In this problem, you are dealing with a situation where a normal approximation to the binomial distribution is being applied. Let's summarize what’s happening in each part:

Part 1:

• You are asked to determine whether the normal approximation is appropriate.
• Conditions for using normal approximation are that both $np \geq 10$ and $n(1-p) \geq 10$, where $n$ is the sample size, and $p$ is the probability of success.
• For the sample of 245 employed adults, $np = 29.4$ and $n(1-p) = 215.6$, both are greater than 10, so the approximation is appropriate.

Part 2:

• You calculate the probability that less than 8.4% of the individuals hold multiple jobs. The probability calculated is $0.0418$, likely using the z-score and cumulative normal distribution table.

Part 3:

• A new sample of 333 is considered, and you're finding the probability that less than 8.4% hold multiple jobs. The result is $0.0217$.

Part 4 (Upcoming):

• You'll likely be asked to find the probability for more than 7.9% in a sample of 333 employed adults. You'll apply the same normal approximation and calculate the z-score accordingly.

Would you like to continue working on Part 4 or need more detailed explanations of any part?

Relative Questions:

1. What is the difference between using normal approximation and the exact binomial distribution?
2. How is the z-score calculated for a given percentage in a normal distribution?
3. Why is it important that both $np$ and $n(1-p)$ are greater than 10 when using normal approximation?
4. How does increasing the sample size affect the accuracy of normal approximation?
5. What would happen if the percentage value (e.g., 8.4%) were closer to 50% instead of 12%? Would normal approximation still be valid?

Tip:

When using the normal approximation to the binomial distribution, remember to apply a continuity correction of $\pm 0.5$ when calculating probabilities for discrete distributions.