Chapter 6.CR, Problem 15CR

To determine

a.

To find:

The probability of random variable less than given $x$-value.

Answer to Problem 15CR

Solution:

The required probability is $0.8665$.

Explanation of Solution

Given:

$\mu =22.0$

$\sigma =1.8$

$x=24.0$

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 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.

The probabilities being calculated using the $x$- and $z$-values are related as below:

$P\left(X\le x\right)=P\left(Z\le z\right)$.

Calculation:

Given value of random variable $\left(x\right)$, mean $\left(\mu \right)$ and standard deviation $\left(\sigma \right)$, then $z$-value $\left(z\right)$ is obtained using:

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

Thereafter, the relation $P\left(X\le x\right)=P\left(Z\le z\right)$ is used to find the required probability.

First, the $z$-value is computed:

$\begin{array}{rll}z& =& \frac{\left(24.0\right)-\left(22.0\right)}{\left(1.8\right)}\\ & =& \frac{2.0}{1.8}\\ & =& 1.11\end{array}$

Then, from the table:

$P\left(Z\le z\right)=0.86650$,

which in turn gives:

$P\left(X\le x\right)=0.8665$.

Conclusion:

The required probability is $0.8665$.

To determine

b.

To find:

Value of the mean.

Answer to Problem 15CR

Solution:

The value of the mean is $116.11$.

Explanation of Solution

Given:

$\sigma =0.24$

$x=116.00$

$P\left(X\le x\right)=0.3228$

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 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.

Calculation:

Given value of probability $P\left(X\le x\right)=P\left(Z\le z\right)$, $x$-value $\left(x\right)$ and standard deviation $\left(\sigma \right)$, then mean $\left(\mu \right)$ is obtained using:

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

before which the $z$-value is looked up from the standard table.

The $z$-value as from the standard normal table is as:

$z=-0.46$

Here, the mean has to be computed:

$\begin{array}{rll}\mu & =& \left(116.00\right)-\left(-0.46\right)\times \left(0.24\right)\\ & =& 116.00-\left(-0.1104\right)\\ & =& 116.00+0.1104\\ & =& 116.1104\end{array}$

Rounding off to two decimals places gives:

$\mu =116.11$.

Conclusion:

The value of the mean is $116.11$.

To determine

c.

To find:

The standard deviation for given data.

Answer to Problem 15CR

Solution:

The required standard deviation is $0.25$.

Explanation of Solution

Given:

$\mu =0.46$

$x=0.72$

$P\left(X\le x\right)=0.85$

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 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.

Calculation:

Given the value of probability $P\left(X\le x\right)=P\left(Z\le z\right)$, value of random variable $\left(x\right)$ and mean $\left(\mu \right)$, then standard deviation $\left(\sigma \right)$ is obtained using:

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

before which the $z$-value is looked up from the standard table.

The $z$-value as from the standard normal table is:

$z=1.04$

Here, the standard deviation is to be calculated:

$\begin{array}{rll}\sigma & =& \frac{\left(0.72\right)-\left(0.46\right)}{\left(1.04\right)}\\ & =& \frac{0.26}{1.04}\\ & =& 0.25\end{array}$

This value rounded off to the $3^{\text{rd}}$ decimal place gives:

$\sigma =0.25$

Conclusion:

The required standard deviation is $0.25$.

To determine

d.

To find:

The $x$-value for given data.

Answer to Problem 15CR

Solution:

The required $x$-value is $972$.

Explanation of Solution

Given:

$\mu =1052$

$\sigma =66$

$P\left(X\le x\right)=0.1131$

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 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.

Calculation:

Given the value of probability $P\left(X\le x\right)=P\left(Z\le z\right)$, standard deviation $\left(\sigma \right)$ and mean $\left(\mu \right)$, then value of random variable $\left(x\right)$ is obtained using:

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

before which the $z$-value is looked up from the standard table.

The $z$-value from the table is:

$z=-1.21$.

Then, the $x$-value, that is the value of the random variable is to be obtained by:

$\begin{array}{rll}x& =& \left(1052\right)+\left(66\right)\times \left(-1.21\right)\\ & =& 1052+(-79.86)\\ & =& 1052-79.86\\ & =& 972.14\end{array}$

This number rounded off to the nearest whole number gives:

$x=972$.

Conclusion:

The required $x$-value is $972$.