Question

A young investment manager tells his client that the probability of making a positive return with his suggested portfolio is 80%. If it is known that returns are normally distributed with a mean of 5.8%, what is the risk, measured by standard deviation, that this investment manager assumes in his calculation?

Answer 1

Answer:

The risk, measured by standard deviation, that this investment manager assumes in his calculation is 0.0689

Step-by-step explanation:

Let X = rate of return

Since his client wants to make a positive return:

X > 0

P(X > 0) = 0.8

P(Z > (0-0.058)/sd) = 0.8

(0-0.058)/sd = -0.8416

-0.058/-0.8416 = standard deviation

standard deviation = 0.0689

Therefore, The risk, measured by standard deviation, that this investment manager assumes in his calculation is 0.0689

Answer 2

Probability of an event shows the chance of occurrence of that outcome.

The risk (standard deviation) that the considered investment manager assumes is 6.86% approximately.

How to get the z scores?

If we've got a normal distribution, then we can convert it to standard normal distribution and its values will give us the z score.

If we have $X \sim N(\mu, \sigma)$

(X is following normal distribution with mean $\mu$ and standard deviation $\sigma$ )

then it can be converted to standard normal distribution as

$Z = \dfrac{X - \mu}{\sigma}, \\\\Z \sim N(0,1)$

(Know the fact that in continuous distribution, probability of a single point is 0, so we can write

$P(Z \leq z) = P(Z < z) )$

Also, know that if we look for Z = z in z tables, the p value we get is

$P(Z \leq z) = \rm p \: value$

For the given case, let X be the random variable tracking the percentage of returns, and let $\sigma$ (standard deviation of distribution of X) be the risk assumed by the investment manager.

Then, by the given data, we get:

$X \sim N(5.8, \sigma)$

Also, since it is given that 80% probability is for the return to be positive.

Or

Values of X > 0 are 80% of the total values recorded.

Using standard normal distribution, this is written as:

$P(Z > \dfrac{0 - 5.8}{\sigma}) = 0.8\\\\1 - P(Z \leq \dfrac{0 - 5.8}{\sigma}) = 0.8\\\\P(Z \leq \dfrac{0 - 5.8}{\sigma}) = 0.2$

Seeing in z-tables, this p value is obtained on Z = -0.845 approximately.

Thus,

$-0.845 = \dfrac{0-5.8}{\sigma}\\\\\sigma = \dfrac{5.8}{0.845} \approx 6.86 \%$

Thus,

The risk (standard deviation) that the considered investment manager assumes is 6.86% approximately.

Learn more about standard normal distribution here:

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