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 0.69. sketch the region. choose the correct graph below. the probability 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 \(P(Z > 0.69)\).
We know that \(P(Z>z)=1 - P(Z\leq z)\).

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

Looking up \(z = 0.69\) 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, \(P(Z\leq0.69)=0.7549\).

Step3: Calculate the probability

Using the formula \(P(Z > 0.69)=1 - P(Z\leq0.69)\), we substitute the value of \(P(Z\leq0.69)\).
\(P(Z > 0.69)=1-0.7549 = 0.2451\)

Answer:

The probability is \(0.2451\)