An online retailer knows that 52.1% of its purchases are made by a credit card.

If a random sample of 500 purchases from that retailer is selected, what is the probability that less than 48.4% purchases in the sample would be by credit cards?

Probability = 


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.

An online retailer knows that 52.1% of its purchases are made by a credit card. If a random sample of 500 purchases from that retailer is selected, what is the probability that less than 48.4% purchases in the sample would be by credit cards?

Discover how to calculate the probability that less than 48.4% of purchases in a sample are made by credit cards using the normal approximation to the binomial distribution. With a sample size of 500 and a known purchase probability of 52.1%, this problem involves calculating a z-score and finding the corresponding probability. Ideal for college-level statistics students.

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