Question

find the indicated area under the curve of the standard normal distribution; then convert it to a percentage and fill in the blank.
about ____ % of the area is between ( z = - 1 ) and ( z = 1 ) (or within 1 standard deviation of the mean).
about ( square % ) of the area is between ( z = - 1 ) and ( z = 1 ) (or within 1 standard deviation of the mean).
(round to two decimal places as needed.)

Explanation:

Step1: Use the standard normal distribution property

The total area under the standard normal curve is \(1\). The area to the left of \(z = 1\) is \(\Phi(1)\), and the area to the left of \(z=-1\) is \(\Phi(-1)\). The area between \(z = - 1\) and \(z = 1\) is \(A=\Phi(1)-\Phi(-1)\).
We know that the standard normal distribution is symmetric about \(z = 0\), so \(\Phi(-z)=1 - \Phi(z)\). Then \(A=\Phi(1)-(1 - \Phi(1))=2\Phi(1)-1\).

Step2: Look up the value of \(\Phi(1)\)

From the standard - normal table (or using a calculator with a normal - distribution function, e.g., for a standard normal random variable \(Z\sim N(0,1)\)), \(\Phi(1)\approx0.8413\).

Step3: Calculate the area

Substitute \(\Phi(1)\approx0.8413\) into the formula \(A = 2\Phi(1)-1\). Then \(A=2\times0.8413-1=1.6826 - 1=0.6826\).

Step4: Convert to a percentage

To convert the area to a percentage, multiply the area by \(100\). So the percentage is \(0.6826\times100 = 68.26\%\).

Answer:

\(68.26\)