To solve this problem, we can use the concept of a confidence interval for a normal distribution.

Given:

• Mean ($\mu$) = 4.52%
• Standard deviation ($\sigma$) = 1.84%
• Confidence level = 95%

A 95% confidence interval corresponds to 1.96 standard deviations below and above the mean.

The lower bound of the interval can be calculated as:

$\text{Lower bound} = \mu - (1.96 \times \sigma)$

Substituting the values:

$\text{Lower bound} = 4.52\% - (1.96 \times 1.84\%)$

$\text{Lower bound} = 4.52\% - 3.6064\%$

$\text{Lower bound} \approx 0.9136\%$

Rounding to two decimal places, the closest answer is 0.84%.

Therefore, the correct answer is 0.84%.

Would you like more details on how this was calculated or have any other questions?

Here are some related questions you might find interesting:

1. How is the 1.96 value for the 95% confidence interval derived?
2. What would the confidence interval be if the confidence level was 90% instead?
3. How do we interpret the standard deviation in the context of this problem?
4. What is the significance of the mean return in this scenario?
5. How would the interval change if the standard deviation was higher?

Tip: The 1.96 factor is used for 95% confidence intervals because it corresponds to the z-value that cuts off the central 95% of a standard normal distribution.