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P Do Homework quindarius shy Google Chrome https//www.mathxlcom/Student/PlayerHomework aspx? homeworkld 404903568. false&cld; 4260534¢erwin; yes &questionld-68flushed; BUSA 2100 IB Spring17 Homework: CH07 Continuous Random Variables Score: 0 of 2 pts 9 of 11 (10 complete) v 7.1.33 In the last quarter of 2007, a group of 64 mutual funds had a mean return of 2 9% with a standard deviation of67w consider the Normal model No 0290067 for the retums of thesemutual funds a) What value represents the 40th percentile of these returns? b) What value represents the 99th percentle? c) What's the IOR, or interquartile range, of the quarterly retums for this group of funds?Explanation / Answer
Mean ( u ) =0.029
Standard Deviation ( sd )=0.067
Normal Distribution = Z= X- u / sd ~ N(0,1)
a.
P ( Z < x ) = 0.4
Value of z to the cumulative probability of 0.4 from normal table is -0.253
P( x-u/s.d < x - 0.029/0.067 ) = 0.4
That is, ( x - 0.029/0.067 ) = -0.25
--> x = -0.25 * 0.067 + 0.029 = 0.012
b.
P ( Z < x ) = 0.01
Value of z to the cumulative probability of 0.01 from normal table is -2.326
P( x-u/s.d < x - 0.029/0.067 ) = 0.01
That is, ( x - 0.029/0.067 ) = -2.33
--> x = -2.33 * 0.067 + 0.029 = -0.1269
c.
Q1 = P ( Z < x ) = 0.25
Value of z to the cumulative probability of 0.25 from normal table is -0.674
P( x-u/s.d < x - 0.029/0.067 ) = 0.25
That is, ( x - 0.029/0.067 ) = -0.67
--> x = -0.67 * 0.067 + 0.029 = -0.0162
Q3 = P ( Z < x ) = 0.75
Value of z to the cumulative probability of 0.75 from normal table is 0.674
P( x-u/s.d < x - 0.029/0.067 ) = 0.75
That is, ( x - 0.029/0.067 ) = 0.67
--> x = 0.67 * 0.067 + 0.029 = 0.0742
IAR = Q3 - Q1 = 0.0742 - -0.0162 = 0.058
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