Question

A normal distribution has a mean of 101 and a standard Deviation of 12. find the probability that a value selected at random is in the following interval.at most 13

Answer 1

84.134%

Step-by-step explanation:

First, determine the value of the z-score.

`\begin{gathered} Z=(X-\mu)/(\sigma) \\ =(113-101)/(12) \\ =(12)/(12) \\ z-score=1 \end{gathered}`

Next, we determine the probability that a value selected at random is at most 113:

`\begin{gathered} P(X\le113)=P(x\le1)_{} \\ =0.84134 \\ =84.134\% \end{gathered}`

Thus, the probability that a value selected at random is in the given interval is 84.134%.