Independent solution

How to solve this Normal Distribution question

Answer in brief

Independence converts the two-year at-least-one probability into a one-year negative-profit probability of 0.20. Hence -100/sigma is the 20th standard-normal quantile, giving sigma=118.818 and choice C.

Setup

Setup

Let p be the probability of a negative result in one year. Use the complement of seeing no negative year in two independent trials.

1−(1−p)2=0.361-(1-p)^2=0.36

Model

Model

Solve for the annual probability, then standardize the zero threshold under the normal model.

1−p=0.64=0.80⟹p=0.201-p=\sqrt{0.64}=0.80\quad\Longrightarrow\quad p=0.20
Φ ⁣(0−100σ)=0.20\Phi\!\left(\frac{0-100}{\sigma}\right)=0.20

Compute

Compute

Use the 20th standard-normal quantile and solve for the positive standard deviation.

Φ−1(0.20)=−0.8416212336…\Phi^{-1}(0.20)=-0.8416212336\ldots
−100σ=−0.8416212336-\frac{100}{\sigma}=-0.8416212336
σ=118.8182950…\sigma=118.8182950\ldots

Answer

Answer

The annual standard deviation rounds to 119.

119(C)\boxed{119\quad\text{(C)}}