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. find the probability that a given score is between -2.09 and 3.59 and draw a sketch of the region. sketch the region. choose the correct graph below. (options a, b, c, d show normal distribution curves with shaded regions between -2.09 and 3.59 or other regions)

Explanation:

Step1: Identify the distribution

The 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 the area between \( z = -2.09 \) and \( z = 3.59 \).

Step2: Recall standard normal table usage

For a standard normal distribution, the probability \( P(-2.09 < Z < 3.59) = P(Z < 3.59) - P(Z < -2.09) \).

Step3: Find \( P(Z < 3.59) \)

From the standard normal table, \( P(Z < 3.59) \approx 0.9998 \) (since 3.59 is a very high z - score, almost all the area is to the left of it).

Step4: Find \( P(Z < -2.09) \)

From the standard normal table, \( P(Z < -2.09) = 1 - P(Z < 2.09) \). Looking up \( z = 2.09 \), we get \( P(Z < 2.09) \approx 0.9817 \), so \( P(Z < -2.09)=1 - 0.9817 = 0.0183 \).

Step5: Calculate the probability

\( P(-2.09 < Z < 3.59)=0.9998 - 0.0183 = 0.9815 \)

For the graph: We need the region between \( z=-2.09 \) and \( z = 3.59 \) in the standard normal curve (bell - shaped curve). The correct graph should have the shaded area between these two z - scores. Looking at the options, option C (assuming the description of option C is the one with the shaded area between - 2.09 and 3.59) is the correct graph.

Answer:

The probability that a given score is between - 2.09 and 3.59 is approximately \( 0.9815 \), and the correct graph is the one (e.g., option C) with the shaded region between \( z=-2.09 \) and \( z = 3.59 \) in the standard normal curve.