下表显示了新加坡 18 岁男孩的年龄对应 BMI（身体质量指数）情况。

年龄（岁）	体重指标	体重不足	超重
		$\le 5^{\text{th}}$ 百分位数	$\ge 90^{\text{th}}$ 百分位数
18		$\le 16.7$	$\ge m$

已知 18 岁男孩的 BMI 服从正态分布 $\text{N}(22.7, \sigma^2)$。

1. Formulate the probability equation using the percentile information given in the table.

Let $X$ be the BMI of a randomly chosen 18-year-old boy.
$X \sim \text{N}(22.7, \sigma^2)$
From the table, the $5^{\text{th}}$ percentile is $16.7$, which means:
$\text{P}(X \le 16.7) = 0.05$

2. Standardise the variable and determine the standard normal z-score.

Standardising $X$ to the standard normal variable $Z$:
$\text{P}\left(Z \le \frac{16.7 - 22.7}{\sigma}\right) = 0.05$
Using the inverse normal function on the calculator for an area of $0.05$:
$\frac{16.7 - 22.7}{\sigma} \approx -1.64485$

3. Solve for the standard deviation $\sigma$.

$-\frac{6}{\sigma} = -1.64485$
$\sigma = \frac{6}{1.64485}$
$\sigma \approx 3.6477$
$\sigma \approx 3.65 \text{ (shown)}$

1. Set up the probability equation for the $90^{\text{th}}$ percentile of the normal distribution based on the given table.

Let $X$ be the BMI of a randomly chosen 18-year-old boy.
$X \sim \text{N}(22.7, 3.65^2)$

Since the overweight indicator starts at the $90^{\text{th}}$ percentile,
$\mathrm{P}(X \le m) = 0.90$

2. Use the inverse normal function on a graphing calculator to evaluate $m$.

$m = 27.377\dots$
$m \approx 27.4 \text{ (to 3 s.f.)}$

1. Define the random variable for the average BMI and find its mean and variance.

Let $X$ and $Y$ be the BMI of a randomly chosen 18-year-old and 15-year-old boy respectively.
$X \sim \text{N}(22.7, 3.65^2)$ and $Y \sim \text{N}(21.6, 3.49^2)$

Let the average BMI be $A = \frac{X+Y}{2}$.
$\text{E}(A) = \frac{\text{E}(X) + \text{E}(Y)}{2} = \frac{22.7 + 21.6}{2} = 22.15$
$\text{Var}(A) = \left(\frac{1}{2}\right)^2 \text{Var}(X+Y) = \frac{1}{4}(3.65^2 + 3.49^2) = \frac{25.5026}{4} = 6.37565$

Hence, $A \sim \text{N}(22.15, 6.37565)$

2. Calculate the required probability.

We need to find the probability that the average is more than $20.1$, which is $\text{P}(A > 20.1)$.

Using a graphing calculator:
$\text{P}(A > 20.1) \approx 0.79156$
$\text{P}(A > 20.1) = 0.792 \text{ (to 3 s.f.)}$

1. State the statistical assumption required to compute the variance of the linear combination of the two random variables.

In part (iii), to find the variance of the average BMI (a linear combination $\frac{X+Y}{2}$), it was assumed that the two variables are independent.

In context: The BMI of a randomly chosen 15-year-old boy and the BMI of a randomly chosen 18-year-old boy are independent.