Question

86% teens own smartlrtpgone.
Random sample 270 drawn

Find the mean u^p

Answer 1

What is the standard deviation

Answer 2

Probability for 90%

Answer 3

To find the mean of the sampling distribution of the sample proportion (\( \hat{p} \)), we use the formula:

\[
\mu_{\hat{p}} = p
\]

Where:
- \( \mu_{\hat{p}} \) is the mean of the sampling distribution of the sample proportion.
- \( p \) is the true proportion of the population.

In this scenario, you mentioned that 86% of teens own a smartphone. Therefore, we have:

\[
p = 0.86
\]

So, the mean of the sample proportion (\( \mu_{\hat{p}} \)) is:

\[
\mu_{\hat{p}} = 0.86
\]

Thus, the mean \( \mu_{\hat{p}} \) of the sample proportion of teenagers who own smartphones is **0.86** or **86%**.

Answer 4

The standard deviation of the sampling distribution of the sample proportion (\( \hat{p} \)) is calculated using the formula:

\[
\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}}
\]

Where:
- \( \sigma_{\hat{p}} \) is the standard deviation of the sample proportion.
- \( p \) is the proportion of the population (0.86 in this case).
- \( n \) is the sample size (270 in this case).

Now, we can plug in the values:

\[
p = 0.86
\]
\[
1 - p = 0.14
\]
\[
n = 270
\]

Now substituting these values into the formula:

\[
\sigma_{\hat{p}} = \sqrt{\frac{0.86 \times 0.14}{270}}
\]

Calculating the values:

\[
0.86 \times 0.14 = 0.1204
\]

Now calculate the standard deviation:

\[
\sigma_{\hat{p}} = \sqrt{\frac{0.1204}{270}} = \sqrt{0.000445 }
\]

Calculating the square root:

\[
\sigma_{\hat{p}} \approx 0.0211
\]

So, the standard deviation of the sample proportion is approximately **0.0211**.

Answer 5

To find the probability that the sample proportion of students who own smartphones is within a certain range, you can use the normal approximation to the binomial distribution. Since we are considering a large sample size (\(n = 270\)), the sample proportion can be approximated by a normal distribution with the mean (\(\mu_{\hat{p}}\)) and standard deviation (\(\sigma_{\hat{p}}\)) that we calculated earlier.

### Given:
- Proportion of teens who own a smartphone (\(p = 0.86\))
- Sample size (\(n = 270\))
- Mean of the sample proportion (\(\mu_{\hat{p}} = 0.86\))
- Standard deviation of the sample proportion (\(\sigma_{\hat{p}} \approx 0.0211\))

### Finding the Probability for a 90% Confidence Interval

1. **Determine the Z-score for 90% Confidence Interval:**
Since we are interested in a 90% confidence interval, we look for the Z-score that corresponds to the middle 90% of the standard normal distribution.

The critical Z-value for a 90% confidence interval can be found using a Z-table or standard normal distribution calculator. The Z-score that corresponds to the 95th percentile (since 90% in the middle means there is 5% in each tail) is approximately:

\[
Z \approx 1.645
\]

2. **Calculate the Margin of Error (ME):**
\[
ME = Z \cdot \sigma_{\hat{p}}
\]

\[
ME = 1.645 \cdot 0.0211 \approx 0.0347
\]

3. **Confidence Interval:**
The confidence interval can be calculated as:

\[
\hat{p} \pm ME
\]

This gives us:

\[
0.86 \pm 0.0347
\]

Therefore, the interval is:

\[
(0.86 - 0.0347, 0.86 + 0.0347) \approx (0.8253, 0.8947)
\]

### Conclusion
The 90% confidence interval for the proportion of teens who own a smartphone based on your sample of 270 is approximately **(0.8253, 0.8947)**, meaning that you can be 90% confident that the true proportion of all teens who own smartphones lies within this interval.

If you are looking for the probability that the sample proportion falls within a specific range, please specify that range, and I can help you compute that probability using the normal approximation as well.