Question

1. what proportion of a normal distribution is located between each of the following z - score boundaries?

a. z = - 1.64 and z = + 1.64
b. z = - 1.96 and z = + 1.96
c. z = - 1.00 and z = + 1.00

Explanation:

Step1: Use z - table property

The total area under the standard - normal curve is 1. The area to the left of a z - score $z$ is given by the cumulative distribution function $\varPhi(z)$ of the standard normal distribution. The area between $z_1$ and $z_2$ ($z_1

Step2: Find areas for part a

For $z_1=-1.64$ and $z_2 = 1.64$, from the standard normal table, $\varPhi(-1.64)=0.0505$ and $\varPhi(1.64)=0.9495$. Then the area between them is $\varPhi(1.64)-\varPhi(-1.64)=0.9495 - 0.0505=0.899$.

Step3: Find areas for part b

For $z_1=-1.96$ and $z_2 = 1.96$, from the standard normal table, $\varPhi(-1.96)=0.025$ and $\varPhi(1.96)=0.975$. Then the area between them is $\varPhi(1.96)-\varPhi(-1.96)=0.975 - 0.025 = 0.95$.

Step4: Find areas for part c

For $z_1=-1.00$ and $z_2 = 1.00$, from the standard normal table, $\varPhi(-1.00)=0.1587$ and $\varPhi(1.00)=0.8413$. Then the area between them is $\varPhi(1.00)-\varPhi(-1.00)=0.8413 - 0.1587=0.6826$.

Answer:

a. $0.899$
b. $0.95$
c. $0.6826$