Chapter 7.4, Problem 15SE

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(−3.00 < Z < 2.00)
3. c. P(Z < 2.00)
4. d. P(Z = 0)

To determine

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

Answer to Problem 15SE

The standard normal area for $P\left(-1.22<Z<2.15\right)$ is 0.873.

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,

$\begin{array}{c}P\left(-1.22<Z<2.15\right)=P\left(Z<2.15\right)-P\left(Z<-1.22\right)\\ =\left[\begin{array}{c}\left[P\left(-\infty <Z<0\right)+P\left(0<Z<2.15\right)\right]\\ -P\left(-\infty <Z<-1.22\right)\end{array}\right]\\ =0.5+P\left(0<Z<2.15\right)-P\left(-\infty <Z<-1.22\right)\end{array}$

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 $P\left(Z<-1.22\right)$:

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 0.02.

Thus, $P\left(Z<-1.22\right)=0.1112$.

Hence,

$\begin{array}{c}P\left(-1.22<Z<2.15\right)=0.5+P\left(0<Z<2.15\right)-P\left(-\infty <Z<-1.22\right)\\ =0.5+0.4842-0.1112\\ =0.873\end{array}$

Hence, the standard normal area for $P\left(-1.22<Z<2.15\right)$ is 0.873.

To determine

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

Answer to Problem 15SE

The standard normal area for $P\left(-3.00<Z<2.00\right)$ is 0.97585.

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

$\begin{array}{c}P\left(-3.00<Z<2.00\right)=P\left(Z<2.00\right)-P\left(Z<-3.00\right)\\ =\left[\begin{array}{c}\left[P\left(-\infty <Z<0\right)+P\left(0<Z<2.00\right)\right]\\ -P\left(-\infty <Z<-3.00\right)\end{array}\right]\\ =0.5+P\left(0<Z<2.00\right)-P\left(-\infty <Z<-3.00\right)\end{array}$

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

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

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

• 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, $P\left(Z<-3.00\right)=0.00135$.

Hence,

$\begin{array}{c}P\left(-3.00<Z<2.00\right)=0.5+P\left(0<Z<2.00\right)-P\left(-\infty <Z<-3.00\right)\\ =0.5+0.4772-0.00135\\ =0.97585\end{array}$

Hence, the standard normal area for $P\left(-3.00<Z<2.00\right)$ is 0.97585.

To determine

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

Answer to Problem 15SE

The standard normal area for $P\left(Z<2.00\right)$ is 0.9772.

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

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

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

Hence,

$\begin{array}{c}P\left(Z<2.00\right)=0.5+P\left(0<Z<2.00\right)\\ =0.5+0.4772\\ =0.9772\end{array}$

Hence, the standard normal area for $P\left(Z<2.00\right)$ is 0.9772.

To determine

Find the standard normal area for $P\left(Z=0\right)$.

Answer to Problem 15SE

The standard normal area for $P\left(Z=0\right)$ is 0.

Explanation of Solution

Calculation:

Calculate the value of $P\left(Z=0\right)$.

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

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

Thus, $P\left(Z=0\right)=\underset{\_}{0}$.

Moreover, it can be said that in case of continuous distribution the probability value for a particular point is 0.

Therefore, the standard normal area for $P\left(Z=0\right)$ is 0.