Chapter 7.4, Problem 14SE

Find the standard normal area for each of the following, showing your reasoning clearly and indicating which table you used.

1. a. P(1.22 < Z < 2.15)
2. b. P(2.00 < Z < 3.00)
3. c. P(−2.00 < Z < 2.00)
4. d. P(Z > 0.50)

To determine

Find the standard normal area for $P\left(1.22<Z<2.15\right)$.

Answer to Problem 14SE

The standard normal area for $P\left(1.22<Z<2.15\right)$ is 0.0954.

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 $P\left(1.22<Z<2.15\right)$ can be written as,

$\begin{array}{c}P\left(1.22<Z<2.15\right)=P\left(Z<2.15\right)-P\left(Z<1.22\right)\\ =\left[\begin{array}{c}\left[P\left(-\infty <Z<0\right)+P\left(0<Z<2.15\right)\right]\\ -\left[P\left(-\infty <Z<0\right)+P\left(0<Z<1.22\right)\right]\end{array}\right]\\ =P\left(0<Z<2.15\right)-P\left(0<Z<1.22\right)\end{array}$

Calculate the value of $P\left(0<Z<2.15\right)$:

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

• Locate the z value 2.1 in column of z.
• Locate the probability value corresponding to z-value 2.1 in the column 0.05.

Thus, $P\left(0<Z<2.15\right)=0.4842$.

Calculate the value of $P\left(0<Z<1.22\right)$:

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 0.02.

Thus, $P\left(0<Z<1.22\right)=0.3888$.

Hence,

$\begin{array}{c}P\left(1.22<Z<2.15\right)=P\left(0<Z<2.15\right)-P\left(0<Z<1.22\right)\\ =0.4842-0.3888\\ =0.0954\end{array}$

Hence, the standard normal area for $P\left(1.22<Z<2.15\right)$ is 0.0954.

To determine

Find the standard normal area for $P\left(2.00<Z<3.00\right)$.

Answer to Problem 14SE

The standard normal area for $P\left(2.00<Z<3.00\right)$ is 0.02145.

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 $P\left(2.00<Z<3.00\right)$ can be written as,

$\begin{array}{c}P\left(2.00<Z<3.00\right)=P\left(Z<3.00\right)-P\left(Z<2.00\right)\\ =\left[\begin{array}{c}\left[P\left(-\infty <Z<0\right)+P\left(0<Z<3.00\right)\right]\\ -\left[P\left(-\infty <Z<0\right)+P\left(0<Z<2.00\right)\right]\end{array}\right]\\ =P\left(0<Z<3.00\right)-P\left(0<Z<2.00\right)\end{array}$

Calculate the value of $P\left(0<Z<3.00\right)$:

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

• Locate the z value 3.0 in column of z.
• Locate the probability value corresponding to z-value 3.0 in the column 0.00.

Thus, $P\left(0<Z<3.00\right)=0.49865$.

Calculate the value of $P\left(0<Z<2.00\right)$:

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

• Locate the z value 2.0 in column of z.
• Locate the probability value corresponding to z-value 2.0 in the column 0.00.

Thus, $P\left(0<Z<2.00\right)=0.4772$.

Hence,

$\begin{array}{c}P\left(2.00<Z<3.00\right)=P\left(0<Z<3.00\right)-P\left(0<Z<2.00\right)\\ =0.49865-0.4772\\ =0.02145\end{array}$

Hence, the standard normal area for $P\left(2.00<Z<3.00\right)$ is 0.02145.

To determine

Find the standard normal area for $P\left(-2.00<Z<2.00\right)$.

Answer to Problem 14SE

The standard normal area for $P\left(-2.00<Z<2.00\right)$ is 0.9544.

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 $P\left(-2.00<Z<2.00\right)$ can be written as,

$\begin{array}{c}P\left(-2.00<Z<2.00\right)=P\left(Z<2.00\right)-P\left(Z<-2.00\right)\\ =\left[\begin{array}{c}\left[P\left(-\infty <Z<0\right)+P\left(0<Z<2.00\right)\right]\\ -P\left(-\infty <Z<-2.00\right)\end{array}\right]\\ =0.5+P\left(0<Z<2.00\right)-P\left(-\infty <Z<-2.00\right)\end{array}$

From part (b), it is found that $P\left(0<Z<2.00\right)=0.4772$.

Calculate the value of $P\left(Z<-2.00\right)$:

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

• Locate the z value –2.0 in column of z.
• Locate the probability value corresponding to z-value –2.0 in the column 0.00.

Thus, $P\left(Z<-2.00\right)=0.0228$.

Hence,

$\begin{array}{c}P\left(-2.00<Z<2.00\right)=0.5+P\left(0<Z<2.00\right)-P\left(-\infty <Z<-2.00\right)\\ =0.5+0.4772-0.0228\\ =0.9544\end{array}$

Hence, the standard normal area for $P\left(-2.00<Z<2.00\right)$ is 0.9544.

To determine

Find the standard normal area for $P\left(Z>0.50\right)$.

Answer to Problem 14SE

The standard normal area for $P\left(Z>0.50\right)$ is 0.3085.

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 $P\left(Z>0.50\right)$ can be written as,

$\begin{array}{c}P\left(Z>0.50\right)=1-P\left(Z<0.50\right)\\ =1-\left[P\left(-\infty <Z<0\right)+P\left(0<Z<0.50\right)\right]\\ =1-0.5-P\left(0<Z<0.50\right)\\ =0.5-P\left(0<Z<0.50\right)\end{array}$

Calculate the value of $P\left(0<Z<0.50\right)$:

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

• Locate the z value 0.5 in column of z.
• Locate the probability value corresponding to z-value 0.5 in the column 0.00.

Thus, $P\left(0<Z<0.50\right)=0.1915$.

Hence,

$\begin{array}{c}P\left(Z>0.50\right)=0.5-P\left(0<Z<0.50\right)\\ =0.5-0.1915\\ =0.3085\end{array}$

Hence, the standard normal area for $P\left(Z>0.50\right)$ is 0.3085.