>>12
average would naturally be halfway between the two bounds of the confidence interval, so add those together and divide by two. once you have that, you can solve for confidence level.
The bounds of a confidence interval for an average (arithmetic mean) are given by mean plus or minus Z*(standard deviation/square root of sample size). You can take either bound of the confidence interval (4.35 using minus or 9.65 using plus) and solve for z. So you could solve: upper bound = mean + z*(standard deviation/square root of sample size) for z. Now that you have z, you can figure out the corresponding area under the standard normal curve from -z to +z by whatever means at your disposal. That area corresponds to the probability involved with the confidence interval.
Let's say that area was 0.95. Then the confidence level would be 1 - 0.95 = 0.05. (Note the terminology -- a 0.95 confidence interval is based upon the confidence level of 0.05.)