#### Fjrodriguez

##### New Member

Doing the question number 24 of GARP mock exam 2020, I found an issue.

Explanation:

*C is correct. The lower bound of the*

– (95% confidence factor * Volatility of Surplus).

Expected surplus:

VA*[1 + E(RA)] – VL*[1 + E(RL)] = 180*1.06 – 140*1.10 = USD 36.80 million.

For a 95% confidence interval, the appropriate

**95% confidence interval**is equal to: Expected Surplus– (95% confidence factor * Volatility of Surplus).

Expected surplus:

VA*[1 + E(RA)] – VL*[1 + E(RL)] = 180*1.06 – 140*1.10 = USD 36.80 million.

For a 95% confidence interval, the appropriate

**z-value is 1.96**. Therefore, the lower bound of the surplus at the 95% confidence level = 36.80 – 1.96*35.76 = USD -33.2896 million.I dont really understand why they took as z-value 1,96 instead of 1,65 with a Surplus at risk 95%.

I thought that surplus at risk is calculated as Var (1 tail) because in other exercises as well as in kaplan books i have always seen like this. Indeed, this same exercise in Garp exam 2017 or 2016 (i cant remember exactly) used 1,65 as zvalue instead of 1,96, but since 2018 exam they use 1,96.

could anyone solve me this doubt? do anyone think the same as i do?

Thank you in advance.