Assume that x has a normal distribution with the specified mean and standard deviation. find the indicated probability. (round your answer to four decimal places.) μ = 15.7; σ = 3.6 p(10 ≤ x ≤ 26) =

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

The first one is 2/3the second is 1/160the third is 4/125