Suppose charlie places $7000 in an account that pays 12% interest compounded each year. assume that no withdrawals are made from the account.

amount at the end of one year
amount at the end of two years

7000/.12=840
840x2=1680

At the end of year one you will have $840 and $7840 total
At the end of year two you will have $1680 and $8680 total

P(10 ≤ x ≤ 26) = 0.94083  
First, let's determine how many standard deviations each of the limits are  
(10 - 15.7) / 3.6 = -1.583333333 
(26 - 15.7) / 3.6 = 2.861111111  
So the probability of x being between 10 and 26 is the area covered by a normal distribution curve from -1.58333 to 2.86111. Using a standard normal table, look up the values -1.58 and 2.86. 
For the curve from infinity on the left to 2.86, you get the value 0.99788 which means that 99.788% of the curve is to the left of that point. 
For the curve from infinity on the left to -1.58, you get the value 0.05705 which means that 5.705% of the curve is to the left of that point. 
Since we want the area between those two points, we can simply subtract.
So: 
0.99788 - 0.05705 = 0.94083 
 So p(10 ≤ x ≤ 26) = 0.94083