Suppose the returns on an asset are normally distributed. The historical average
ID: 3134697 • Letter: S
Question
Suppose the returns on an asset are normally distributed. The historical average annual return for the asset was 6.7 percent and the standard deviation was 12.6 percent. What is the probability that your return on this asset will be less than –10.1 percent in a given year? Use the NORMDIST function in Excel® to answer this question. (Do not round intermediate calculations. Enter your answer as a percent rounded to 2 decimal places, e.g., 32.16.)
What range of returns would you expect to see 95 percent of the time? (Enter your answers for the range from lowest to highest. Negative amounts should be indicated by a minus sign. Do not round intermediate calculations. Enter your answers as a percent rounded to 2 decimal places, e.g., 32.16.)
What range would you expect to see 99 percent of the time? (Enter your answers for the range from lowest to highest. Negative amounts should be indicated by a minus sign. Do not round intermediate calculations. Enter your answers as a percent rounded to 2 decimal places, e.g., 32.16.)
Suppose the returns on an asset are normally distributed. The historical average annual return for the asset was 6.7 percent and the standard deviation was 12.6 percent. What is the probability that your return on this asset will be less than –10.1 percent in a given year? Use the NORMDIST function in Excel® to answer this question. (Do not round intermediate calculations. Enter your answer as a percent rounded to 2 decimal places, e.g., 32.16.)
Explanation / Answer
a)
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = -10.1
u = mean = 6.7
s = standard deviation = 12.6
Thus,
z = (x - u) / s = -1.333333333
Thus, using a table/technology, the left tailed area of this is
P(z < -1.333333333 ) = 0.09121122 = 9.12% [ANSWER]
[You can also do =NORMDIST(-10.1, 6.7, 12.6, 1) to get the answer.]
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b)
As the middle area is
Middle Area = P(x1<x<x2) = 0.95
Then the left tailed area of the left endpoint is
P(x<x1) = (1-P(x1<x<x2))/2 = 0.025
Thus, the z score corresponding to the left endpoint, by table/technology, is
z1 = -1.959963985
By symmetry,
z2 = 1.959963985
As
u = mean = 6.7
s = standard deviation = 12.6
Then
x1 = u + z1*s = -17.99554621%
x2 = u + z2*s = 31.39554621% [ANSWERS]
[You can also do =NORMINV(0.025, 6.7, 12.6) and =NORMINV(1-0.025, 6.7, 12.6).]
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c)
As the middle area is
Middle Area = P(x1<x<x2) = 0.99
Then the left tailed area of the left endpoint is
P(x<x1) = (1-P(x1<x<x2))/2 = 0.005
Thus, the z score corresponding to the left endpoint, by table/technology, is
z1 = -2.575829304
By symmetry,
z2 = 2.575829304
As
u = mean = 6.7
s = standard deviation = 12.6
Then
x1 = u + z1*s = -25.75544922%
x2 = u + z2*s = 39.15544922% [ANSWER]
[You can also do =NORMINV(0.005, 6.7, 12.6) and =NORMINV(1-0.005, 6.7, 12.6).]
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