Assume that a randomly selected subject is given a bone density test. Those test scores are normally distributed with a mean of 0 and a standard deviation of 1. Find the probability that a given score is between negative 2.13 and 3.88 and draw a sketch of the region.

Answer: 0.9834Step-by-step explanation:Given : The test scores are normally distributed withMean : [tex]\mu=\ 0[/tex]Standard deviation :[tex]\sigma= 1[/tex]The formula to calculate the z-score :-[tex]z=\dfrac{x-\mu}{\sigma}[/tex]For x = -2.13[tex]z=\dfrac{-2.13-0}{1}=-2.13[/tex]For x = 3.88[tex]z=\dfrac{3.88-0}{1}=3.88[/tex]The p-value = [tex]P(-2.13<z<3.88)=P(z<3.88)-P(z<-2.13)[/tex][tex]0.9999477-0.0165858=0.9833619\approx0.9834[/tex]Hence, the probability that a given score is between negative 2.13 and 3.88 = 0.9834