在每年举行的一次标准化评估测试中，结果发现，平均每 5 名参加测试的学生中就有 1 名获得优秀（distinction）。 此外还发现，平均有 93% 的参加测试的学生通过了测试。

1. Calculate the expected number of students awarded a distinction using the formula for the mean of a binomial distribution.

Let $X$ be the number of students awarded a distinction.
$X \sim \mathrm{B}\left(23, \frac{1}{5}\right)$

$\mathrm{E}(X) = np = 23 \times \frac{1}{5} = 4.6$

1. Define the binomial random variable and state the required probability inequality.

Let $n$ be the number of students sitting for the test.
Let $X$ be the number of students awarded distinction.
$X \sim \mathrm{B}(n, 0.2)$
We require $\mathrm{P}(X \ge 9) > 0.7$.

2. Rewrite the inequality in terms of the cumulative probability.

$1 - \mathrm{P}(X \le 8) > 0.7$
$\mathrm{P}(X \le 8) < 0.3$

3. Use a graphing calculator to find the probability values for integers around the expected minimum $n$.

Using a graphing calculator to evaluate $\mathrm{P}(X \le 8)$ for integer values of $n$:
When $n = 50$, $\mathrm{P}(X \le 8) \approx 0.307 > 0.3$ (so $\mathrm{P}(X \ge 9) \approx 0.693$)
When $n = 51$, $\mathrm{P}(X \le 8) \approx 0.284 < 0.3$ (so $\mathrm{P}(X \ge 9) \approx 0.716$)

4. Conclude the final answer from the evaluated probabilities.

The minimum number of students is $51$.

1. Define the random variable and its exact binomial distribution.

Let $Y$ be the number of students who pass the test out of 60.
$Y \sim \mathrm{B}(60, 0.93)$

2. Compute the mean and variance to state the approximate normal distribution.

Expected value, $\mu = E(Y) = np = 60 \times 0.93 = 55.8$
Variance, $\sigma^2 = \mathrm{Var}(Y) = np(1-p) = 60 \times 0.93 \times 0.07 = 3.906$
Since $n$ is reasonably large, we approximate $Y$ with a normal distribution:
$Y \sim \mathrm{N}(55.8, 3.906)$ approximately.

3. Apply continuity correction and find the required probability.

We require the probability that more than 55 students pass, which is $\mathrm{P}(Y > 55) = \mathrm{P}(Y \ge 56)$.
Applying continuity correction for the continuous approximation, we evaluate $\mathrm{P}(Y_{\text{cont}} > 55.5)$.
Using a graphing calculator:
$\mathrm{P}(Y_{\text{cont}} > 55.5) \approx 0.56032 \approx 0.560$ (to 3 s.f.)