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 - 1.86 and 1.96. sketch the region. choose the correct graph below.

Explanation:

Step1: Recall the standard - normal distribution property

We know that for a standard - normal distribution \(Z\sim N(0,1)\), the probability \(P(-1.86<Z<1.86)\) can be found using the standard - normal table. The standard - normal table gives \(P(Z < z)\) for different values of \(z\).

Step2: Use the symmetry of the standard - normal distribution

We know that \(P(-1.86<Z<1.86)=P(Z < 1.86)-P(Z < - 1.86)\). Since the standard - normal distribution is symmetric about \(z = 0\), \(P(Z < -1.86)=1 - P(Z < 1.86)\). So \(P(-1.86<Z<1.86)=2P(Z < 1.86)-1\).
Looking up in the standard - normal table, \(P(Z < 1.86)=0.9686\).

Step3: Calculate the probability

Substitute \(P(Z < 1.86)=0.9686\) into the formula \(P(-1.86<Z<1.86)=2\times0.9686 - 1\).

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For the graph, the correct graph is the one where the area between \(z=-1.86\) and \(z = 1.86\) under the standard - normal curve is shaded. That is graph A.

Answer:

A. The graph with the area between \(z=-1.86\) and \(z = 1.86\) shaded; Probability \(=0.9372\)