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 - 2.03 and 3.79 and draw a sketch of the region. sketch the region. choose the correct graph below. a. b. c. d.

Explanation:

Step1: Recall the properties of standard - normal distribution

We know that if \(Z\) is a standard - normal random variable (\(Z\sim N(0,1)\)), and we want to find \(P(-2.03<Z<3.79)\). We use the standard - normal table (or z - table) which gives \(P(Z < z)\) for a given \(z\) value.

Step2: Use the property of normal distribution probabilities

\(P(-2.03<Z<3.79)=P(Z < 3.79)-P(Z < - 2.03)\).
From the standard - normal table, \(P(Z < 3.79)\approx1\) (since for \(z = 3.79\), the area to the left is very close to 1). And \(P(Z < - 2.03)=1 - P(Z < 2.03)\). Looking up \(P(Z < 2.03)\) in the z - table, we get \(P(Z < 2.03)=0.9788\), so \(P(Z < - 2.03)=1 - 0.9788 = 0.0212\).

Step3: Calculate the probability

\(P(-2.03<Z<3.79)=P(Z < 3.79)-P(Z < - 2.03)=1 - 0.0212=0.9788\).

The correct graph is the one where the area between \(z=-2.03\) and \(z = 3.79\) under the standard - normal curve is shaded. That is option B.

Answer:

The probability is \(0.9788\). The correct graph is B.