The time it takes a supercomputer to perform a task is norma

The time it takes a supercomputer to perform a task is normally distributed with mean 10 milliseconds and standard deviation 4 milliseconds. What is the probability that it takes more than 18.2 milliseconds to perform the task? (use the normal table)

(a) 0.979818 (b) 0.845632 (c) 0.020182 (d) 0.223578 (e) none of the preceding

Please show step by step, thank you

Solution

Normal Distribution
Mean ( u ) =10
Standard Deviation ( sd )=4
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
P(X > 18.2) = (18.2-10)/4
= 8.2/4 = 2.05
= P ( Z >2.05) From Standard Normal Table
= 0.020182