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=(x-\mu)/(\sigma)=(x-266)/(12)`

doing this inside the probability, we get the following:

`\begin{gathered} P(x\ge242)=P((x-\mu)/(\sigma)\ge(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