(a) $z = \pm 0.84$ gives marks limits $19$ and $61$. (b) The forward-difference polynomial is $x^3 - 2x^2 + 3x + 2$, whose derivative is $3x^2 - 4x + 3$.

Part (a) — Central 60% of marks

Mean $\mu = 40$, S.D. $\sigma = 25$. The central $60\%$ leaves $30\%$ on each side of the mean. Since $P(0 < z < 0.84) = 0.30$, the limits correspond to $z = \pm 0.84$.

$x = \mu \pm z\sigma = 40 \pm 0.84(25) = 40 \pm 21.$

Lower limit $= 40 - 21 = 19$; upper limit $= 40 + 21 = 61$.

So the central $60\%$ of the candidates score between 19 and 61 marks.

Part (b) — Newton's forward interpolation, then derivative

Data: $x = 0,1,2,3$; $f(x) = 2,4,8,20$. Difference table:

With $x_0 = 0$, $h = 1$, $p = \dfrac{x - x_0}{h} = x$, Newton's forward formula:

$f(x) = f_0 + p\,\Delta f_0 + \frac{p(p-1)}{2!}\Delta^2 f_0 + \frac{p(p-1)(p-2)}{3!}\Delta^3 f_0.$ …