The Wall Street Journal reported that long term Treasury bonds had a mean return of 24.03% in 2008. Assume that the returns for the long term Treasury bonds were distributed as a normal random...

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The Wall Street Journal reported that long term Treasury bonds had a mean return of 24.03% in 2008. Assume that the returns for the long term Treasury bonds were distributed as a normal random variable, with a mean of 24.03 and a standard deviation of 10. If you select an individual Treasury bond from this population, what is the probability that it would have a return of...


a. less than 0 (a loss)?
b. between 10 and 20 ?
c. greater than 10 ?


If you select a random sample of 4 Treasure bonds from this population, what is the probability that the sample would have a mean return of...


d. less than 0 (a loss)?
e. between 10 and 20 ?
f. greater than 10 ?
g. Compare your results in parts (d) through (f) to those in parts (a) through (c)

Answered Same DayApr 22, 2021BUS201

Answer To: The Wall Street Journal reported that long term Treasury bonds had a mean return of 24.03% in 2008....

Pooja answered on Apr 23 2021
106 Votes
1)
a)
P( X<0)    = ?
    
I know that, z = (X-mean)/(sd)    
z1 = (0-24.03)/10)    -2.4030
hence,    
P( X<
0)    = P(Z<-2.403)
P( X<0)    = NORMSDIST(-2.403)
P( X<0)    = 0.0081
b)
P(10 < X < 20) = P(X<20) - P(X<10) = ?
    
I know that, z = (X-mean)/(sd)    
z1 = (10-24.03)/10) =    -1.4030
z2 = (20-24.03)/10) =    -0.4030
    
hence,    
P(10 < X < 20)=    P(Z<-0.403) - P(Z<-1.403)
P(10 < X < 20)=    NORMSDIST(-0.403) - NORMSDIST(-1.403)
P(10 < X < 20)=    0.2632
c)
P( X>10) = 1 - P(X<10) =...
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