Question

Suppose the random variable Y has a normal distribution with an expected value equal to 12 and variance equal to 16. What is the value of k such as that P(X<k)= 0.25

Answer 1

$$P(X<k)=P(Z<\dfrac{k-12}{\sqrt{16}})=0.25$$

$$\dfrac{k-12}{4}\approx-0.67449$$

$$k\approx9.30204$$