The sample size is $$ n=200 $$ and $$ \sigma=15 $$, and the confidence level is $$ 95 \% $$. (Round to 3 decimals) $$ E=\square $$

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To calculate the margin of error $$ E $$ for a confidence interval, we can use the formula:

$$
E = z \cdot \frac{\sigma}{\sqrt{n}}
$$

where:
- $$ z $$ is the z-score corresponding to the desired confidence level,
- $$ \sigma $$ is the population standard deviation,
- $$ n $$ is the sample size.

For a 95% confidence level, the z-score is approximately $$ 1.96 $$.

Given:
- $$ n = 200 $$
- $$ \sigma = 15 $$

Now, we can calculate $$ E $$:

1. Calculate $$ \frac{\sigma}{\sqrt{n}} $$:

$$
\frac{\sigma}{\sqrt{n}} = \frac{15}{\sqrt{200}} = \frac{15}{14.1421} \approx 1.0607
$$

2. Now, multiply by the z-score:

$$
E = 1.96 \cdot 1.0607 \approx 2.082
$$

Rounding to three decimal places, we find:

$$
E \approx 2.082
$$

Thus, the margin of error $$ E $$ is approximately $$ \boxed{2.082} $$.
The margin of error $$ E $$ is approximately 2.082.

The sample size is \( n=200 \) and \( \sigma=15 \), and the confidence level is \( 95 \% \). (Round to 3 decimals) \( E=\square \)

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