X is a normally distributed random variable with a mean of 1

X is a normally distributed random variable with a mean of 10.0 and a standard deviation of 3.00. Find the value x such that P(X < x) is equal to 0.877. Round your answer to two decimal places.

Solution

Normal Distribution
Mean ( u ) =10
Standard Deviation ( sd )=3
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
P ( Z < x ) = 0.877
Value of z to the cumulative probability of 0.877 from normal table is 1.16
P( x-u/s.d < x - 10/3 ) = 0.877
That is, ( x - 10/3 ) = 1.16
--> x = 1.16 * 3 + 10 = 13.48