In the last quarter of 2007, a group of 64 mutual funds had a mean return of 1.2

Question

In the last quarter of 2007, a group of 64 mutual funds had a mean return of 1.2% with a standard deviation of 6.6%. Consider the Normal model N (0.012.0.066) for the returns of these mutual funds. a) What value represents the 40th percentile of these returns? b) What value represents the 99th percentile? c) What's the IQR, or interquartile range, of the quarterly returns for this group of funds? a) The value that represents the 40th percentile is %. (Round to two decimal places as needed.) b) The value that represents the 99th percentile is %. (Round to two decimal places as needed.) c) The IQR is %. (Round to two decimal places as needed.)

Explanation / Answer

Solution:

In the last quarter of 2007 a group of 64 mutual funds had a mean return of 1.2% with a standard deviation of 6.6%. Consider the normal model N(0.012,.066) for the returns of these mutual funds.

Formula use to solve:

f(x;,) = [1/((2)* )]*e^-[(x-)²/(2²)]
--
z = (x - ) /
z = z-score
x = raw score or observation to be standardized
= mean of the population
= standard deviation of the population

===

a) What value represents the 40th percentile of these returns?
P(Z<x) = 0.4
--> P((Z-mean)/s <(x-0.012)/0.065) = 0.4
--> P(Z<(x - 0.012)/0.065)=0.4
--> (x - 0.012)/0.065= -0.25 (check standard normal table)
--> x = 0.012-0.25*0.065= -0.00425

b) What value represents the 99th percentile?
P(Z < x) = 0.99
--> P((Z-mean)/s <(x-0.012)/0.065) = 0.99
--> P(Z<(x - 0.012)/0.065)=0.99
--> (x - 0.012)/0.065= 2.33 (check standard normal table)
--> x = 0.012 + 2.33*0.065= 0.16345

c) What is the IQR, or interquartile range of the quarterly range of this group of funds?
IQR = z(75) - z(25) = 0.06785 - (-0.0198) = 0.08765

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