Question
Assuming that the returns from holding small-company stocks are normally distributed, what is the approximate probability that your money will double in value in a single year? What about triple in value.
Many of my classmates have an expected return of 17.6% and a standard deviation of 34.8. Then they have the formula Prob(d>(200-17.6)/34.8 = prob. d>2.37
1-d>2.37=1-.991 = .009 chance of doubling.
I just need this problem explained to me. I'm not sure where all the numbers and information are coming from. Any help would be greatly appreciated.
Many of my classmates have an expected return of 17.6% and a standard deviation of 34.8. Then they have the formula Prob(d>(200-17.6)/34.8 = prob. d>2.37
1-d>2.37=1-.991 = .009 chance of doubling.
I just need this problem explained to me. I'm not sure where all the numbers and information are coming from. Any help would be greatly appreciated.
Answers
normal distribution problem
I do not know where your mean (expected return) and sigma are coming from. I assume some text book.
I do not know where your mean (expected return) and sigma are coming from. I assume some text book.
Now if we assume a mean return of 17.6%
and a sigma of 34.8%
doubling is 100% which is 82.4% above mean and tripling is 200% which is 182.4% above mean
how many sigmas is 82.4?
82.4/34.8 =2.37 sigmas to double
how many sigmas is 182.4?
182.4/34.8 = 5.24 sigmas to triple
(note you used 200 where you should have used 100 in interpreting classmate results)
and a sigma of 34.8%
doubling is 100% which is 82.4% above mean and tripling is 200% which is 182.4% above mean
how many sigmas is 82.4?
82.4/34.8 =2.37 sigmas to double
how many sigmas is 182.4?
182.4/34.8 = 5.24 sigmas to triple
(note you used 200 where you should have used 100 in interpreting classmate results)
Now go to tables for normal distribution:
to be 2.37 sigmas above mean:
well I only have a rough table here. For z = 2.3, F(z) = .989
for z = 2.4, F(z) = .992
so about 99% is below 2.37 sigma above mean and we only have about a 100-99 or a 1 percent chance of doubling. (you probably have more accurate tables). You will need a very accurate table though to find any probability of exceeding 5.24 sigmas above mean
to be 2.37 sigmas above mean:
well I only have a rough table here. For z = 2.3, F(z) = .989
for z = 2.4, F(z) = .992
so about 99% is below 2.37 sigma above mean and we only have about a 100-99 or a 1 percent chance of doubling. (you probably have more accurate tables). You will need a very accurate table though to find any probability of exceeding 5.24 sigmas above mean
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