SECTION B (40 MARKS) QUESTION FIVE In a competition to grow the tallest hollyhock, the heights recorded by 50 primary school children were as follows. Height were measured to the nearest centimetre.

$$\int_{-\infty}^{\infty} f(x)\,dx = 1$$ $$\int_{0}^{1} (a+bx^2)\,dx = 1$$ $$\left[ ax + \frac{bx^3}{3} \right]_{0}^{1} = 1$$ $$\left( a(1) + \frac{b(1)^3}{3} \right) - \left( a(0) + \frac{b(0)^3}{3} \right) = 1$$ $$a + \frac{b}{3} = 1 \quad \text{(Equation 1)}$$

$$E(X) = \int_{-\infty}^{\infty} x f(x)\,dx = \frac{3}{5}$$ $$\int_{0}^{1} x(a+bx^2)\,dx = \frac{3}{5}$$ $$\int_{0}^{1} (ax+bx^3)\,dx = \frac{3}{5}$$ $$\left[ \frac{ax^2}{2} + \frac{bx^4}{4} \right]_{0}^{1} = \frac{3}{5}$$ $$\left( \frac{a(1)^2}{2} + \frac{b(1)^4}{4} \right) - \left( \frac{a(0)^2}{2} + \frac{b(0)^4}{4} \right) = \frac{3}{5}$$ $$\frac{a}{2} + \frac{b}{4} = \frac{3}{5} \quad \text{(Equation 2)}$$

From Equation 1, $a = 1 - \frac{b}{3}$. Substitute this into Equation 2:

$$\frac{1}{2}\left( 1 - \frac{b}{3} \right) + \frac{b}{4} = \frac{3}{5}$$ $$\frac{1}{2} - \frac{b}{6} + \frac{b}{4} = \frac{3}{5}$$

To combine the terms with $b$, find a common denominator (12):

$$\frac{1}{2} + \frac{-2b+3b}{12} = \frac{3}{5}$$ $$\frac{1}{2} + \frac{b}{12} = \frac{3}{5}$$ $$\frac{b}{12} = \frac{3}{5} - \frac{1}{2}$$

To subtract the fractions on the right, find a common denominator (10):

$$\frac{b}{12} = \frac{6}{10} - \frac{5}{10}$$ $$\frac{b}{12} = \frac{1}{10}$$ $$b = \frac{12}{10} = \frac{6}{5}$$

Now substitute $b = \frac{6}{5}$ back into $a = 1 - \frac{b}{3}$:

$$a = 1 - \frac{6/5}{3}$$ $$a = 1 - \frac{6}{15}$$ $$a = 1 - \frac{2}{5}$$ $$a = \frac{3}{5}$$

The values are $\boxed{a = \frac{3}{5}, b = \frac{6}{5}}$.

Part ii) Determine the variance of $X$.

The variance is given by $Var(X) = E(X^2) - [E(X)]^2$. We know $E(X) = \frac{3}{5}$.

$$E(X^2) = \int_{-\infty}^{\infty} x^2 f(x)\,dx$$ $$E(X^2) = \int_{0}^{1} x^2 \left( \frac{3}{5} + \frac{6}{5}x^2 \right)\,dx$$ $$E(X^2) = \int_{0}^{1} \left( \frac{3}{5}x^2 + \frac{6}{5}x^4 \right)\,dx$$ $$E(X^2) = \left[ \frac{3}{5}\frac{x^3}{3} + \frac{6}{5}\frac{x^5}{5} \right]_{0}^{1}$$ $$E(X^2) = \left[ \frac{x^3}{5} + \frac{6x^5}{25} \right]_{0}^{1}$$ $$E(X^2) = \left( \frac{1^3}{5} + \frac{6(1)^5}{25} \right) - \left( \frac{0^3}{5} + \frac{6(0)^5}{25} \right)$$ $$E(X^2) = \frac{1}{5} + \frac{6}{25}$$

To combine the fractions, find a common denominator (25):

$$E(X^2) = \frac{5}{25} + \frac{6}{25} = \frac{11}{25}$$

$$Var(X) = E(X^2) - [E(X)]^2$$ $$Var(X) = \frac{11}{25} - \left(\frac{3}{5}\right)^2$$ $$Var(X) = \frac{11}{25} - \frac{9}{25}$$ $$Var(X) = \frac{2}{25}$$

The variance of $X$ is $\boxed{\frac{2}{25}}$.