Question

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.

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 random variable \(Z\), then \(\Phi(-1.57)=1-\Phi(1.57)\). So \(P(-1.57<Z<1.57)=\Phi(1.57)-(1 - \Phi(1.57)) = 2\Phi(1.57)-1\).

Step3: Look up the value in the standard - normal table

Looking up the value of \(\Phi(1.57)\) in the standard - normal table, we find that \(\Phi(1.57)=0.9418\). Then \(P(-1.57<Z<1.57)=2\times0.9418 - 1=0.8836\).
The correct graph is the one that has 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; Probability is \(0.8836\)