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 - 3.66. sketch the region. choose the correct graph below. o a. 3.66 o b. - 3.66 o c. - 3.66 o d. 3.66 the probability of a bone density test score greater than - 3.66 is (round to four decimal places as needed.)

Explanation:

Step1: Use the standard normal distribution property

For a standard normal distribution \(Z\sim N(0,1)\), the probability \(P(Z > - 3.66)\) can be found using the property \(P(Z>z)=1 - P(Z\leq z)\).

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

Looking up \(z=-3.66\) in the standard - normal table (the cumulative distribution function of the standard normal distribution \(\varPhi(z)\)). The value of \(P(Z\leq - 3.66)\) is approximately \(0.0001\).

Step3: Calculate the probability

Using the formula \(P(Z > - 3.66)=1 - P(Z\leq - 3.66)\). Substitute \(P(Z\leq - 3.66) = 0.0001\) into the formula, we get \(P(Z>-3.66)=1 - 0.0001=0.9999\).

Answer:

\(0.9999\)