Marked out of 1.00 For $$ C=95 \% $$ If $$ \sigma=5.34 $$ and $$ \mathrm{n}=45 $$ then $$ E=1.56 $$

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To determine if the given information is correct, we can calculate the margin of error $$ E $$ using the formula for the margin of error in a confidence interval for a population mean:

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

Where:
- $$ E $$ is the margin of error,
- $$ z $$ is the z-score corresponding to the desired confidence level,
- $$ \sigma $$ is the population standard deviation,
- $$ n $$ is the sample size.

### Step 1: Find the z-score for $$ C = 95\% $$

For a 95% confidence level, the z-score can be found using standard normal distribution tables or calculators. The z-score for 95% confidence is approximately:

$$
z \approx 1.96
$$

### Step 2: Substitute the known values into the formula

Given:
- $$ \sigma = 5.34 $$
- $$ n = 45 $$

We can now calculate $$ E $$:

$$
E = 1.96 \cdot \frac{5.34}{\sqrt{45}}
$$

### Step 3: Calculate $$ \sqrt{n} $$

First, we calculate $$ \sqrt{45} $$:

$$
\sqrt{45} \approx 6.708
$$

### Step 4: Calculate $$ \frac{\sigma}{\sqrt{n}} $$

Now, we calculate $$ \frac{5.34}{\sqrt{45}} $$:

$$
\frac{5.34}{6.708} \approx 0.796
$$

### Step 5: Calculate the margin of error $$ E $$

Now we can calculate $$ E $$:

$$
E = 1.96 \cdot 0.796 \approx 1.56
$$

### Conclusion

The calculated margin of error $$ E $$ is approximately 1.56, which matches the given value. Therefore, the information provided is correct.
The margin of error $$ E $$ is correctly calculated as 1.56 based on the given values of $$ C = 95\% $$, $$ \sigma = 5.34 $$, and $$ n = 45 $$.

Marked out of 1.00 For \( C=95 \% \) If \( \sigma=5.34 \) and \( \mathrm{n}=45 \) then \( E=1.56 \)

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