Question

Suppose a normal distribution has a mean of 266 and a standard deviation of
12. What is P(x >= 242)

Answer 1

If we standarize the variable x, we get:

$z=\frac{x-\mu}{\sigma}=\frac{x-266}{12}$

doing this inside the probability, we get the following:

$\begin{gathered} P(x\ge242)=P(\frac{x-\mu}{\sigma}\ge\frac{242-266}{12})=P(z\ge-2) \\ By\text{ the property of the complement, we have:} \\ P(z\ge-2)=1-P(z<-2)=1-0.0228=0.9772 \\ \Rightarrow P(x\ge242)=0.97 \end{gathered}$

therefore, the probability is 0.97