Alice uses a computer program to simulate 80 tosses of a biased coin. Let $Y$ be the random variable denoting the number of heads obtained and $p$ be the probability of obtaining a head. It is given that $80 + E(Y) = 6\,\mathrm{Var}(Y).$

1. Express the expected value and variance of $Y$ in terms of $p$.

Since $Y$ represents the number of heads in 80 independent coin tosses, $Y$ follows a binomial distribution, $Y \sim \mathrm{B}(80, p)$.
We then have:
$\mathrm{E}(Y) = np = 80p$
$\mathrm{Var}(Y) = np(1-p) = 80p(1-p)$

2. Substitute the expressions into the given equation to form a quadratic equation in $p$.

Given that $80 + \mathrm{E}(Y) = 6\,\mathrm{Var}(Y)$:
$80 + 80p = 6(80p(1-p))$
Dividing the entire equation by 80:
$1 + p = 6p(1-p)$
$1 + p = 6p - 6p^2$
$6p^2 - 5p + 1 = 0$

3. Solve for $p$ and reject the invalid value.

Factorising the quadratic equation:
$(3p - 1)(2p - 1) = 0$
$p = \frac{1}{3} \quad \text{or} \quad p = \frac{1}{2}$
Since the question states that the coin is biased, $p \neq 0.5$. Therefore, we reject $p = \frac{1}{2}$.
$p = \frac{1}{3}$

1. Identify the probability of obtaining a head, $p$, and set up the distribution for the remaining tosses.

From part (i), solving the given relation $80 + 80p = 6(80p(1-p))$ yields $p = \frac{1}{3}$ or $p = \frac{1}{2}$.
Since the question states it is a biased coin, $p \neq \frac{1}{2}$, so $p = \frac{1}{3}$.

Let $W$ be the random variable denoting the number of heads obtained in the remaining 75 tosses.
Since the coin tosses are independent, $W \sim \mathrm{B}\left(75, \frac{1}{3}\right)$.

2. Formulate the required conditional probability in terms of $W$.

We require the probability of obtaining at least 30 heads in total, given that the first 5 tosses are heads.
Because the trials are independent, the outcome of the first 5 tosses does not affect the remaining 75 tosses.
To reach a total of at least 30 heads, we need at least $30 - 5 = 25$ heads in the remaining 75 tosses.

$\text{Required probability} = \mathrm{P}(W \ge 25)$

3. Evaluate the probability using a graphing calculator.

$\mathrm{P}(W \ge 25) = 1 - \mathrm{P}(W \le 24)$
Using a graphing calculator,
$\mathrm{P}(W \ge 25) = 1 - 0.44855$
$\mathrm{P}(W \ge 25) \approx 0.55145$

$\text{Required probability} = 0.551 \text{ (to 3 s.f.)}$

1. Deduce the probability $p$ from part (i) and find the population mean and variance.

From part (i), using $Y \sim \text{B}(80, p)$:
$80 + E(Y) = 6\text{Var}(Y)$
$80 + 80p = 6(80p(1-p))$
$6p^2 - 5p + 1 = 0$
$(3p-1)(2p-1) = 0$
Since the coin is biased, $p \neq \frac{1}{2}$, so $p = \frac{1}{3}$.
The population mean and variance are:
$E(Y) = 80\left(\frac{1}{3}\right) = \frac{80}{3}$
$\text{Var}(Y) = 80\left(\frac{1}{3}\right)\left(\frac{2}{3}\right) = \frac{160}{9}$

2. Apply the Central Limit Theorem to find the distribution of the sample mean.

Since the sample size $n = 50$ is large ($n \ge 30$), by the Central Limit Theorem, the sample mean $\bar{Y}$ is approximately normally distributed:
$\bar{Y} \sim \text{N}\left(\mu, \frac{\sigma^2}{n}\right)$
$\bar{Y} \sim \text{N}\left(\frac{80}{3}, \frac{160/9}{50}\right)$
$\bar{Y} \sim \text{N}\left(\frac{80}{3}, \frac{16}{45}\right)$

3. Calculate the probability that the sample mean is less than 25.

Using a graphing calculator:
$\text{P}(\bar{Y} < 25) \approx 0.0025946$
$\text{P}(\bar{Y} < 25) = 0.00259 \text{ (to 3 s.f.)}$