So we can start with the full of possibilities and eliminate them one by one.
The full set is {0,1,2,3,4,5,6,7,8,9}.
Now we know that any prime greater than 2 is odd as otherwise it would have 2 as a factor, so we can eliminate all of these digits that would be an even number, leaving:
{1,3,5,7,9}
We also know that any prime greater than 5 cannot be a multiple of 5 and that all numbers with 5 in the digits are a multiple of 5, so we can eliminate 5.
{1,3,7,9}
We know that 11,13,17 and 19 are all primes, so we cannot eliminate any more of these, leaving the set:
The confidence interval would be (10.44, 12.16). This means that if we take repeated samples, the true mean lies in 90% of these intervals.
To find the confidence interval, we use:
We first find the z-value associated with this. To do this: Convert 90% to a decimal: 90% = 90/100 = 0.9 Subtract from 1: 1-0.9 = 0.1 Divide by 2: 0.1/2 = 0.05 Subtract from 1: 1-0.05 = 0.95
Using a z-table (http://www.z-table.com) we see that this is directly between two z-scores, 1.64 and 1.65; we will use 1.645: