一家苗圃向长青私人有限公司（Evergreen Pte. Ltd.）供应植物。所供应植物的高度（以厘米为单位）可以认为服从分布 $N(\mu,\sigma^{2})$。 多年来，长青公司因植物太矮退回了 6% 的供应植物，因太高退回了 4%。已知长青公司仅接受高度在 25 cm 与 91 cm 之间的植物。 证明 $\sigma = 20$（精确到最接近的整数）。由此求出被长青公司接受的植物高度的中位数。

1. Set up a system of linear equations by standardising the given probabilities.

Let $H \sim \mathrm{N}(\mu, \sigma^2)$ be the height of a plant in cm.
Given $\mathrm{P}(H < 25) = 0.06$ and $\mathrm{P}(H > 91) = 0.04$.

Standardising the variables:
$\mathrm{P}\left(Z < \frac{25 - \mu}{\sigma}\right) = 0.06 \implies \frac{25 - \mu}{\sigma} \approx -1.55477$
$\mathrm{P}\left(Z > \frac{91 - \mu}{\sigma}\right) = 0.04 \implies \frac{91 - \mu}{\sigma} \approx 1.75069$

This gives the system of equations:
$25 - \mu = -1.55477\sigma \quad \text{--- (1)}$
$91 - \mu = 1.75069\sigma \quad \text{--- (2)}$

2. Solve the system of equations to find the unrounded values of $\sigma$ and $\mu$, and show the rounded value of $\sigma$.

Subtracting (1) from (2):
$91 - 25 = (1.75069 - (-1.55477))\sigma$
$66 = 3.30546\sigma$
$\sigma = 19.9669 \approx 20 \text{ (shown to nearest integer)}$

Substituting $\sigma = 19.9669$ into (1):
$\mu = 25 + 1.55477(19.9669) = 56.044$

3. Formulate an equation for the median height of the accepted plants using conditional probability.

Let $m$ be the median height of the accepted plants. Evergreen only accepts plants between 25 cm and 91 cm, which means we are dealing with a conditional probability:
$\mathrm{P}(H < m \mid 25 < H < 91) = 0.5$
$\frac{\mathrm{P}(25 < H < m)}{\mathrm{P}(25 < H < 91)} = 0.5$

Since $\mathrm{P}(25 < H < 91) = 1 - 0.06 - 0.04 = 0.90$:
$\mathrm{P}(25 < H < m) = 0.5 \times 0.90 = 0.45$
$\mathrm{P}(H < m) - \mathrm{P}(H \le 25) = 0.45$
$\mathrm{P}(H < m) - 0.06 = 0.45 \implies \mathrm{P}(H < m) = 0.51$

4. Solve for the median using the inverse normal distribution.

Using the inverse normal distribution with $\mu = 56.044$ and unrounded $\sigma = 19.9669$:
$m \approx 56.545$

The median height of the accepted plants is $56.5\text{ cm}$ (to 3 s.f.).

1. Determine the probability of rejecting a plant and define the binomial distribution for two batches.

The probability of a plant being rejected is the sum of the probabilities of it being too short or too tall:
$p = 0.06 + 0.04 = 0.10$

The total number of plants inspected in two batches is:
$n = 2 \times 10 = 20$

Let $X$ be the number of plants rejected out of the 20 plants.
Assuming the height of each plant is independent, $X$ follows a binomial distribution:
$X \sim \text{B}(20, 0.10)$

2. Calculate the probability that at most 4 plants are rejected.

We require the probability $\text{P}(X \le 4)$.
Using a graphing calculator:
$\text{P}(X \le 4) = 0.956825\dots \approx 0.957 \text{ (to 3 s.f.)}$

1. Find the unrounded mean $\mu$ and standard deviation $\sigma$ of the plant heights from the given rejection percentages.

Let $X \sim N(\mu, \sigma^2)$ be the height of a plant.
Given $\text{P}(X < 25) = 0.06$ and $\text{P}(X > 91) = 0.04$, we have:
$\frac{25 - \mu}{\sigma} = \Phi^{-1}(0.06) \approx -1.55477$
$\frac{91 - \mu}{\sigma} = \Phi^{-1}(0.96) \approx 1.75069$
Subtracting the first equation from the second gives:
$66 = 3.30546 \sigma \implies \sigma \approx 19.967$
Substitute $\sigma$ back to find $\mu$:
$\mu = 91 - 1.75069(19.967) \approx 56.044$

2. Formulate the inequality comparing the average heights.

Let $X_A, X_B, X_C$ be the independent heights of the plants chosen by Anna, Bella and Cinderella respectively.
The condition "average height of the three plants is at least twice the average height of the plants chosen by Anna and Bella" translates to:
$\frac{X_A + X_B + X_C}{3} \ge 2 \left( \frac{X_A + X_B}{2} \right)$
$\frac{X_A + X_B + X_C}{3} \ge X_A + X_B$
$X_C - 2X_A - 2X_B \ge 0$

3. Define the linear combination $Y = X_C - 2X_A - 2X_B$ and find its expected value and variance.

Let $Y = X_C - 2X_A - 2X_B$. Since $X_A, X_B, X_C$ are independent normal variables drawn from the same distribution:
$\text{E}(Y) = \text{E}(X_C) - 2\text{E}(X_A) - 2\text{E}(X_B)$
$\text{E}(Y) = \mu - 2\mu - 2\mu = -3\mu \approx -3(56.044) = -168.13$
$\text{Var}(Y) = \text{Var}(X_C) + (-2)^2\text{Var}(X_A) + (-2)^2\text{Var}(X_B)$
$\text{Var}(Y) = \sigma^2 + 4\sigma^2 + 4\sigma^2 = 9\sigma^2 \approx 9(19.967)^2 = 3588.1$
Thus, $Y \sim N(-168.13, 3588.1)$.

4. Calculate the required probability using the normal distribution.

We want to find $\text{P}(Y \ge 0)$.
Using a graphing calculator with lower bound $0$, upper bound $\infty$, mean $-168.13$, and standard deviation $\sqrt{3588.1} \approx 59.901$:
$\text{P}(Y \ge 0) \approx 0.0025021 \approx 0.00250$ (to 3 s.f.)