What is the minimum sample size needed in order to use normal approximations for an estimate about a population, given a probability of success of 10 percent using the success-failure conditions of np≥10 and n(1−p)≥10.

Answer 1

To determine the minimum sample size needed for using normal approximations in a binomial distribution, we can use the success-failure condition:

1. \( np \geq 10 \)
2. \( n(1 - p) \geq 10 \)

where:

• \( p \) is the probability of success,
• \( n \) is the sample size, and
• \( (1 - p) \) is the probability of failure.

In this case, we have \( p = 0.10 \).

Let's apply the conditions:

1. For the first condition \( np \geq 10 \): \[ n \cdot 0.10 \geq 10 \] To solve for \( n \): \[ n \geq \frac{10}{0.10} = 100 \]

2. For the second condition \( n(1 - p) \geq 10 \): \[ n \cdot (1 - 0.10) \geq 10 \] Thus, \[ n \cdot 0.90 \geq 10 \] To solve for \( n \): \[ n \geq \frac{10}{0.90} \approx 11.11 \rightarrow \text{take the ceiling value, } n \geq 12 \]

Now, both these conditions need to be satisfied simultaneously. The more stringent condition is \( n \geq 100 \).

Therefore, the minimum sample size required is:

\[ \boxed{100} \]