Question

find the area of the shaded region. the graph depicts the standard normal distribution with mean 0 and standard deviation 1. round to four decimal places.
a. 0.9656
b. 0.0344
c. - 0.0344
d. 0.4656

Explanation:

Step1: Use the property of the standard normal distribution

The standard normal distribution is symmetric about \(z = 0\). The area to the left of \(z=- 1.82\) is \(P(Z<-1.82)\), and the area to the right of \(z = 1.82\) is \(P(Z>1.82)\). Since \(P(Z < -a)=P(Z>a)\) for a standard - normal variable \(Z\) and \(a>0\).
We know that the total area under the standard - normal curve is \(1\). The area of the un - shaded regions is \(P(Z < - 1.82)+P(Z>1.82)\). And \(P(Z < - 1.82)=P(Z>1.82)\).

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

From the standard normal table (or using a calculator with a normal - distribution function, e.g., \(P(Z\leq z)\) for a standard normal \(Z\)), \(P(Z\leq1.82)=0.9656\), \(P(Z < - 1.82)=1 - P(Z\leq1.82)\).
\(P(Z < - 1.82)=1 - 0.9656=0.0344\), and \(P(Z>1.82)=0.0344\).

Step3: Calculate the area of the shaded region

The area of the shaded region \(A = 1-(P(Z < - 1.82)+P(Z>1.82))\).
Substitute \(P(Z < - 1.82)=0.0344\) and \(P(Z>1.82)=0.0344\) into the formula: \(A=1-(0.0344 + 0.0344)=1 - 0.0688=0.9312\) (using the wrong - method above is a mistake).

Another way:
The area of the shaded region can also be calculated as \(P(-1.82<Z<1.82)\).
Since \(P(-1.82 < Z < 1.82)=\Phi(1.82)-\Phi(-1.82)\), and \(\Phi(-z)=1-\Phi(z)\) for the cumulative - distribution function \(\Phi(z)\) of the standard normal distribution.
\(P(-1.82 < Z < 1.82)=\Phi(1.82)-(1 - \Phi(1.82))=2\Phi(1.82)-1\).
Looking up \(\Phi(1.82)\) in the standard - normal table: \(\Phi(1.82)=0.9656\).
\(P(-1.82 < Z < 1.82)=2\times0.9656-1=1.9312 - 1=0.9656\).

Answer:

A. \(0.9656\)