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?

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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

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 [tex]X \sim N(\mu, \sigma)[/tex]

(X is following normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] )

then it can be converted to standard normal distribution as

[tex]Z = \dfrac{X - \mu}{\sigma}, \\\\Z \sim N(0,1)[/tex]

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

[tex]P(Z \leq z) = P(Z < z) )[/tex]

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

[tex]P(Z \leq z) = \rm p \: value[/tex]

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

Then, by the given data, we get:

[tex]X \sim N(5.8, \sigma)\\\\[/tex]

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:

[tex]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[/tex]

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

Thus,

[tex]-0.845 = \dfrac{0-5.8}{\sigma}\\\\\sigma = \dfrac{5.8}{0.845} \approx 6.86 \%[/tex]

Thus,

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

Learn more about standard normal distribution here:

https://brainly.com/question/10984889

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