Chapter 7.4, Problem 17SE

Find the standard normal area for each of the following. Sketch the normal curve and shade in the area represented below.

1. a. P(Z < −1.28)
2. b. P(Z > 1.28)
3. c. P(−1.96 < Z < 1.96)
4. d. P(−1.65 < Z < 1.65)

To determine

Find the standard normal area for $P\left(Z<-1.28\right)$.

Answer to Problem 17SE

The standard normal area for $P\left(Z<-1.28\right)$ is 0.1003.

Explanation of Solution

Calculation:

Normal distribution:

A continuous random variable X is said to follow normal distribution if the probability density function of X is,

$f\left(x\right)=\frac{1}{\sigma \sqrt{2\pi }}{e}^{-\frac{{\left(x-\mu \right)}^2}{2{\sigma }^2}},-\infty <x<\infty$, with mean $\mu$ and standard deviation $\sigma$.

Standard normal distribution:

A continuous random variable Z is said to follow standard normal distribution if the probability density function of Z is,

$f\left(z\right)=\frac{1}{\sqrt{2\pi }}{e}^{-\frac{z^2}{2}},-\infty <z<\infty$, with mean 0 and standard deviation 1 for the standardized random variable $z=\frac{x-\mu }{\sigma }$ .

For a standard normal variable it is known that,

$\begin{array}{c}P\left(-\infty <Z<0\right)=0.5\\ =P\left(0<Z<\infty \right)\end{array}$

The probability $P\left(Z<-1.28\right)$ can be written as,

$P\left(Z<-1.28\right)=P\left(-\infty <Z<-1.28\right)$.

Calculate the value of $P\left(Z<-1.28\right)$:

From Appendix C-2: Table “CUMULATIVE STANDARD NORMAL DISTRIBTION”,

• Locate the z value –1.2 in column of z.
• Locate the probability value corresponding to z-value –1.2 in the column 0.08.

Thus, $P\left(Z<-1.28\right)=0.1003$.

Hence, the standard normal area for $P\left(Z<-1.28\right)$ is 0.1003.

To determine

Find the standard normal area for $P\left(Z>1.28\right)$.

Answer to Problem 17SE

The standard normal area for $P\left(Z>1.28\right)$ is 0.1003.

Explanation of Solution

Calculation:

For a standard normal variable it is known that,

$\begin{array}{c}P\left(-\infty <Z<0\right)=0.5\\ =P\left(0<Z<\infty \right)\end{array}$

The probability $P\left(Z>1.28\right)$ can be written as,

$\begin{array}{c}P\left(Z>1.28\right)=1-P\left(Z<1.28\right)\\ =1-\left[P\left(-\infty <Z<0\right)+P\left(0<Z<1.28\right)\right]\\ =1-0.5-P\left(0<Z<1.28\right)\\ =0.5-P\left(0<Z<1.28\right)\end{array}$

Calculate the value of $P\left(0<Z<1.28\right)$:

From Appendix C-1: Table “STANDARD NORMAL AREAS”,

• Locate the z value 1.2 in column of z.
• Locate the probability value corresponding to z-value 1.2 in the column 0.08.

Thus, $P\left(0<Z<1.28\right)=0.3997$.

Hence,

$\begin{array}{c}P\left(Z>1.28\right)=0.5-P\left(0<Z<1.28\right)\\ =0.5-0.3997\\ =0.1003\end{array}$

Hence, the standard normal area for $P\left(Z>1.28\right)$ is 0.1003.

To determine

Find the standard normal area for $P\left(-1.96<Z<1.96\right)$.

Answer to Problem 17SE

The standard normal area for $P\left(-1.96<Z<1.96\right)$ is 0.95.

Explanation of Solution

Calculation:

For a standard normal variable it is known that,

$\begin{array}{c}P\left(-\infty <Z<0\right)=0.5\\ =P\left(0<Z<\infty \right)\end{array}$

The probability $P\left(-1.96<Z<1.96\right)$ can be written as,

$\begin{array}{c}P\left(-1.96<Z<1.96\right)=P\left(Z<1.96\right)-P\left(Z<-1.96\right)\\ =\left[\begin{array}{c}\left[P\left(-\infty <Z<0\right)+P\left(0<Z<1.96\right)\right]\\ -P\left(-\infty <Z<-1.96\right)\end{array}\right]\\ =0.5+P\left(0<Z<1.96\right)-P\left(-\infty <Z<-1.96\right)\end{array}$

Calculate the value of $P\left(0<Z<1.96\right)$:

From Appendix C-1: Table “STANDARD NORMAL AREAS”,

• Locate the z value 1.9 in column of z.
• Locate the probability value corresponding to z-value 1.9 in the column 0.06.

Thus, $P\left(0<Z<1.96\right)=0.4750$.

Calculate the value of $P\left(Z<-1.96\right)$:

From Appendix C-2: Table “CUMULATIVE STANDARD NORMAL DISTRIBTION”,

• Locate the z value –1.9 in column of z.
• Locate the probability value corresponding to z-value –1.9 in the column 0.06.

Thus, $P\left(Z<-1.96\right)=0.0250$.

Hence,

$\begin{array}{c}P\left(-1.96<Z<1.96\right)=0.5+P\left(0<Z<1.96\right)-P\left(-\infty <Z<-1.96\right)\\ =0.5+0.4750-0.0250\\ =0.95\end{array}$

Hence, the standard normal area for $P\left(-1.96<Z<1.96\right)$ is 0.95.

To determine

Find the standard normal area for $P\left(-1.65<Z<1.65\right)$.

Answer to Problem 17SE

The standard normal area for $P\left(-1.65<Z<1.65\right)$ is 0.901.

Explanation of Solution

Calculation:

For a standard normal variable it is known that,

$\begin{array}{c}P\left(-\infty <Z<0\right)=0.5\\ =P\left(0<Z<\infty \right)\end{array}$

The probability $P\left(-1.65<Z<1.65\right)$ can be written as,

$\begin{array}{c}P\left(-1.65<Z<1.65\right)=P\left(Z<1.65\right)-P\left(Z<-1.65\right)\\ =\left[\begin{array}{c}\left[P\left(-\infty <Z<0\right)+P\left(0<Z<1.65\right)\right]\\ -P\left(-\infty <Z<-1.65\right)\end{array}\right]\\ =0.5+P\left(0<Z<1.65\right)-P\left(-\infty <Z<-1.65\right)\end{array}$

Calculate the value of $P\left(0<Z<1.65\right)$:

From Appendix C-1: Table “STANDARD NORMAL AREAS”,

• Locate the z value 1.6 in column of z.
• Locate the probability value corresponding to z-value 1.6 in the column 0.05.

Thus, $P\left(0<Z<1.65\right)=0.4505$.

Calculate the value of $P\left(Z<-1.65\right)$:

From Appendix C-2: Table “CUMULATIVE STANDARD NORMAL DISTRIBTION”,

• Locate the z value –1.6 in column of z.
• Locate the probability value corresponding to z-value –1.6 in the column 0.05.

Thus, $P\left(Z<-1.65\right)=0.0495$.

Hence,

$\begin{array}{c}P\left(-1.65<Z<1.65\right)=0.5+P\left(0<Z<1.65\right)-P\left(-\infty <Z<-1.65\right)\\ =0.5+0.4505-0.0495\\ =0.901\end{array}$

Hence, the standard normal area for $P\left(-1.65<Z<1.65\right)$ is 0.901.