A survey of the US population found that 26% of consumers shop for groceries online.  Suppose you select a sample of 200 consumers. Using what you know about the distribution of the sample proportion answer the following question.

Suppose instead you take a sample of 800 consumers. What is the probability that between 23% and 29% of consumers in your sample shop for groceries online?

(Round answers to four decimal places.)

To solve this problem, we can use the Central Limit Theorem (CLT) for proportions. The CLT tells us that, for a sufficiently large sample size, the sampling distribution of the sample proportion can be approximated by a normal distribution with mean and standard deviation given by:

- **Mean of the sample proportion** \( \mu_{\hat{p}} = p \)
- **Standard deviation of the sample proportion** \( \sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}} \)

where:
- \( p = 0.26 \) (the population proportion of consumers who shop for groceries online),
- \( n = 800 \) (the sample size).

### Step 1: Calculate the mean and standard deviation of the sample proportion

Given:
- \( p = 0.26 \)
- \( n = 800 \)

Calculate the standard deviation:
\[
\sigma_{\hat{p}} = \sqrt{\frac{0.26(1 - 0.26)}{800}} = \sqrt{\frac{0.26 \times 0.74}{800}} = \sqrt{\frac{0.1924}{800}} = \sqrt{0.0002405} \approx 0.0155
\]

### Step 2: Convert the percentages to proportions and find the corresponding \( z \)-scores

We want the probability that the sample proportion \( \hat{p} \) is between 0.23 and 0.29. We convert these values to their corresponding \( z \)-scores using the formula:
\[
z = \frac{\hat{p} - \mu_{\hat{p}}}{\sigma_{\hat{p}}}
\]

Calculating the \( z \)-scores:
- For \( \hat{p} = 0.23 \):
  \[
  z_1 = \frac{0.23 - 0.26}{0.0155} \approx \frac{-0.03}{0.0155} \approx -1.9355
  \]
  
- For \( \hat{p} = 0.29 \):
  \[
  z_2 = \frac{0.29 - 0.26}{0.0155} \approx \frac{0.03}{0.0155} \approx 1.9355
  \]

### Step 3: Find the probability using the standard normal distribution

We need to find the probability that \( z \) is between \(-1.9355\) and \(1.9355\). Using the standard normal table or a calculator, we find the probabilities:

- \( P(Z < 1.9355) \approx 0.9736 \)
- \( P(Z < -1.9355) \approx 0.0264 \)

Therefore, the probability that the sample proportion is between 0.23 and 0.29 is:
\[
P(-1.9355 < Z < 1.9355) = P(Z < 1.9355) - P(Z < -1.9355) = 0.9736 - 0.0264 = 0.9472
\]

### Final Answer

The probability that between 23% and 29% of consumers in the sample of 800 shop for groceries online is approximately **0.9472** (rounded to four decimal places).

Would you like more details on this, or have any questions?

Here are some related questions you might find useful:

1. How do we determine the sample size required for a given level of confidence in a proportion problem?
2. What is the impact of increasing the sample size on the standard deviation of the sample proportion?
3. How does the Central Limit Theorem apply to sample proportions?
4. What are the conditions for using the normal approximation for a binomial distribution?
5. How do we calculate probabilities for values not directly in the standard normal table?

**Tip:** Always check if the sample size is large enough for the normal approximation to be valid using the rule \( np \geq 10 \) and \( n(1-p) \geq 10 \).

A survey of the US population found that 26% of consumers shop for groceries online. Suppose you select a sample of 200 consumers. Using what you know about the distribution of the sample proportion answer the following question. Suppose instead you take a sample of 800 consumers. What is the probability that between 23% and 29% of consumers in your sample shop for groceries online? (Round answers to four decimal places.)

Find the probability that between 23% and 29% of consumers in a sample of 800 shop for groceries online, using the Central Limit Theorem. This problem involves concepts like normal distribution, sampling proportions, and z-scores. Suitable for high school and college students studying probability and statistics.

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