Question

In the last quarter of​ 2007, a group of 64 mutual funds had a mean return of 0.8​% with a standard deviation of 4.9​%. Consider the Normal model ​N(0.008​,0.049​) 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?

Answer 1

Solution:
Given in the question
Mean return() = 0.008
Standard devition() = 0.049
No. of sample(n) = 64
Solution(a)
We need to calculate 40th percentile of these returns
So we will use standard normal distribution to calculate return, Z-score from Z table at p-value = 0.40 is -00.2534
So return can be calculated as
Z = (X-)//sqrt(n)
-0.2534 = (X-0.008)/0.049/sqrt(64)
X = 0.008 - 0.0016 = 0.0064
So 40th percentile is 0.64%
Solution(b)
In this we need to calculate 99th percentile, which can be calculated as
P-value = 0.99, Z-score = 2.3264
So return can be calculated as
Z = (X-)//sqrt(n)
2.3264 = (X-0.008)/0.049/sqrt(64)
X = 0.008 + 0.0142 = 0.0222
So 99th percentile is 2.22%
Solution(c)
For IQR Range we will calculate 25 th percentile and 75 th percentile
For which Z-score for 25th percentile is -0.6745
Z = (X-)//sqrt(n)
-0.6745 = (X-0.008)/0.049/sqrt(64)
X = 0.008 - 0.0041 = 0.0039
For which Z-score for 75th percentile is 0.6745
Z = (X-)//sqrt(n)
0.6745 = (X-0.008)/0.049/sqrt(64)
X = 0.008 + 0.0041 = 0.0121
So IQR = 75th percentile - 25th Percentile = 0.0121 - 0.0039 = 0.0082
So IQR = 0.82%