亚历克斯（Alex）和本（Ben）进行了一场由十局组成的乒乓球比赛，在每一局中，本输掉的概率均为 0.7。

1. State one assumption required for the binomial model in the context of the question.

For a binomial model to be valid, the trials must be independent or have a constant probability of success. In this context, it assumes that the outcome of each game (whether Ben loses) is independent of the outcome of any other game.

1. Define the random variable and formulate the required probability expression

Let $X$ be the number of games Ben loses out of $10$.
$X \sim \text{B}(10, 0.7)$
Since he plays 10 games, 'more than half' means he loses strictly more than 5 games.
We need to find $\mathrm{P}(X > 5)$.

2. Evaluate the probability using a graphing calculator

$\mathrm{P}(X > 5) = 1 - \mathrm{P}(X \le 5)$
$\mathrm{P}(X > 5) = 1 - 0.150268$
$\mathrm{P}(X > 5) \approx 0.850 \text{ (to 3 s.f.)}$

1. Formulate the probability inequality using the given binomial distribution.

Let $Y$ be the number of games Ben loses out of $n$ games.
Given $Y \sim \text{B}(n, 0.3)$.
We require $\mathrm{P}(Y > 8) \le 0.01$.
This is equivalent to $1 - \mathrm{P}(Y \le 8) \le 0.01$
$\Rightarrow \mathrm{P}(Y \le 8) \ge 0.99$

2. Use a graphing calculator to test consecutive integer values of $n$.

Using a graphing calculator to tabulate values of $\mathrm{P}(Y \le 8)$ for different values of $n$:
When $n = 14$, $\mathrm{P}(Y \le 8) = 0.9917 \ge 0.99$
When $n = 15$, $\mathrm{P}(Y \le 8) = 0.9848 < 0.99$

3. Conclude the greatest value of $n$.

Since $\mathrm{P}(Y \le 8)$ decreases as $n$ increases, the greatest value of $n$ that satisfies the inequality is $14$.

1. Calculate the mean and variance of the binomial distribution to use for the approximation.

Let $X$ be the number of games Ben loses out of $50$.
From the question, $X \sim \text{B}(50, 0.3)$.
Mean, $\mu = np = 50 \times 0.3 = 15$
Variance, $\sigma^2 = np(1-p) = 50 \times 0.3 \times (1 - 0.3) = 10.5$

2. State the suitable approximation and its parameters.

Since $n$ is large, $np = 15 > 5$ and $n(1-p) = 35 > 5$, a normal approximation is suitable.
Thus, $X$ is approximately normally distributed: $X \sim \text{N}(15, 10.5)$.
The parameters of the normal distribution used are $\mu = 15$ and $\sigma^2 = 10.5$.

3. Apply the continuity correction and calculate the probability.

We need to find $P(10 \le X \le 20)$.
Applying continuity correction for the continuous normal approximation:
$P(10 \le X \le 20) \approx P(10 - 0.5 < X < 20 + 0.5)$
$= P(9.5 < X < 20.5)$
Using a graphing calculator with $\mu = 15$ and $\sigma = \sqrt{10.5}$:
$P(9.5 < X < 20.5) \approx 0.91037 \approx 0.910$ (to 3 s.f.)