Question

In the last quarter of​ 2007, a group of 64 mutual funds had a mean return of 3.3​% with a standard deviation of 4.6​%. Consider the Normal model ​N(0.033​,0.046​) 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

a) 0.021362

b) 0.139996

c) 0.062008

Explanation:

We are given the following information in the question:

Mean, μ = 3.3​% = 0.033

Standard Deviation, σ = 4.6​% = 0.046

We are given that the distribution of mutual funds is a bell shaped distribution that is a normal distribution.

a) We have to find the value of x such that the probability is 0.4.

`P( X < x) = P( z < \displaystyle(x - 0.033)/(0.046))=0.4`

Calculation the value from standard normal z table, we have,

`P( z< -0.253) = 0.4`

`\displaystyle(x - 0.033)/(0.046) = -0.253\\x = 0.021362`

b) We have to find the value of x such that the probability is 0.99

`P( X < x) = P( z < \displaystyle(x - 0.033)/(0.046))=0.99`

Calculation the value from standard normal z table, we have,

`P( z< 2.326) = 0.99`

`\displaystyle(x - 0.033)/(0.046) = 2.326\\x =0.139996`

c) IQR =
`Q_3- Q_1 = 75^(th)\text{Percentile} - 25^(th)\text{Percentile}`

We have to find the value of x such that the probability is 0.75

`P( X < x) = P( z < \displaystyle(x - 0.033)/(0.046))=0.75`

Calculation the value from standard normal z table, we have,

`P( z< 0.674) = 0.75`

`\displaystyle(x - 0.033)/(0.046) = 0.674\\x =0.064004`

We have to find the value of x such that the probability is 0.25

`P( X < x) = P( z < \displaystyle(x - 0.033)/(0.046))=0.25`

Calculation the value from standard normal z table, we have,

`P( z< -0.674) = 0.25`

`\displaystyle(x - 0.033)/(0.046) = -0.674\\x =0.001996`

IQR =
`0.064004 - 0.001996 = 0.062008`