Chapter 6.1, Problem 22E

To determine

(a)

To find:

The missing value in the given table.

Answer to Problem 22E

Solution:

$z=2$

Explanation of Solution

In statistics, a measurable function that is used to represent outcomes or occurrences of random phenomenon as real numbers is defined as a random variable.

The normal distribution is the most prevalent continuous probability distribution function and the real world hosts a milieu of the variables that are normally distributed..

In the normal distribution (also known as Gaussian or Gauss-Laplacian) distribution, the probability density function is given by:

$f\left(x|\mu ,{\sigma }^2\right)=\frac{1}{\sqrt{2\pi {\sigma }^2}}{e}^{-\frac{{\left(x-\mu \right)}^2}{2{\sigma }^2}}$

Where, $x$ is the value of the random variable while $\mu$ is the mean and the variance of the random variable which is normally distributed as ${\sigma }^2$.

The normal Gaussian distribution with the following characteristics:

$\mu =0$, and

$\sigma =1$

is called standard normal distribution.

The appropriate probability density function transforms to

$f\left(x|0,1\right)=\frac{1}{\sqrt{2\pi }}{e}^{-\frac{x^2}{2}}$

A normal distribution can be converted to standard normal distribution by defining the $z$-score that is represented by using the following transformation:

$z=\frac{x-\mu }{\sigma }$

where $x$ is the value of random variable and $\mu$ and $\sigma$ are the mean and standard deviation of original normal distribution respectively.

Given:

$x=2.87$, $\mu =2.45$, $\sigma =0.21$

Now, given any three of where value of random variable $\left(x\right)$, $z$-value $\left(z\right)$, mean $\left(\mu \right)$ and standard deviation $\left(\sigma \right)$, the fourth can be obtained by algebraically manipulating the relation mentioned above using the following relations:

$z=\frac{x-\mu }{\sigma }$

Calculation:

Here, the $z$-value has to be computed as,

$\begin{array}{rll}z& =& \frac{\left(2.87\right)-\left(2.45\right)}{\left(0.21\right)}\\ & =& \frac{0.42}{0.21}\\ & =& 2\end{array}$

To determine

b.

To find:

The missing value in the given table.

Answer to Problem 22E

Solution:

$\mu =602.8$

Explanation of Solution

Given:

$z=1.40$, $\mu =89.10$, $\sigma =7.45$

Calculation:

Here, the mean has to be computed,

$\begin{array}{rll}\mu & =& \left(579.0\right)-\left(-2.80\right)\times \left(8.5\right)\\ & =& 579.0-\left(-23.8\right)\\ & =& 579.0+23.8\\ & =& 602.8\end{array}$

To determine

c.

To find:

The missing value in the given table.

Answer to Problem 22E

Solution:

$x=99.53$

Explanation of Solution

Given:

$z=1.40$, $\mu =89.10$, $\sigma =7.45$

Calculation:

In this subpart, the $x$-value, that is the value of the random variable is to be obtained:

$\begin{array}{rll}x& =& \left(89.10\right)+\left(7.45\right)\times \left(1.40\right)\\ & =& 89.10+10.43\\ & =& 99.53\end{array}$

To determine

d.

To find:

The missing value in the given table.

Answer to Problem 22E

Solution:

$\sigma =0.45$

Explanation of Solution

Given:

$z=-3.20$, $x=13.67$, $\mu =15.11$

Calculation:

Here, the standard deviation is to be calculated:

$\begin{array}{rll}\sigma & =& \frac{\left(13.67\right)-\left(15.11\right)}{\left(-3.20\right)}\\ & =& \frac{\left(-1.44\right)}{\left(-3.20\right)}\\ & =& \frac{1.44}{3.20}\\ & =& 0.45\end{array}$