Chapter 6.1, Problem 21E

To determine

(a)

To find:

The missing value in the given table.

Answer to Problem 21E

Solution:

$z=-1.25$

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:

$x=35.0$, $\mu =37.0$, $\sigma =1.6$

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,

$\begin{array}{rll}z& =& \frac{\left(35.0\right)-\left(37.0\right)}{\left(1.6\right)}\\ & =& \frac{\left(-2.0\right)}{1.6}\\ & =& -\frac{2.0}{1.6}\\ & =& -1.25\end{array}$

To determine

(b)

To find:

The missing value in the given table.

Answer to Problem 21E

Solution:

$\mu =7.4$

Explanation of Solution

Given:

$z=1.75$, $x=12.3$, $\sigma =2.8$

Calculation:

Here, the mean has to be computed as,

$\begin{array}{rll}\mu & =& \left(12.3\right)-\left(1.75\right)\times \left(2.8\right)\\ & =& 12.3-4.9\\ & =& 7.4\end{array}$

To determine

(c)

To find:

The missing value in the given table.

Answer to Problem 21E

Solution:

$x=2.78$

Explanation of Solution

Given:

$z=-3.10$, $\mu =3.40$, $\sigma =0.20$

Calculation:

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

$\begin{array}{rll}x& =& \left(3.40\right)+\left(0.20\right)\times \left(-3.10\right)\\ & =& 3.40+\left(-0.62\right)\\ & =& 3.40-0.62\\ & =& 2.78\end{array}$

To determine

(d)

To find:

The missing value in the given table.

Answer to Problem 21E

Solution:

$\sigma =20$

Explanation of Solution

Given:

$z=2.15$, $x=479$, $\mu =436$

Calculation:

Here, the standard deviation is to be calculated as,

$\begin{array}{rll}\sigma & =& \frac{\left(479\right)-\left(436\right)}{\left(2.15\right)}\\ & =& \frac{43}{2.15}\\ & =& 20\end{array}$