Answer:
The mutual friend in in the top 28.8%. So X% = 28.8%.
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
[tex]\mu = 0.09, \sigma = 0.16[/tex]
One mutual fund in this group managed to earn a rate of return that was double that of this group's average that year. This performance would put the fund in the top X% of those funds in that year.Calculate X%.
Here we have
[tex]X = 2\mu = 2*0.09 = 0.18[/tex]
Top X%
X is 1 subtracted by the pvalue of Z when X = 0.18. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{0.18 - 0.09}{0.16}[/tex]
[tex]Z = 0.56[/tex]
[tex]Z = 0.56[/tex]has a pvalue of 0.712.
100 - 71.2 = 28.8%
The mutual friend in in the top 28.8%. So X% = 28.8%.
