Chapter 6.2, Problem 15E

a.

To determine

To find: The probability that more than 100,000 jobs are available for RNs.

Answer to Problem 15E

The probability that More than 100,000 jobs are available for RNs is 0.6862.

Explanation of Solution

Given info:

The annual number of jobs for registered nurses follow a normal distribution with mean 103,900 and standard deviation 8,040.

Calculations:

Normal distribution:

A random variable x is said to follow a Normal distribution, if the probability distribution is

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

Where,

$\mu =\text{Mean of normal variable }X$

$\sigma =\text{Standard deviation of normal variable }X$.

$\pi \text{ = 3}\text{.1415}\mathrm{...}$.

$e=\text{ 2}\text{.71828}$

Standard Normal Distribution:

A Random variable z is said to a standard normal variable if z follows normal distribution with mean $\mu =0$ and variance ${\sigma }^2=1$.

Consider a random variable X define as the number of jobs available for registered nurses.

Software procedure:

Step-by-step procedure to obtain the probability using the MINITAB software:

• Choose Graph > Probability Distribution Plot choose View Probability> OK.
• From Distribution, choose ‘Normal’ distribution.
• Enter the Mean as 103,900 and Standard deviation as 8,040.
• Click the Shaded Area tab.
• Choose X value and Right Tail for the region of the curve to shade.
• Enter the X value as 100,000.
• Click OK.

Output using the MINITAB software is given below:

From MINITAB, the probability that More than 100,000 jobs are available for RNs is 0.6862.

b.

To determine

To find: The probability that more than 80,000 but less than 95,000 jobs are available for RNs.

Answer to Problem 15E

The probability that more than 80,000 but less than 95,000 jobs are available for RNs is 0.1327.

Explanation of Solution

Calculations:

Normal distribution:

A random variable x is said to follow a Normal distribution, if the probability distribution is

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

Where,

$\mu =\text{Mean of normal variable }X$

$\sigma =\text{Standard deviation of normal variable }X$.

$\pi \text{ = 3}\text{.1415}\mathrm{...}$.

$e=\text{ 2}\text{.71828}$

Standard Normal Distribution:

A Random variable z is said to a standard normal variable if z follows normal distribution with mean $\mu =0$ and variance ${\sigma }^2=1$.

Consider a random variable X define as the number of jobs available for registered nurses.

Software procedure:

Step-by-step procedure to obtain the probability using the MINITAB software:

• Choose Graph > Probability Distribution Plot choose View Probability> OK.
• From Distribution, choose ‘Normal’ distribution.
• Enter the Mean as 103,900 and Standard deviation as 8,040.
• Click the Shaded Area tab.
• Choose X value and Both Tail for the region of the curve to shade.
• Enter the X value 1 as 80,000.
• Enter the X value 2 as 95,000.
• Click OK.

Output using the MINITAB software is given below:

From MINITAB, the probability that more than 80,000 but less than 95,000 jobs are available for RNs is 0.1327.

c.

To determine

To find: The X­value if the probability is 0.1977 that more than X amount of jobs are available.

Answer to Problem 15E

The required X-value is 110,733.

Explanation of Solution

Calculations:

Normal distribution:

A random variable x is said to follow a Normal distribution, if the probability distribution is

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

Where,

$\mu =\text{Mean of normal variable }X$

$\sigma =\text{Standard deviation of normal variable }X$.

$\pi \text{ = 3}\text{.1415}\mathrm{...}$.

$e=\text{ 2}\text{.71828}$

Standard Normal Distribution:

A Random variable z is said to a standard normal variable if z follows normal distribution with mean $\mu =0$ and variance ${\sigma }^2=1$.

Consider a random variable X define as the number of jobs available for registered nurses.

Software procedure:

Step-by-step procedure to obtain the probability using the MINITAB software:

• Choose Graph > Probability Distribution Plot choose View Probability> OK.
• From Distribution, choose ‘Normal’ distribution.
• Enter the Mean as 103,900 and Standard deviation as 8,040.
• Click the Shaded Area tab.
• Choose X value and Both Tail for the region of the curve to shade.
• Enter the X value 1 as 80,000.
• Enter the X value 2 as 95,000.
• Click OK.

Output using the MINITAB software is given below:

From MINITAB, the required X-value is 110,733.