In each large batch of SIM cards, 15% of the cards are defective. From each batch, a random sample of 10 cards is drawn for inspection.

1. State the distribution with its parameters.

Let $X$ be the number of defective SIM cards in a sample of 10.
Since there is a fixed sample size of $10$ and a constant probability of $0.15$ of finding a defective card, the distribution is Binomial.
$X \sim \mathrm{B}(10, 0.15)$

2. State the assumption for the binomial distribution to be a valid model.

The assumption is that whether any one SIM card is defective is independent of whether the other SIM cards in the sample are defective.
(Alternatively, the probability of any SIM card being defective remains constant at $0.15$.)

1. Find the probability of a single batch being 'good'.

Let $X$ be the number of defective SIM cards in a sample of 10.
$X \sim \mathrm{B}(10, 0.15)$
A batch is 'good' if $X < 2$, which means $X \le 1$.
$\mathrm{P}(\text{good}) = \mathrm{P}(X \le 1)$
$= \binom{10}{0}(0.15)^0(0.85)^{10} + \binom{10}{1}(0.15)^1(0.85)^9$
$= 0.19687 + 0.34743 \approx 0.54430$

2. Define the distribution for the number of good batches and state the Normal approximation.

Let $Y$ be the number of good batches produced out of 33.
$Y \sim \mathrm{B}(33, 0.54430)$
Since $n=33$ is sufficiently large such that $np = 33(0.54430) \approx 17.96 > 5$ and $n(1-p) = 33(1 - 0.54430) \approx 15.04 > 5$, $Y$ can be approximated by a Normal distribution.
$\mu = np = 17.9619$
$\sigma^2 = np(1-p) = 17.9619(0.45570) = 8.1852$
Thus, $Y \sim \mathrm{N}(17.9619, 8.1852)$ approximately.

3. Determine the required number of good batches based on the 'more than 80%' condition.

We require more than 80% of the 33 daily batches to be good.
$80\% \text{ of } 33 = 0.8 \times 33 = 26.4$
Thus, we need $\mathrm{P}(Y > 26.4)$.
Since $Y$ is a discrete count of batches, this strictly translates to $Y \ge 27$.

4. Apply the continuity correction and calculate the probability using the Normal distribution.

Applying a continuity correction to translate the discrete threshold $Y \ge 27$ to our continuous Normal distribution:
$\mathrm{P}(Y \ge 27) \approx \mathrm{P}(Y_{\text{norm}} > 26.5)$
Using the Normal distribution $\mathrm{N}(17.9619, 8.1852)$ on a graphing calculator:
$\mathrm{P}(Y_{\text{norm}} > 26.5) \approx 0.00142$ (to 3 s.f.)