Using the normal distribution, it is found that:
The z-score of a measure X of a normally distributed variable that has mean represented by [tex]\mu[/tex] and standard deviation represented by [tex]\sigma[/tex] is given by the following rule:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
In the context of this problem, the mean and the standard deviation are given as follows:
[tex]\mu = 14.7, \sigma = 4.7[/tex]
For a return of at least 20%, the probability is one subtracted by the p-value of Z when X = 20, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
Z = (20 - 14.7)/4.7
Z = 1.13
Z = 1.13 has a p-value of 0.8708
1 - 0.8708 = 0.1292.
For the return of 10% or less, the probability is the p-value of Z when X = 10, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
Z = (10 - 14.7)/4.7
Z = -1
Z = -1 has a p-value of 0.1587.
To be in the top 10%, the return has to be of at least the 90th percentile, which is X when Z = 1.28, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
1.28 = (X - 14.7)/4.7
X - 14.7 = 1.28 x 4.7
X = 20.72%.
More can be learned about the normal distribution at https://brainly.com/question/4079902
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