Question

Risk taking is an important part of investing. In order to make suitable investment decisions on behalf of their customers, portfolio managers give a questionnaire to new customers to measure their desire to take financial risks. The scores on the questionnaire are approximately normally distributed with a mean of 49.5 and a standard deviation of 15. The customers with scores in the bottom 10% are described as "risk averse." What is the questionnaire score that separates customers who are considered risk averse from those who are not? Carry your intermediate computations to at least four decimal places. Round your answer to one decimal place.

Answer 1

Answer:

$a=49.5 -1.28*15=30.3$

So the value of height that separates the bottom 10% of data from the top 90% is 30.3.  

If the score is lower than 30.3 we consider this score as risk averse

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 scores of a population, and for this case we know the distribution for X is given by:

$X \sim N(49.5,15)$  

Where $\mu=49.5$ and $\sigma=15$

We are interested in the bottom 10% of the data.

For this part we want to find a value a, such that we satisfy this condition:

$P(X>a)=0.9$   (a)

$P(X<a)=0.1$   (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

As we can see on the figure attached the z value that satisfy the condition with 0.1 of the area on the left and 0.9 of the area on the right it's z=-1.28. On this case P(Z<-1.28)=0.1 and P(z>-1.28)=0.9

If we use condition (b) from previous we have this:

$P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.1$  

$P(z<\frac{a-\mu}{\sigma})=0.1$

But we know which value of z satisfy the previous equation so then we can do this:

$z=-1.28<\frac{a-49.5}{15}$

And if we solve for a we got

$a=49.5 -1.28*15=30.3$

So the value of height that separates the bottom 10% of data from the top 90% is 30.3.  

If the score is lower than 30.3 we consider this score as risk averse