Chapter 6.CR, Problem 16CR

To determine

a.

To find:

The first and third quartile for given mean and standard deviation values.

Answer to Problem 16CR

Solution:

The first quartile is $Q_1=13.593$ and the third quartile is $Q_3=16.449$.

Explanation of Solution

Given:

$\mu =15.0$

$\sigma =2.1$

Description:

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 (also known as Gaussian or Gauss-Laplacian) distribution is the most prevalent continuous probability distribution function and the real world hosts a milieu of the variables that are normally distributed and 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 and vice versa using a relation that is represented by using the following transformation:

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

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

The first quartile is defined as the middle point between smallest numbers and median that has the $z$-value $\left(z_a\right)$ corresponding to the area:

$P\left(z\le z_a\right)=0.25$.

The third quartile is defined as the mid-point between median and largest number that has $z$-value $\left(z_b\right)$ corresponding to the area:

$P\left(z\le z_b\right)=0.75$.

Initially, the $z$-values corresponding to first and third quartiles are looked up from table after which $x$-values are computed using the relation:

$x=z\times \sigma +\mu$.

Calculation:

The first quartile is obtained at the position in the table where the value is 0.25 as:

$z_a=-0.67$

Whose equivalent $x$-value is:

$\begin{array}{rll}{Q}_1& =& \left(-0.67\right)\times 2.1+15.0\\ & =& 13.593\end{array}$

The third quartile is obtained at the position in the table where the value is 0.75 as:

$z_b=0.69$

Whose equivalent $x$-value is:

$\begin{array}{rll}{Q}_3& =& \left(0.69\right)\times 2.1+15.0\\ & =& 16.449\end{array}$

Conclusion:

The first quartile is $Q_1=13.593$ and the third quartile is $Q_3=16.449$.

To determine

b.

To find:

Value of the mean and third quartile for given standard deviation and first quartile.

Answer to Problem 16CR

Solution:

The value of the mean is $3434$ and third quartile is $3572$.

Explanation of Solution

Given:

$\sigma =200$

First quartile $Q_1=3300$

Description:

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 (also known as Gaussian or Gauss-Laplacian) distribution is the most prevalent continuous probability distribution function and the real world hosts a milieu of the variables that are normally distributed and 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 and vice versa using a relation that is represented by using the following transformation:

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

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

The first quartile is defined as the middle point between smallest numbers and median that has the $z$-value $\left(z_a\right)$ corresponding to the area:

$P\left(z\le z_a\right)=0.25$.

The third quartile is defined as the mid-point between median and largest number that has $z$-value $\left(z_b\right)$ corresponding to the area:

$P\left(z\le z_b\right)=0.75$.

Using the first quartile, the corresponding $z$-value is obtained and the mean is computed using:

$\mu =x-z\times \sigma$.

Thereafter, the third quartile $x$-value is calculated using the relation:

$x=z\times \sigma +\mu$.

Calculation:

The $z$-value corresponding to first quartile from the standard normal table is as:

$z=-0.67$ is calculated above

Then, the mean is:

$\begin{array}{rll}\mu & =& \left(3300\right)-\left(-0.67\right)\times \left(200\right)\\ & =& 3434\end{array}$

Thereafter, the $z$-value corresponding third quartile from the standard normal table is as:

$z=0.69$

Then, the third quartile $x$-value is:

$\begin{array}{rll}{Q}_3& =& \left(0.69\right)\times \left(200\right)+\left(3434\right)\\ & =& 3572\end{array}$

Conclusion:

The value of the mean is $3434$ and third quartile is $3572$.

To determine

c.

To find:

The mean and standard deviation for given data.

Answer to Problem 16CR

Solution:

The required mean is $69.78$ and standard deviation is $22.06$.

Explanation of Solution

Given:

First quartile $Q_1=55$

Third quartile $Q_3=85$

Description:

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 (also known as Gaussian or Gauss-Laplacian) distribution is the most prevalent continuous probability distribution function and the real world hosts a milieu of the variables that are normally distributed and 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 $\mu$ $\sigma$ are the mean and standard deviation of original normal distribution respectively, specifically, the $x$-value can be represented as:

$x=z\times \sigma +\mu$.

The first quartile is defined as the middle point between smallest numbers and median that has the $z$-value $\left(z_a\right)$ corresponding to the area:

$P\left(z\le z_a\right)=0.25$.

The third quartile is defined as the mid-point between median and largest number that has $z$-value $\left(z_b\right)$ corresponding to the area:

$P\left(z\le z_b\right)=0.75$.

The $z$-values corresponding to the first and third quartiles are looked up in the standard tables and then using the transformation relations, two equations are obtained from where the mean and standard deviation values are computed.

Calculation:

The $z$-value corresponding to first quartile from the standard normal table is as:

$z_1=-0.67,z_3=0.69$

$\begin{array}{l}{Q}_3-Q_1=(z_3-z_1)\sigma \\ 85-55=(0.69+0.67)\sigma \end{array}$

$\begin{array}{l}\sigma =\frac{30}{1.36}\\ =22.06\end{array}$

Conclusion:

The required mean is $69.78$ and standard deviation is $22.06$.

To determine

d.

To find:

The values of standard deviation and first quartile for given data.

Answer to Problem 16CR

Solution:

The value of standard deviation is $0.58$ and the first quartile is $1.4114$.

Explanation of Solution

Given:

$\mu =1.8$

$Q_3=2.2$

Description:

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 (also known as Gaussian or Gauss-Laplacian) distribution is the most prevalent continuous probability distribution function and the real world hosts a milieu of the variables that are normally distributed and 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 $\mu$, $\sigma$ are the mean and standard deviation of original normal distribution respectively, specifically, the $x$-value can be represented as:

$x=z\times \sigma +\mu$.

The first quartile $\left(Q_1\right)$ is defined as the middle point between smallest numbers and median that has the $z$-value $\left(z_a\right)$ corresponding to the area:

$P\left(z\le z_a\right)=0.25$.

The third quartile $\left(Q_3\right)$ is defined as the mid-point between median and largest number that has $z$-value $\left(z_b\right)$ corresponding to the area:

$P\left(z\le z_b\right)=0.75$.

The relation between first quartile, mean and third quartile is:

$\frac{Q_1+Q_3}{2}=\mu$.

The $z$-value corresponding to third quartile is computed and then the standard deviation is obtained is computed using the relation above.

Since mean and third quartile are given, the first quartile is computed using the relation above.

Calculation:

The $z$-value corresponding to third quartile from the standard normal table is as:

$z_3=0.69$

Then, the value of standard deviation is:

$\sigma =\frac{Q_3-\mu }{z}$

$\begin{array}{rll}\sigma & =& \frac{2.2-1.8}{0.69}\\ & =& 0.58\end{array}$

Z-score of the first quartile is -0.67

$\begin{array}{l}{Q}_1=\mu +z\sigma \\ =1.8+(-0.67)(0.58)\\ =1.4114\end{array}$

Conclusion:

The value of standard deviation is $0.58$ and the first quartile is $1.4114$.