An investment website can tell what devices are used to access the site. The site managers wonder whether they should enhance the facilities for trading via​ "smart phones," so they want to estimate the proportion of users who access the site that way​ (even if they also use their computers​ sometimes). They draw a random sample of 200200 investors from their customers. Suppose that the true proportion of smart phone users is 4343​%. Complete parts a through c below.

Given that the true proportion of smart phone users is 43%, let's proceed with each part of the problem.

### **Part (a): Describe the sampling distribution model of the proportion of smart phone users in samples of 200 investors.**

The sampling distribution of a sample proportion follows a normal distribution when the sample size is large enough, according to the Central Limit Theorem. To model this, we need to calculate the mean and standard deviation.

- **Mean (\(\mu_p\))**: The mean of the sampling distribution is equal to the true proportion of smartphone users.
  \[
  \mu_p = p = 0.43
  \]

- **Standard deviation (\(\sigma_p\))**: The standard deviation of the sample proportion is given by:
  \[
  \sigma_p = \sqrt{\frac{p(1 - p)}{n}}
  \]
  where:
  - \( p = 0.43 \) (true proportion),
  - \( n = 200 \) (sample size).

Let's calculate the standard deviation.

\[
\sigma_p = \sqrt{\frac{0.43(1 - 0.43)}{200}} = \sqrt{\frac{0.43 \times 0.57}{200}} \approx \sqrt{\frac{0.2451}{200}} \approx \sqrt{0.0012255} \approx 0.035
\]

Thus, the standard deviation is approximately \( \sigma_p = 0.035 \).

**Summary of the sampling distribution:**
- Mean: \( 0.43 \)
- Standard deviation: \( 0.035 \)
- Since the sample size \( n = 200 \) is sufficiently large and the conditions for normality are satisfied (both \( np \geq 10 \) and \( n(1 - p) \geq 10 \)), we can approximate the sampling distribution with a normal distribution: \( N(0.43, 0.035) \).

### **Part (b): What is the probability that the sample proportion is between 38% and 48%?**

To find this probability, we need to standardize the values and use the normal distribution. The z-score formula is:

\[
z = \frac{\hat{p} - \mu_p}{\sigma_p}
\]

1. For \( \hat{p} = 0.38 \):

\[
z = \frac{0.38 - 0.43}{0.035} = \frac{-0.05}{0.035} \approx -1.43
\]

2. For \( \hat{p} = 0.48 \):

\[
z = \frac{0.48 - 0.43}{0.035} = \frac{0.05}{0.035} \approx 1.43
\]

Now, we use the z-scores to find the probabilities:

- From standard normal distribution tables, \( P(Z \leq -1.43) \approx 0.0764 \).
- \( P(Z \leq 1.43) \approx 0.9236 \).

The probability that the sample proportion is between 38% and 48% is:

\[
P(0.38 \leq \hat{p} \leq 0.48) = P(Z \leq 1.43) - P(Z \leq -1.43) \approx 0.9236 - 0.0764 = 0.8472
\]

So, the probability is approximately **0.8472** or **84.72%**.

### **Part (c): What is the probability that the sample proportion is less than 40%?**

Again, we'll standardize the value of \( \hat{p} = 0.40 \):

\[
z = \frac{0.40 - 0.43}{0.035} = \frac{-0.03}{0.035} \approx -0.86
\]

From standard normal distribution tables, \( P(Z \leq -0.86) \approx 0.1949 \).

Thus, the probability that the sample proportion is less than 40% is approximately **0.1949** or **19.49%**.

---

### Summary:
- **(a)** The sampling distribution is approximately normal with mean \( 0.43 \) and standard deviation \( 0.035 \).
- **(b)** The probability that the sample proportion is between 38% and 48% is approximately **84.72%**.
- **(c)** The probability that the sample proportion is less than 40% is approximately **19.49%**.

---

Would you like more details or have any further questions? Here are 5 related questions you might find useful:

1. How do we calculate the z-score for a specific sample proportion?
2. Why is the sample proportion distribution approximated by a normal curve?
3. What happens to the standard deviation if the sample size increases?
4. How can we interpret probabilities in the context of this problem?
5. How would the results change if the true proportion of smartphone users was different?

**Tip:** Larger sample sizes lead to a smaller standard deviation of the sampling distribution, making estimates more precise.

An investment website can tell what devices are used to access the site. The site managers wonder whether they should enhance the facilities for trading via​ 'smart phones,' so they want to estimate the proportion of users who access the site that way​ (even if they also use their computers​ sometimes). They draw a random sample of 200 investors from their customers. Suppose that the true proportion of smart phone users is 43%. Complete parts a through c below.

Explore the sampling distribution of smart phone users on an investment website. Learn how to calculate the mean and standard deviation for a sample size of 200, and find probabilities for sample proportions. This problem uses the Central Limit Theorem, Z-scores, and normal distribution approximations.

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