Chapter 14, Problem 2ICA

For the following pressure data, recorded in units of pound-force per square inch, answer the following questions

a. What is the mean of the data?

b. What is the median of the data?

c. What is the variance of the data?

d. What is the standard deviation of the data?

To determine

Calculate the mean of the data provided.

Answer to Problem 2ICA

The mean of the data provided is 23.3 psi.

Explanation of Solution

Formula used:

Write the expression for mean.

$\text{Mean}=\frac{\text{ Sum of all the data provided}}{\text{Total number of data}}$ (1)

Calculation:

Find the sum of all the data provided.

$\begin{array}{c}\text{Sum of all the data provided}=\left(\begin{array}{l}3\text{psi}+42\text{psi}+6\text{psi}+45\text{psi}+30\text{psi}\\ +18\text{psi}+9\text{psi}+3\text{psi}+54\text{psi}\end{array}\right)\\ =210\text{psi}\end{array}$

Substitute $210\text{psi}$ for Sum of all the data provided and 9 for Total number of data in equation (1) to calculate the mean.

$\begin{array}{c}\text{Mean}=\frac{\text{210}\text{psi}}{\text{9}}\\ =23.3\text{ psi}\end{array}$

Conclusion:

Thus, the mean of the data provided is 23.3 psi.

To determine

Calculate the median of the data provided.

Answer to Problem 2ICA

The median of the data provided is $\underset{\_}{18\text{psi}}$.

Explanation of Solution

Calculation:

Median of the data set can be found out by arranging the data in ascending order and choosing the center number.

Arranging in ascending order yields the series.

$3\text{psi},3\text{psi},6\text{psi},9\text{psi},18\text{psi},30\text{psi},42\text{psi},45\text{psi},54\text{psi}$

The center value in an ascending order series is $18\text{psi}$ so the median is $18\text{psi}$.

Conclusion:

Thus, the median of the data provided is $\underset{\_}{18\text{psi}}$.

To determine

Calculate the variance of the data provided.

Answer to Problem 2ICA

The variance of the data provided is $\underset{\_}{395.5{\text{psi}}^2}$.

Explanation of Solution

Formula used:

Write the expression for variance.

$\text{Variance}=\frac{1}{N-1}\times {\displaystyle \sum _{i=1}^N{\left(\text{Mean}-{\text{Data}}_i\right)}^2}$ (2)

Calculation:

Substitute $9$ for $N$ in equation (2) to expand the expression.

$\begin{array}{c}\text{Variance}=\frac{1}{9-1}\times {\displaystyle \sum _{i=1}^9{\left(\text{Mean}-{\text{Data}}_i\right)}^2}\\ =\frac{1}{\text{8}}\left(\begin{array}{l}{\left(\text{Mean}-{\text{Data}}_1\right)}^2+{\left(\text{Mean}-{\text{Data}}_2\right)}^2\\ +{\left(\text{Mean}-{\text{Data}}_3\right)}^2+{\left(\text{Mean}-{\text{Data}}_4\right)}^2\\ +{\left(\text{Mean}-{\text{Data}}_5\right)}^2+{\left(\text{Mean}-{\text{Data}}_6\right)}^2\\ +{\left(\text{Mean}-{\text{Data}}_7\right)}^2+{\left(\text{Mean}-{\text{Data}}_8\right)}^2\\ +{\left(\text{Mean}-{\text{Data}}_9\right)}^2\end{array}\right)\end{array}$

Substitute 23.3 psi for Mean, 3 psi for ${\text{Data}}_1$, 42 psi for ${\text{Data}}_2$, 6 psi for ${\text{Data}}_3$, 45 psi for ${\text{Data}}_4$, 30 psi for ${\text{Data}}_5$, 18 psi for ${\text{Data}}_6$, 9 psi for ${\text{Data}}_7$, 3 psi for ${\text{Data}}_8$ and 54 psi for ${\text{Data}}_9$ to calculate the variance.

$\begin{array}{c}\text{Variance}=\frac{1}{8}\times \left(\begin{array}{l}{\left(\text{23}\text{.3}\text{psi}-3\text{psi}\right)}^2+{\left(\text{23}\text{.3}\text{psi}-42\text{psi}\right)}^2+{\left(\text{23}\text{.3}\text{psi}-6\text{psi}\right)}^2\\ +{\left(\text{23}\text{.3}\text{psi}-45\text{psi}\right)}^2+{\left(\text{23}\text{.3}\text{psi}-30\text{psi}\right)}^2+{\left(\text{23}\text{.3}\text{psi}-18\text{psi}\right)}^2\\ +{\left(\text{23}\text{.3}\text{psi}-9\text{psi}\right)}^2+{\left(\text{23}\text{.3}\text{psi}-3\text{psi}\right)}^2+{\left(\text{23}\text{.3}\text{psi}-54\text{psi}\right)}^2\end{array}\right)\\ =\frac{1}{8}\times \left(\begin{array}{l}412.09{\text{psi}}^2+349.69{\text{psi}}^2+299.29{\text{psi}}^2+470.89{\text{psi}}^2+44.89{\text{psi}}^2\\ +28.09{\text{psi}}^2+204.49{\text{psi}}^2+412.09{\text{psi}}^2+942.49{\text{psi}}^2\end{array}\right)\\ =395.5{\text{psi}}^2\end{array}$

Conclusion:

Thus, the variance of the data provided is $\underset{\_}{395.5{\text{psi}}^2}$.

To determine

Calculate the standard deviation of the data provided.

Answer to Problem 2ICA

The standard deviation of the data provided is $\underset{\_}{19.88\text{psi}}$.

Explanation of Solution

Formula used:

Write the expression for standard deviation.

$\text{Standard deviation}=\sqrt{\text{Variance}}$ (3)

Calculation:

Substitute $395.5{\text{psi}}^2$ for Variance in equation (3) to obtain the value of standard deviation.

$\begin{array}{c}\text{Standard deviation}=\sqrt{395.5{\text{psi}}^2}\\ =19.88\text{psi}\end{array}$

Conclusion:

Thus, the standard deviation of the data provided is $\underset{\_}{19.88\text{psi}}$.