Chapter 7.1, Problem 29E

Find the area under the standard normal curve that lies outside the interval between

1. $z=-0.38$ and $z=1.02$
2. $z=-1.42$ and $z=1.78$
3. $z=0.01$ and $z=2.67$
4. $z=-2.45$ and $z=-0.34$

To determine

To find area outside the -0.38 and 1.02

Answer to Problem 29E

Area outside the -0.38 and 1.02 is 0.5058

Explanation of Solution

Given:

  $\begin{array}{l}{z}_1=-0.38\\ z_2=1.02\end{array}$

Shaded region is outside of above z-scores

That is to find $P\left(Z<-0.38\right)+P\left(Z>1.02\right)$

Calculation:

Using standard normal table,

  $\begin{array}{l}P\left(Z<-0.38\right)=0.3520\hfill \\ P\left(Z<1.02\right)=0.8461\hfill \\ So,\hfill \\ P\left(Z>1.02\right)=1–0.8461=0.1539\hfill \\ Therefore,\hfill \\ P\left(Z<-0.38\right)+P\left(Z>1.02\right)=0.3520+0.1539=0.5058\hfill \end{array}$

To determine

To find area outside the -1.42 and 1.78

Answer to Problem 29E

Area outside the -1.42 and 1.78 is 0.1153

Explanation of Solution

Given:

  $\begin{array}{l}{z}_1=-1.42\\ z_2=1.78\end{array}$

Shaded region is outside of above z-scores

That is to find $P\left(Z<-1.42\right)+P\left(Z>1.78\right)$

Calculation:

Using standard normal table,

  $\begin{array}{l}P\left(Z<-1.42\right)=0.0778\hfill \\ P\left(Z<1.78\right)=0.9625\hfill \\ So,\hfill \\ P\left(Z>1.78\right)=1–0.9625=0.0375\hfill \\ Therefore,\hfill \\ P\left(Z<-1.42\right)+P\left(Z>1.78\right)=0.0778+0.0375=0.1153\hfill \end{array}$

To determine

To find area outside the 0.01 and 2.67

Answer to Problem 29E

Area outside the 0.01 and 2.67 is 0.5078

Explanation of Solution

Given:

  $\begin{array}{l}{z}_1=0.01\\ z_2=2.67\end{array}$

Shaded region is outside of above z-scores

That is to find $P\left(Z<0.01\right)+P\left(Z>2.67\right)$

Calculation:

Using standard normal table,

  $\begin{array}{l}P\left(Z<0.01\right)=0.5040\hfill \\ P\left(Z<2.67\right)=0.9962\hfill \\ So,\hfill \\ P\left(Z>2.67\right)=1–0.9962=0.0038\hfill \\ Therefore,\hfill \\ P\left(Z<0.01\right)+P\left(Z>2.67\right)=0.5040+0.0038=0.5078\hfill \end{array}$

To determine

To find area outside the -2.45 and -0.34

Answer to Problem 29E

Area outside the -2.45 and -0.34 is 0.6402

Explanation of Solution

Given:

  $\begin{array}{l}{z}_1=-2.45\\ z_2=-0.34\end{array}$

Shaded region is outside of above z-scores

That is to find $P\left(Z<-2.45\right)+P\left(Z>-0.34\right)$

Calculation:

Using standard normal table,

  $\begin{array}{l}P\left(Z<-2.45\right)=0.0071\hfill \\ P\left(Z<-0.34\right)=0.3669\hfill \\ So,\hfill \\ P\left(Z>-0.34\right)=1–0.3669=0.6331\hfill \\ Therefore,\hfill \\ P\left(Z<-2.45\right)+P\left(Z>-0.34\right)=0.0071+0.6331=0.6402\hfill \end{array}$