To solve this problem, we apply the rule for using the normal approximation for a binomial distribution. The rule is as follows:

The normal approximation can be used if both:

1. $np \geq 5$, and
2. $n(1-p) \geq 5$,

where:

• $n$ is the sample size,
• $p$ is the probability of success.

Step 1: Set up the inequality

We are given $n = 135$ and need to find the minimum $p$ such that both conditions are satisfied.

Condition 1: $np \geq 5$

$135p \geq 5 \implies p \geq \frac{5}{135} \approx 0.037 \, (\text{or } 3.7\%).$

Condition 2: $n(1-p) \geq 5$

$135(1-p) \geq 5 \implies 135 - 135p \geq 5 \implies 135p \leq 130 \implies p \leq \frac{130}{135} \approx 0.963 \, (\text{or } 96.3\%).$

From these, the limiting factor for minimum $p$ is the first condition: $p \geq 0.037$ (3.7%).

Step 2: Round to the nearest integer

The nearest integer to 3.7% is 4%. However, since the answer choices don't include 4%, we look for the closest higher option, which is 5%.

Thus, the minimum probability needed is 5%.

Would you like me to provide further details or verify any steps?

Here are 5 related questions you can explore:

1. Why is the condition $np \geq 5$ important for using the normal approximation?
2. What happens when $np$ or $n(1-p)$ is less than 5?
3. How does increasing the sample size $n$ affect the minimum $p$?
4. What is the general formula for finding the cutoff probability $p$ for a given $n$?
5. How can we adjust the normal approximation for small sample sizes?

Tip: Always ensure that both $np$ and $n(1-p)$ are checked before using the normal approximation for a binomial distribution!