Question

Assume that a randomly selected subject is given a bone density test. Those test scores are normally distributed with a mean of O and a standard deviation of 1. Find the probability that a given score is less than -1.91

Answer 1

Given the following parameters

$\begin{gathered} \mu=0 \\ \sigma=1 \\ x=-1.91 \end{gathered}$

The z-score formula is given below

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

Calculate the z-score value using the above formula by substituting the parameters above

$z=\frac{-1.91-0}{1}=\frac{-1.91}{1}=-1.91$

The probability that the score is less than -1.91 can be diagramatically represented as

The probability can calculated as;

[tex]\begin{gathered} Pr(z<-1.91)\Rightarrow Pr(0\le z)-Pr(0Hence, the probability that a given score is less than -1.91 is 0.0281