Question

In the last quarter of​ 2007, a group of 64 mutual funds had a mean return of 2.2 2.2​% with a standard deviation of 6.3 6.3​%. Consider the Normal model ​N( 0.022 0.022​, 0.063 0.063​) 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 nothing ​%. ​(Round to two decimal places as​ needed.)

Answer 1

a) 0.625; b) 16.879; c) 8.442

Explanation:

Since this is a normal distribution, we want the value of the mean, μ: 2.2; and the value of the standard deviation, σ: 6.3.

For the 40th percentile, we look in a z chart. We want to find the value as close to 0.40 as we can get; this is 0.4013, and it corresponds to a z score of z = -0.25.

Our formula for a z score is
`z=(X-\mu)/(\sigma)`. Using our values, we have

-0.25 = (X-2.2)/6.3

Multiply both sides by 6.3:

6.3(-0.25) = X-2.2

-1.575 = X-2.2

Add 2.2 to each side:

-1.575+2.2 = X-2.2+2.2

0.625 = X

For the 99th percentile, the value in the z chart closest to 0.99 is 0.9901, which corresponds to a z score of z = 2.33:

2.33 = (X-2.2)/6.3

Multiply both sides by 6.3:

6.3(2.33) = X-2.2

14.679 = X-2.2

Add 2.2 to each side:

14.679+2.2 = X-2.2+2.2

16.879 = X

For the IQR, we find the values for the 75th percentile (Q3) and the 25th percentile (Q1). The value in a z chart closest to 0.75 is 0.7486, which corresponds to a z score of z = 0.67:

0.67 = (X-2.2)/6.3

Multiply both sides by 6.3:

6.3(0.67) = X-2.2

4.221 = X-2.2

Add 2.2 to each side:

4.221+2.2 = X-2.2+2.2

6.421 = X

The value in a z chart closest to 0.25 is 0.2514, which corresponds to a z score of z = -0.67:

-0.67 = (X-2.2)/6.3

Multiply both sides by 6.3:

6.3(-0.67) = X-2.2

-4.221 = X-2.2

Add 2.2 to each side:

-4.221+2.2 = X-2.2+2.2

-2.021 = X

This makes the interquartile range

6.421--2.021 = 8.442