A young investment manager tells his client that the probability of making a positive return with his suggested portfolio is 81%. What is the risk (standard deviation) that this investment manager has assumed in his calculation if it is known that returns are normally distributed with a mean of 6.6%? Use Table 1. (Round "z" value to 2 decimal places and final answer to 3 decimal places.)
What is the Standard deviation ?
I have found the z which satisfies this equation, it is .88, however I'm a bit puzzled as to what to do next.
Risk and standard deviation are synonymous here.
What the investment manager is saying is that returns follow a normal distribution with mean 6.6 and some unknown variance $\sigma^2$. If we denote the return by $X$ we have $$ X \sim N(6.6, \sigma^2) $$
The investment manage also tells us that the probability that $X$ is greater than 0 is 81%. Thus we also have $$ P(X>0) = .81 = 1 - P(X\leq 0) = 1- F_X(0)$$ where $F_x(\cdot)$ is the cumulative distribution function (CDF). There are many tables for the CDF of a standard normal distribution (mean=0 and variance=1) so we look to standarize $X$ to follow a $N(0,1)$ distribution.
We can standardize $X$ by subtracting the mean and dividing by the standard deviation. Thus: $$ Y=\frac{X-6.6}{\sigma} \sim N(0,1)$$
But recall we know that $P(X>0)=.81 \Rightarrow P(\frac{X-6.6}{\sigma} > \frac{0-6.6}{\sigma})=.81$. Thus we now have that $P(Y > \frac{-6.6}{\sigma}) = .81$ which we can solve for $\sigma$ using the statistical tables :)