Answer:
P(X > 112) = 0.21186.
Step-by-step explanation:
We are given that the random variable X is normally distributed, with mean  = 100 and standard deviation
 = 100 and standard deviation  = 15.
 = 15. 
Let X = <u><em>a random variable</em></u>
The z-score probability distribution for the normal distribution is given by;
                                Z  =   ~ N(0,1)
  ~ N(0,1)
where,  = population mean = 100
 = population mean = 100
              = standard deviaton = 15
 = standard deviaton = 15
Now, the probability that the random variable X is greater than 112 is given by = P(X > 112)
       P(X > 112) = P(  >
 >  ) = P(Z > 0.80) = 1- P(Z
 ) = P(Z > 0.80) = 1- P(Z  0.80)
 0.80)
                                                        = 1 - 0.78814 = <u>0.21186</u>  
The above probability is calculated by looking at the value of x = 0.80 in the z table which has an area of 0.78814.