assume that a randomly selected subject is given a bone density test. those test scores are normally…
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. draw a graph and find the probability of a bone density test score between - 1.57 and 1.57. sketch the region. choose the correct graph below. a. b. c. d.
Answer
Explanation:
Step1: Recall the properties of standard - normal distribution
We know that if (Z) is a standard - normal random variable ((Z\sim N(0,1))), and we want to find (P(-1.57<Z<1.57)).
Step2: Use the symmetry of the standard - normal distribution
The standard - normal distribution is symmetric about (z = 0). We know that (P(-1.57<Z<1.57)=\Phi(1.57)-\Phi(-1.57)), where (\Phi(z)) is the cumulative - distribution function of the standard - normal distribution. Since (\Phi(-z)=1 - \Phi(z)) for a standard - normal variable (Z), then (P(-1.57<Z<1.57)=\Phi(1.57)-(1 - \Phi(1.57)) = 2\Phi(1.57)-1).
Step3: Look up the value in the standard - normal table
From the standard - normal table, (\Phi(1.57)=0.9418). Then (P(-1.57<Z<1.57)=2\times0.9418 - 1=0.8836). The correct graph is the one that shows the area between (z=-1.57) and (z = 1.57) shaded. That is graph B.
Answer:
B. The graph with the area between (z=-1.57) and (z = 1.57) shaded.