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
= 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)
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 = <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.