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 - 1.93 and 1.93. sketch the region. choose the correct graph below.

Explanation:

Step1: Recall the standard - normal distribution property

Let \(Z\) be a standard - normal random variable with mean \(\mu = 0\) and standard deviation \(\sigma=1\). We want to find \(P(-1.93<Z<1.93)\).
We know that \(P(-1.93 < Z<1.93)=P(Z < 1.93)-P(Z < - 1.93)\).

Step2: Use the standard - normal table

From the standard - normal table (z - table), the cumulative distribution function \(\varPhi(z)\) gives \(P(Z < z)\).
We know that the standard - normal distribution is symmetric about \(z = 0\), so \(P(Z < -z)=1 - P(Z < z)\). For \(z = 1.93\), from the z - table, \(P(Z < 1.93)=0.9732\) and \(P(Z < - 1.93)=1 - 0.9732 = 0.0268\).

Step3: Calculate the probability

\(P(-1.93 < Z<1.93)=P(Z < 1.93)-P(Z < - 1.93)=0.9732-(1 - 0.9732)=0.9732-0.0268 = 0.9464\).

The correct graph is the one where the area between \(z=-1.93\) and \(z = 1.93\) under the standard - normal curve is shaded. Looking at the options, it should be the graph where the middle part between \(-1.93\) and \(1.93\) is shaded.

Answer:

The probability is \(0.9464\). The correct graph is the one with the area between \(z=-1.93\) and \(z = 1.93\) shaded (it is likely option A based on the general description of the shaded region in the problem - but without seeing the exact details of each option's shading precision, we can't be 100% sure of the letter of the correct graph option from the description given).