Statistic mean normal distribution. HELP!!!?

Andrew plans to retire in 60 years. He is thinking of investing his retirement funds in stocks, so he seeks out information on past returns. He learns that over the 101 years from 1900 to 2000, the real (that is, adjusted for inflation) returns on U.S. common stocks had mean 8.3% and standard deviation 20.8%. The distribution of annual returns on common stocks is roughly symmetric, so the mean return over even a moderate number of years is close to Normal.

What is the probability (assuming that the past pattern of variation continues) that the mean annual return on common stocks over the next 60 years will exceed 10%? (±0.0001)

What is the probability (±0.0001) that the mean return will be less than 6%

I tried every possible method to solve this but i still dont think the answers are right. Please help me solve this. Thank you.

The s.d. of the mean will be s.d.of the sample/sqrt(60).

set z=(10.0-8.3)/ [(20.8)/sqrt(60)] to get the z score for exceeding 10% and find the probability (the area under the standard normal density beyond that z.

z = 0.63308
prob = .263 this is from tables, use a caluclator for better accuracy.

Next, z = (6-8.3)/2.68526 same thing
z= -0.8565
The prob = 0.1957