RI 2025 H2 Math Prelim Paper 2 — Normal Distribution (worked solution)

\textbf{(a)} $\sigma > 0.842$ (3\,s.f.) \textbf{(b)} $\mathrm{P}(S > 550) \approx 0.891$ \textbf{(c)} Minimum value of $n$ is $\mathbf{20}$.

Free practice / JC2 H2 Maths Prelim

Normal Distribution: Multi-part normal — RI 2025 H2 Math Prelim Paper 2

Attempt the question on paper, then open the worked solution to check your method.

Raffles Institution2025 PrelimPaper 2●●○ Standard9 marks

What this question tests

Multi-part normal.

Question

In this question you should state the parameters of any distributions you use.

A snack company produces two types of potato chips, Rays and Luffles. The mass, in grams, of a regular packet of Rays follows the distribution $N(100, \sigma^2)$ and the mass, in grams, of a regular packet of Luffles follows the distribution $N(120, 16)$. The masses of all regular packets of potato chips are independent of one another.

If the probability of the mass of a randomly chosen packet of Rays not exceeding $(100 - \sigma^2)$ grams is less than 0.2, find the possible range of values of $\sigma$.

It is given that $\sigma = 3$ for the rest of the question.

(b) Find the probability that 2 randomly chosen packets of Rays and 3 randomly chosen packets of Luffles have a combined mass greater than 0.55 kg.

(c) The snack company decides to launch a new product called Mega Jumbo Pack, which consists of $(24-n)$ regular packets of Rays and $n$ regular packets of Luffles. It is given that the probability of a Mega Jumbo Pack exceeding 20 times the mass of a regular packet of Luffles by more than 500 g is at least 0.1. Find the minimum value of $n$.

Show full worked solution▾

Let $R\sim N(100,\sigma^2)$, $L\sim N(120,16)$.

Find range of $\sigma$ such that $\mathrm{P}(R \leq 100-\sigma^2) < 0.2$. \[\begin{aligned} \mathrm{P}\!\left(Z \leq \frac{(100-\sigma^2)-100}{\sigma}\right) &< 0.2\\[4pt] \mathrm{P}\!\left(Z \leq -\sigma\right) &< 0.2\\[4pt] \Phi(-\sigma) &< 0.2\\[4pt] -\sigma &< -0.8416\\[4pt] \sigma &> 0.842 \end{aligned}\]

Figure for this question

With $\sigma=3$: $R\sim N(100,9)$, $L\sim N(120,16)$.

Let $S = R_1+R_2+L_1+L_2+L_3$ (2 Rays + 3 Luffles). \[\begin{aligned} \mathrm{E}(S) &= 2(100)+3(120) = 200+360 = 560\\[4pt] \mathrm{Var}(S) &= 2(9)+3(16) = 18+48 = 66 \end{aligned}\] $S\sim N(560,66)$. \[\begin{aligned} \mathrm{P}(S > 550) &= \mathrm{P}\!\left(Z > \frac{550-560}{\sqrt{66}}\right) = \mathrm{P}(Z > -1.231) = 0.891 \text{ (3\,s.f.)} \end{aligned}\]

Mega Jumbo Pack: $M = (24-n)R + nL$ where $24-n$ packets of Rays and $n$ packets of Luffles. \[\begin{aligned} \mathrm{E}(M) &= (24-n)(100)+n(120) = 2400-100n+120n = 2400+20n\\[4pt] \mathrm{Var}(M) &= (24-n)(9)+n(16) = 216-9n+16n = 216+7n \end{aligned}\]

We need $\mathrm{P}(M - 20L_0 > 500) \geq 0.1$, where $L_0\sim N(120,16)$ is a single regular Luffles packet.

Let $D = M - 20L_0$: \[\begin{aligned} \mathrm{E}(D) &= (2400+20n) - 20(120) = 2400+20n-2400 = 20n\\[4pt] \mathrm{Var}(D) &= (216+7n)+400 = 616+7n \end{aligned}\]

$\mathrm{P}(D > 500) \geq 0.1$: \[\begin{aligned} \mathrm{P}\!\left(Z > \frac{500-20n}{\sqrt{616+7n}}\right) &\geq 0.1\\[4pt] \frac{500-20n}{\sqrt{616+7n}} &\leq 1.2816\\[4pt] 500-20n &\leq 1.2816\sqrt{616+7n} \end{aligned}\]

From GC, minimum value of $n$ is $n = 20$.

Answer: (a) $\sigma > 0.842$ (3 s.f.) (b) $\mathrm{P}(S > 550) \approx 0.891$ (c) Minimum value of $n$ is $\mathbf{20}$.

← All Normal Distribution questionsMath Academy — JC H2 Maths tuition