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Sagot :
Answer:
a) 0.0062 = 0.62% probability that the return will exceed 55%.
b) 0.3085 = 30.85% probability that the return will be less than 25%
c) 30%.
d) The 75th percentile of returns is 36.75%.
Step-by-step explanation:
Normal Probability Distribution
Problems of normal distributions can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
Mean 30% and standard deviation 10%.
This means that [tex]\mu = 30, \sigma = 10[/tex]
(a) Find the probability that the return will exceed 55%.
This is 1 subtracted by the p-value of Z when X = 55. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{55 - 30}{10}[/tex]
[tex]Z = 2.5[/tex]
[tex]Z = 2.5[/tex] has a p-value of 0.9938
1 - 0.9938 = 0.0062
0.0062 = 0.62% probability that the return will exceed 55%.
(b) Find the probability that the return will be less than 25%
p-value of Z when X = 25. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{25 - 30}{10}[/tex]
[tex]Z = -0.5[/tex]
[tex]Z = -0.5[/tex] has a p-value of 0.3085
0.3085 = 30.85% probability that the return will be less than 25%.
(c) What is the expected value of the return?
The mean, that is, 30%.
(d) Find the 75th percentile of returns.
X when Z has a p-value of 0.75, so X when Z = 0.675.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]0.675 = \frac{X - 30}{10}[/tex]
[tex]X - 30 = 0.675*10[/tex]
[tex]X = 36.75[/tex]
The 75th percentile of returns is 36.75%.
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