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 −1.88. sketch the region. choose the correct graph below. (options a, b, c, d with normal curve graphs and shaded regions)

Explanation:

Step1: Understand the Normal Distribution

The bone density test scores are normally distributed with mean $\mu = 0$ and standard deviation $\sigma = 1$, so it's a standard normal distribution ($Z$-distribution). We need to find the probability that a score is greater than $-1.88$, i.e., $P(Z > -1.88)$.

Step2: Recall the Property of Normal Distribution

For a standard normal distribution, the total area under the curve is 1. The probability $P(Z > -1.88)$ is equal to the area to the right of $z = -1.88$. We can use the symmetry of the normal distribution or the standard normal table.

Step3: Use the Standard Normal Table or Calculator

Using the standard normal table, we know that $P(Z \leq -1.88)$ is the area to the left of $z = -1.88$. Looking up $z = -1.88$ in the standard normal table, we find that $P(Z \leq -1.88) \approx 0.0301$. Then, $P(Z > -1.88) = 1 - P(Z \leq -1.88) = 1 - 0.0301 = 0.9699$.

For the graph, since we want the region where $Z > -1.88$, the shaded area should be to the right of $z = -1.88$. Looking at the options, option A has the shaded area to the right of $-1.88$ (since it's a single tail at $-1.88$ and the rest to the right is shaded), option B has two tails (which is for a different case, like $|Z| > 1.88$), option C has a small shaded area to the left (which would be $P(Z < -1.88)$), and option D has shaded area to the left of $1.88$ (not relevant here). So the correct graph is option A.

Answer:

The correct graph is A. The probability that a bone density test score is greater than -1.88 is approximately 0.9699.