2020 H2 Math Paper 2 Answers

{y = \frac{5}{2} x^2 - 5 x + \frac{1}{2}.}

(a)

(i)

(A) The sequence increases and diverges.
(B) The sequence is a constant sequence

{u_n = 5}

for all

{n \in \mathbb{ Z }^+.}

(ii)

{p=11.}

(b)

(i)

{b=7}

(ii)

{v_5 = 5 a + 28}

(c)

(i)

{3 n^2 - 25 n + 16.}

(ii)

{m = 10.}

(a)

{2 x + y = 39.}

(b)

{\frac{2057}{18} \textrm{ units}^2.}

(c)

(i)

{\frac{2057}{3} \textrm{ units}^2.}

(ii)

{x = \frac{1}{54} \left( y + 3 \right)^{ 2 } + 4.}

(b)

{144 \sqrt{5} \textrm{ cm}^3.}

(c)

(i)

{a = 15.}

(ii)

The net forms a square with sides

{15 \sqrt{2} \textrm{ cm}.}

(a)

{0, 4, 10, 25.}

(b)

{\textrm{Var}(S) = \frac{360 r^2 + 300 r - 340}{(3 r + 1)^2}.}

(c)

{r=3.}

(a)

(b)

{\textrm{P}(T>6) = 0.202.}

(c)

{\textrm{P}(T+W > 30) = 0.108.}

(d)

{\textrm{P}(\textrm{fine} \mid \textrm{late}) = 0.606.}

To be updated

(a)

\textrm{P}(R=1)=\frac{1}{1789515} \allowbreak {< \frac{88}{1789515}} = \textrm{P}(R=2).

(b)

{r=6.}

(a)

Each pen from the box has an equal probability of being selected into the sample of 10.

(b)

{\textrm{P}(X \leq 2) = 0.981.}

(c)

{0.0535.}

(d)

{0.983.}

(e)

Both testing procedures have a similar probability of having a randomly chosen box accepted for sale, but the alternative testing procedure tests less pens on average so it may be cheaper to implement.

(a)

\{ \overline{x} \in \mathbb{R}: \allowbreak {\overline{x} \leq 1.47 \textrm{ or } \overline{x} \geq 1.53 \}.}

(b)

In the earlier test, the percentage of carbon in the steel bars,

{X,}

are known to be normally distributed so

{\overline{X}}

will still be normally distributed.
In the new test, we no longer know if the amount of carbon in each bar is normally distributed. Moreover, the population standard deviation is not know. Hence a large sample size of 40 ensures that the sample mean amount of carbon in each bar

{\overline{Y}}

will be normally distributed approximately by the Central Limit Theorem.

(c)

{\overline{x} = 0.254, s^2 = 0.000146.}

(d)

{p \textrm{-value} = 0.0182 \leq 0.025} \allowbreak \, \Rightarrow \, \allowbreak {H_0 \textrm{ rejected}.}

Hence there is sufficient evidence at the 2.5% level of significance to conclude that the mean amount of carbon in the flat bars is more than 0.25%.

Solutions

Full solutions (with working)