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 greater than - 3.65. sketch the region. choose the correct graph below. the probability of a bone density test score greater than - 3.65 is (round to four decimal places as needed.)

Explanation:

Step1: Use the standard normal distribution property

For a standard normal distribution \(Z\sim N(0,1)\), we want to find \(P(Z > - 3.65)\).
We know that \(P(Z>z)=1 - P(Z\leq z)\)

Step2: Look up the \(z -\)value in the standard - normal table

Looking up \(z=-3.65\) in the standard - normal table (the cumulative distribution function of the standard normal distribution \(\varPhi(z)=P(Z\leq z)\)).
From the standard - normal table, \(P(Z\leq - 3.65)\approx0.0001\)

Step3: Calculate the probability

Using the formula \(P(Z > - 3.65)=1 - P(Z\leq - 3.65)\)
Substitute \(P(Z\leq - 3.65) = 0.0001\) into the formula:
\(P(Z > - 3.65)=1-0.0001 = 0.9999\)

Answer:

The probability is \(0.9999\) and the correct graph is the one with the value \(-3.65\) (Option B for the graph part, assuming the graph labeled with \(-3.65\) is the correct region for \(P(Z > - 3.65)\))