This problem can be approached using the normal approximation to the binomial distribution, because the sample size (500) is large enough, and we're dealing with proportions.

Step 1: Define the variables

• $n = 500$ (sample size)
• $p = 0.521$ (probability of purchasing with a credit card)
• $\hat{p} = 0.484$ (sample proportion given in the question)

We want to calculate the probability that the sample proportion is less than 48.4% (i.e., $P(\hat{p} < 0.484)$).

Step 2: Compute the mean and standard deviation

For the normal approximation, the mean $\mu$ and standard deviation $\sigma$ for the binomial distribution can be calculated as follows:

• $\mu = p = 0.521$
• $\sigma = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.521(1-0.521)}{500}} = \sqrt{\frac{0.521 \times 0.479}{500}} \approx 0.0223$

Step 3: Convert to a z-score

Now, we convert the given sample proportion $\hat{p} = 0.484$ to a z-score using the formula:

$z = \frac{\hat{p} - p}{\sigma}$

$z = \frac{0.484 - 0.521}{0.0223} \approx \frac{-0.037}{0.0223} \approx -1.659$

Step 4: Find the probability

Using the z-score of -1.659, we can now find the corresponding probability using the standard normal distribution table (or a calculator):

$P(z < -1.659) \approx 0.0484$

Final Answer:

The probability that less than 48.4% of the purchases in the sample are made by credit card is approximately 0.0484, or 4.84%.

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Would you like more details or have any questions?

Here are 5 related questions to explore:

1. How would the probability change if the sample size were 1,000?
2. What is the probability that more than 55% of the purchases in the sample are made by credit card?
3. How does the central limit theorem justify the use of the normal approximation here?
4. What is the z-score for a sample proportion of 50%?
5. How would you calculate the probability that between 50% and 55% of the purchases are made by credit card?

Tip: When using the normal approximation to the binomial distribution, ensure that both $np$ and $n(1-p)$ are greater than 5 to ensure accuracy.