Question

According to the Normal model N(0 063,0.022) describing mutual fund returns in the 1st quarter of 2013, determine what percentage of this group of funds you would expect to have the following returns, Complete parts (a) through (d) below a) Over 6 8%? c) More than 1%? b) Between 0% and 76%? d) Less than 0%? is□% a) The expected percentage of returns that are over 68% Type an integer or a decimal rounded to one decimal place as needed)

Answer 1

Answer)

Normal models are defined by N(mean, s.d)

Therefore, mean = 0.063

S.d= 0.022

A)

Over 6.8% that is over 0.068

P(x>0.068)

First we need to estimate the standard z score

Z = (x-mean)/s.d

Z = (0.068-0.063)/0.022

Z = 0.23

From z table, p(z>0.23) = 0.4090

Or 40.9%

B)

P(0<x<0.076) = p(x<0.076)-p(x<0)

P(x<0.076) => z = (0.076-0.063)/0.022 = 0.59

From z table, p(z<0.59) = 0.7224

P(x<0) => z = (0-0.063)/0.022 = -2.86

From z table, P(z<-2.86) = 0.0021

Required probability = 0.7224-0.0021

= 0.7203

= 72.03%

= 72.0%

C)

P(x>0.01)

Z = (0.01-0.063)/0.022

= -2.41

From z table, p(z>-2.41) = 0.992

= 99.2%

D)

P(x<0)

Z = (0-0.063)/0.022

Z = -2.86

From z table, p(z<-2.86)

= 0.0021

= 0.2%