Question

An investment has an expected return of 11 percent per year with a standard deviation of 26 percent. Assuming that the returns on this investment are at least roughly normally distributed, how often do you expect to earn less than -15 percent?

Answer 1

`P(X<-15)=P((X-\mu)/(\sigma)<(-15-\mu)/(\sigma))=P(Z<(-15-11)/(26))=P(Z<-1)`

And we can find this probability using the normal standard distribution table or excel and we got:

`P(Z<-1)=0.159`

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the expected return, and for this case we know the distribution for X is given by:

`X \sim N(11,26)`

Where
`\mu=11` and
`\sigma=26`

We are interested on this probability

`P(X<-15)`

And the best way to solve this problem is using the normal standard distribution and the z score given by:

`z=(x-\mu)/(\sigma)`

If we apply this formula to our probability we got this:

`P(X<-15)=P((X-\mu)/(\sigma)<(-15-\mu)/(\sigma))=P(Z<(-15-11)/(26))=P(Z<-1)`

And we can find this probability using the normal standard distribution table or excel and we got:

`P(Z<-1)=0.159`