绿地工厂（Green Earth Factory）生产太阳能电池板。平均而言，生产的太阳能电池板中有 2% 被发现存有缺陷。太阳能电池板以每批 10 块的形式包装。 已知每批缺陷太阳能电池板的数量服从二项分布。

1. State the assumption of constant probability in context

The probability that any individual solar panel is defective remains constant.

2. State the assumption of independence in context

Whether a solar panel is defective is independent of whether any other solar panel is defective.

1. Define the random variable and state its distribution.

Let $X$ be the number of defective solar panels in a batch of 10.
$X \sim \mathrm{B}(10, 0.02)$

2. Express the required probability using the complement method.

$\mathrm{P}(X \ge 2) = 1 - \mathrm{P}(X \le 1)$
$= 1 - (\mathrm{P}(X = 0) + \mathrm{P}(X = 1))$

3. Calculate the probability.

$\mathrm{P}(X \ge 2) = 1 - \left[ \binom{10}{0}(0.02)^0(0.98)^{10} + \binom{10}{1}(0.02)^1(0.98)^9 \right]$
$= 1 - (0.81707 + 0.16675)$
$= 0.016177 \approx 0.0162 \text{ (to 3 s.f.)}$

1. Calculate the probability that a single school lodges a complaint.

Let $X$ be the number of defective solar panels in a school's allocation of 10.
$X \sim \text{B}(10, 0.02)$
$P(X \ge 2) = 1 - P(X \le 1)$
$= 1 - \left( 0.98^{10} + 10(0.02)(0.98)^9 \right)$
$= 0.0161776 \text{ (to 6 s.f.)}$

2. Define the binomial distribution for the number of schools complaining and evaluate the required probability.

Let $Y$ be the number of schools that lodge a complaint out of 100.
$Y \sim \text{B}(100, 0.0161776)$
We require $P(Y \le 5)$.
Using a graphing calculator,
$P(Y \le 5) \approx 0.99264$
$P(Y \le 5) = 0.993 \text{ (to 3 s.f.)}$

1. Find the expected value and variance for the total number of defective solar panels and LED lights.

Let $X$ be the number of defective solar panels out of $1000$ ($100 \times 10$).
$X \sim \mathrm{B}(1000, 0.02)$
$\mathrm{E}(X) = 1000(0.02) = 20$
$\mathrm{Var}(X) = 1000(0.02)(0.98) = 19.6$

Let $Y$ be the number of defective LED lights out of $2500$ ($100 \times 25$).
$Y \sim \mathrm{B}(2500, 0.01)$
$\mathrm{E}(Y) = 2500(0.01) = 25$
$\mathrm{Var}(Y) = 2500(0.01)(0.99) = 24.75$

2. Use an approximation to find the distribution of the total number of defective items.

Since the sample sizes $n = 1000$ and $n = 2500$ are sufficiently large, by the Central Limit Theorem, $X$ and $Y$ can be well-approximated by Normal distributions.
Let $W = X + Y$ be the total number of defective items across all 100 schools.
Assuming the number of defective solar panels and LED lights are independent:
$\mathrm{E}(W) = \mathrm{E}(X) + \mathrm{E}(Y) = 20 + 25 = 45$
$\mathrm{Var}(W) = \mathrm{Var}(X) + \mathrm{Var}(Y) = 19.6 + 24.75 = 44.35$
Thus, the total number of defective items follows an approximate Normal distribution:
$W \sim \mathrm{N}(45, 44.35)$

3. Calculate the probability that the factory receives a fine (total defective items exceeds 50).

The factory receives a fine if $W > 50$.
Using a graphing calculator, we find:
$\mathrm{P}(W > 50) \approx 0.22638$
$\mathrm{P}(W > 50) = 0.226 \text{ (to 3 s.f.)}$