Chapter 4.5, Problem 59E

a.

To determine

Find the value of z₀ such that $P\left(Z>z_0\right)=0.5$

Answer to Problem 59E

The value of z₀ such that $P\left(Z>z_0\right)=0.5$ is 0.0000.

Explanation of Solution

Calculation:

The value of z₀ such that $P\left(Z>z_0\right)$= 0.5 is obtained below:

In general, the standard normal distribution has $\mu =0$ and $\sigma =1$.

$\begin{array}{c}P\left(Z>z_0\right)=A\left(z_0\right)\\ A(z_0)=A\left(0.50\right)\end{array}$

From Table 4: Appendix 3, Standard Normal Probabilities (Cumulative z Curve Areas), the following is obtained:

• Locate the area 0.5000 From Table 4: Appendix 3.
• Move left until the first column and note the values as 0.0.
• Move upward until the top row is reached and note the values as 0.00.

The value of z₀ such that $P\left(Z>z_0\right)$= 0.5 is 0.000.

b.

To determine

Find the value of z₀ such that $P\left(Z<z_0\right)=0.8643$

Answer to Problem 59E

The value of z₀ such that $P\left(Z<z_0\right)=0.8643$ is 1.10.

Explanation of Solution

Calculation:

In general, the standard normal distribution has $\mu =0$ and $\sigma =1$.

$\begin{array}{c}P\left(Z<z_0\right)=0.8643\\ 1-P\left(Z\ge z_0\right)=0.8643\end{array}$

Here, the following is obtained:

$\begin{array}{c}P\left(Z>z_0\right)=A\left(z_0\right)\\ 1-A\left(z_0\right)=0.8643\\ A\left(z_0\right)=0.1357\end{array}$

Use Table 4: Appendix 3, Standard Normal Probabilities (Cumulative z Curve Areas) to obtain the following:

• Locate the area 0.8643 in Table 4: Appendix 3.
• Move left until the first column is reached and note the value as 1.1
• Move upward until the top row is reached and note the value as 0.0.

The value of z₀ such that $P\left(Z<z_0\right)=0.8643$ is 1.10.

c.

To determine

Find the value of z₀ such that $P\left(-z_0<Z<z_0\right)=0.90$.

Answer to Problem 59E

The value of z₀ such that $P\left(-z_0<Z<z_0\right)=0.90$ is 1.645.

Explanation of Solution

Calculation:

In general, the standard normal distribution has $\mu =0$ and $\sigma =1$.

$P\left(Z\le z_0\right)+P\left(-z_0<Z<z_0\right)+P\left(Z\ge z_0\right)=1$

Here,

$\begin{array}{c}P\left(-z_0<Z<z_0\right)+2P\left(Z\ge z_0\right)=1\\ P\left(-z_0<Z<z_0\right)=1-2P\left(Z\ge z_0\right)\end{array}$

$\begin{array}{c}P\left(-z_0<Z<z_0\right)=0.90\\ P\left(Z>z_0\right)=A\left(z_0\right)\\ 1-2A\left(z_0\right)=0.90\\ A\left(z_0\right)=0.05\end{array}$

Use Table 4: Appendix 3, Standard Normal Probabilities (Cumulative z Curve Areas) to obtain the following:

• Locate the area 0.05 in Table 4: Appendix 3.
• Move left until the first column is reached and note the value as 1.6
• Move upward until the top row is reached and note the value as 0.04 and 0.05

The required value is as follows:

$\begin{array}{c}{z}_0=\frac{1.64+1.65}{2}\\ =1.645\end{array}$

The value of z₀ such that  $P\left(-z_0<Z<z_0\right)=0.90$ is 1.645.

d.

To determine

Find the value of z₀ such that $P\left(-z_0<Z<z_0\right)=0.99$

Answer to Problem 59E

The value of z₀ such that $P\left(-z_0<Z<z_0\right)=0.99$ is 2.575.

Explanation of Solution

Calculation:

In general, the standard normal distribution has $\mu =0$ and $\sigma =1$.

$P\left(Z\le z_0\right)+P\left(-z_0<Z<z_0\right)+P\left(Z\ge z_0\right)=1$

Here,

$\begin{array}{c}P\left(-z_0<Z<z_0\right)+2P\left(Z\ge z_0\right)=1\\ P\left(-z_0<Z<z_0\right)=1-2P\left(Z\ge z_0\right)\end{array}$

$\begin{array}{c}P\left(-z_0<Z<z_0\right)=0.99\\ P\left(Z>z_0\right)=A\left(z_0\right)\\ 1-2A\left(z_0\right)=0.99\\ A\left(z_0\right)=0.005\end{array}$

Use Table 4: Appendix 3, Standard Normal Probabilities (Cumulative z Curve Areas) to obtain the following:

• Locate the area 0.005 in Table 4: Appendix 3.
• Move left until the first column is reached and note the value as 2.5
• Move upward until the top row is reached and note the value as 0.07 and 0.08

The required value is as follows:

$\begin{array}{c}{z}_0=\frac{2.57+2.58}{2}\\ =2.575\end{array}$

The value of z₀ such that  $P\left(-z_0<Z<z_0\right)=0.99$ is 2.575.