to the curve, then a vertical line to the marks axis. Read the value.
  • Analytical estimation (linear interpolation): The 50th student falls in the 41-50 class. $$Q_1 = L + \left( \frac{\frac{N}{4} - cf_b}{f_{Q1}} \right) c$$ $$Q_1 = 40.5 + \left( \frac{50 - 29}{31} \right) 10$$ $$Q_1 = 40.5 + \left( \frac{21}{31} \right) 10 \approx 40.5 + 6.77 \approx 47.27$$

  The lower quartile is approximately $\boxed{47.3}$.

• Upper Quartile ($Q_3$): Position = $\frac{3N}{4} = \frac{3 \times 200}{4} = 150$-th student.

  • Graphical estimation: Locate 150 on the cumulative frequency axis, draw a horizontal line to the curve, then a vertical line to the marks axis. Read the value.
  • Analytical estimation (linear interpolation): The 150th student falls in the 61-70 class. $$Q_3 = L + \left( \frac{\frac{3N}{4} - cf_b}{f_{Q3}} \right) c$$ $$Q_3 = 60.5 + \left( \frac{150 - 102}{50} \right) 10$$ $$Q_3 = 60.5 + \left( \frac{48}{50} \right) 10 \approx 60.5 + 9.6 \approx 70.1$$

  The upper quartile is approximately $\boxed{70.1}$.

(iii) Semi-inter quartile range (SIQR)

The semi-inter quartile range is $\boxed{11.4}$.