A nursery supplies plants to Evergreen Pte. Ltd. The heights, in centimetres, of plants supplied may be taken to follow the distribution $N(\mu,\sigma^{2})$. Over the years, Evergreen rejects 6% of the plants supplied as too short and 4% as too tall. It is known that Evergreen only accepts plants of heights between 25 cm and 91 cm. Show that $\sigma = 20$ correct to the nearest integer. Hence find the median height of the plants accepted by Evergreen.

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.)