Assume that the random variable X is normally distributed, with mean p = 100 and standard deviation o = 15. Compute the probability P(X > 112).

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 = 15.

Let X = a random variable

The z-score probability distribution for the normal distribution is given by;

                               Z  =    ~ N(0,1)

where, = population mean = 100

            = 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 0.80)

                                                       = 1 - 0.78814 = 0.21186  

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.

the correct answer is x/3+10≥32

hope this !