2017 H2 Maths Paper 2 Q10 — Worked Solution

Part (i)

1 mark

Let $X$ be the mass, in grams, of a randomly selected sphere. Then $X \sim N\left( 20,\ {0.5}^{2} \right)$ .

Standardising,

P(X > 20.2)

\Rightarrow P\left( Z > \frac{20.2 - 20}{0.5} \right)

\Rightarrow P(Z > 0.4)

= 0.345\ \left( 3\ \text{s.f.} \right)\ \ \text{(by GC)}

Part (ii)

3 marks

Approach

Coating increases the mass of each sphere by 10%, so the coated mass is a linear transformation, $1.1X$ , of the uncoated mass, not $X + 10\%\ \text{of a fixed amount}$ .

Let $Y$ be the mass, in grams, of a coated sphere, so $Y = 1.1X$ . Since $X \sim N\left( 20,\ {0.5}^{2} \right)$ ,

Y \sim N\left( 1.1(20),\ (1.1 \times 0.5)^{2} \right) = N\left( 22,\ {0.55}^{2} \right)

P(21.5 < Y < 22.45) = 0.612\ \left( 3\ \text{s.f.} \right)\ \ \text{(by GC)}

$\therefore$ $Y \sim N(22,\ 0.3025)$ , and $P(21.5 < Y < 22.45) = 0.612$ (3 s.f.).

Part (iii)

4 marks

Let $B$ be the mass, in grams, of a randomly selected metal bar, and let $B \sim N\left( \mu,\ \sigma^{2} \right)$ .

P(B > 12.2) = 0.6\quad\quad\text{and}\quad\quad P(B < 12) = 0.25

By GC (inverse normal),

\frac{12.2 - \mu}{\sigma} = - 0.253347\ldots\quad\quad\text{and}\quad\quad\frac{12 - \mu}{\sigma} = - 0.674490\ldots

12.2 = \mu - 0.253347\sigma\quad\quad\text{...(1)}

12 = \mu - 0.674490\sigma\quad\quad\text{...(2)}

Solving (1) and (2) simultaneously (by GC),

\mu = 12.3\ \left( 3\ \text{s.f.} \right),\quad\quad\sigma = 0.475\ \left( 3\ \text{s.f.} \right)

To check your answer: with

\mu = 12.320

and

\sigma = 0.47490

,

P(B > 12.2) = 0.600

and

P(B < 12) = 0.250

, which match. ✓

$\therefore$ the mean mass of the bars is $12.3\ g$ and the standard deviation is $0.475\ g$ (3 s.f.).

Part (iv)

4 marks

Approach

The total mass of a component is a sum of three independent normal random variables (two coated spheres and one bar), so it is itself normal, with mean and variance found by adding the individual means and variances.

Let $T = Y_{1} + Y_{2} + B$ , where $Y_{1},Y_{2}$ are the masses of the two coated spheres and $B$ is the mass of the bar, be the total mass, in grams, of the component. Since $Y_{1},Y_{2} \sim N(22,\ 0.3025)$ independently and $B \sim N\left( 12.320,\ {0.47490}^{2} \right)$ independently of $Y_{1},Y_{2}$ ,

T \sim N\left( 2(22) + 12.320,\ 2(0.3025) + {0.47490}^{2} \right) = N(56.320,\ 0.83053)\ \text{(5 s.f.)}

Since $P(T > k) = 0.75$ , by GC (inverse normal),

k = 55.7\ \left( 3\ \text{s.f.} \right)

2017 H2 Maths Paper 2, Question 10

Part (i)

Part (ii)

Part (iii)

Part (iv)

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