9. what proportion of a normal distribution is located between each of the following z - score…

Explanation:

Step1: Use z - table property

The proportion of a normal distribution between $z_1$ and $z_2$ is $P(z_1<Z<z_2)=\Phi(z_2)-\Phi(z_1)$, where $\Phi(z)$ is the cumulative - distribution function of the standard normal distribution. Since the standard normal distribution is symmetric about $z = 0$, we know that $\Phi(-z)=1 - \Phi(z)$.

Step2: Calculate for part a

For $z_1=-1.64$ and $z_2 = 1.64$, we have $P(-1.64<Z<1.64)=\Phi(1.64)-\Phi(-1.64)$. Since $\Phi(-1.64)=1 - \Phi(1.64)$, then $P(-1.64<Z<1.64)=\Phi(1.64)-(1 - \Phi(1.64))=2\Phi(1.64)-1$. Looking up in the z - table, $\Phi(1.64)=0.9495$, so $P(-1.64<Z<1.64)=2\times0.9495 - 1=0.899$.

Step3: Calculate for part b

For $z_1=-1.96$ and $z_2 = 1.96$, we have $P(-1.96<Z<1.96)=\Phi(1.96)-\Phi(-1.96)$. Since $\Phi(-1.96)=1 - \Phi(1.96)$, then $P(-1.96<Z<1.96)=2\Phi(1.96)-1$. Looking up in the z - table, $\Phi(1.96)=0.975$, so $P(-1.96<Z<1.96)=2\times0.975 - 1 = 0.95$.

Step4: Calculate for part c

For $z_1=-1.00$ and $z_2 = 1.00$, we have $P(-1.00<Z<1.00)=\Phi(1.00)-\Phi(-1.00)$. Since $\Phi(-1.00)=1 - \Phi(1.00)$, then $P(-1.00<Z<1.00)=2\Phi(1.00)-1$. Looking up in the z - table, $\Phi(1.00)=0.8413$, so $P(-1.00<Z<1.00)=2\times0.8413 - 1=0.6826$.

Answer:

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