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 -1.57. sketch the region. choose the correct graph below. the probability is . (round to four decimal places as needed.)

Explanation:

Step1: Use the property of the standard normal distribution

The standard normal distribution is symmetric about \(x = 0\). The probability \(P(Z>z)=1 - P(Z\leq z)\). Here \(z=- 1.57\).

Step2: Look up the value in the standard - normal table

We know that for a standard normal random variable \(Z\) (with mean \(\mu = 0\) and standard deviation \(\sigma=1\)), we can use the standard - normal table (z - table). The value of \(P(Z\leq - 1.57)\) is found by looking up the row corresponding to \(-1.5\) and the column corresponding to \(0.07\) in the standard - normal table. \(P(Z\leq - 1.57)=0.0582\).

Step3: Calculate \(P(Z > - 1.57)\)

Using the formula \(P(Z>z)=1 - P(Z\leq z)\), when \(z=-1.57\), we have \(P(Z>-1.57)=1 - P(Z\leq - 1.57)\). Substitute \(P(Z\leq - 1.57) = 0.0582\) into the formula, so \(P(Z>-1.57)=1-0.0582 = 0.9418\).

Answer:

\(0.9418\)