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The problem asks us to determine the probability that the sample proportion of first-time customers is greater than 0.3. Here's how we can approach this step-by-step:

### Given information:
- The population proportion \( p = 0.25 \) (25% of orders come from first-time customers).
- Sample size \( n = 73 \).
- The sample proportion \( \hat{p} = 0.3 \).
- We are asked to find the probability that the sample proportion is greater than 0.3, i.e., \( P(\hat{p} > 0.3) \).

### Solution Approach:
We will use the **normal approximation to the binomial distribution** for large sample sizes. According to the **Central Limit Theorem**, for large \( n \), the sampling distribution of the sample proportion \( \hat{p} \) will be approximately normally distributed with the following properties:

- **Mean of the sampling distribution**: \( \mu_{\hat{p}} = p = 0.25 \)
- **Standard deviation (standard error) of the sampling distribution**:
  \[
  \sigma_{\hat{p}} = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.25(1 - 0.25)}{73}} = \sqrt{\frac{0.25 \times 0.75}{73}} \approx 0.0504
  \]

Now, we standardize the sample proportion \( \hat{p} = 0.3 \) to convert it to a standard normal variable \( Z \).

- **Z-score formula**:
  \[
  Z = \frac{\hat{p} - \mu_{\hat{p}}}{\sigma_{\hat{p}}} = \frac{0.3 - 0.25}{0.0504} \approx 0.9921
  \]

Using the Z-score of 0.9921, we now find the probability from standard normal distribution tables (or using a calculator).

- **P-value for Z = 0.9921** is approximately 0.8397 (from Z-tables or using a normal distribution calculator).

Thus, the probability that \( \hat{p} \) is greater than 0.3 is:
\[
P(\hat{p} > 0.3) = 1 - P(Z \leq 0.9921) = 1 - 0.8397 = 0.1603
\]

### Final Answer:
The probability that the sample proportion is greater than 0.3 is approximately **0.1603**.

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Would you like further details on any step? Here are some related questions for further exploration:

1. How would the probability change if the sample size was larger?
2. What is the effect of increasing the population proportion \( p \)?
3. Why do we use the normal approximation in this problem?
4. How do we calculate the Z-score for different sample proportions?
5. What is the interpretation of a Z-score?

#### Tip:
Always check if the sample size \( n \) is large enough (typically \( np \geq 5 \) and \( n(1-p) \geq 5 \)) before using the normal approximation to the binomial distribution.

Based on historical data, your manager believes that 25% of the company's orders come from first-time customers. A random sample of 73 orders will be used to estimate the proportion of first-time customers. What is the probability that the sample proportion is greater than 0.3?

Calculate the probability that the sample proportion of first-time customers is greater than 0.3 using the normal approximation and Z-scores. This problem uses a sample size of 73, with an assumed population proportion of 25%. Learn how to apply the Central Limit Theorem and calculate Z-scores in this context.

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