Chapter 14, Problem 2ICA

(a)

To determine

Calculate the mean of the data provided.

Answer to Problem 2ICA

The mean of the data provided is 7.78 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.

$\begin{array}{c}\text{Sum of all the data provided}=\left(\begin{array}{l}1\text{psi}+14\text{psi}+2\text{psi}+15\text{psi}+\\ 10\text{psi}+6\text{psi}+3\text{psi}+1\text{psi}+18\text{psi}\end{array}\right)\\ =70\text{psi}\end{array}$

Substitute 70 psi for Sum of all the data provided and 9 for Total number of data in equation (1) to calculate the mean.

$\begin{array}{c}\text{Mean}=\frac{\text{70}\text{psi}}{\text{9}}\\ =7.777\text{ psi}\\ \cong 7.78\text{ psi}\end{array}$

Conclusion:

Thus, the mean of the data provided is 7.78 psi.

(b)

To determine

Calculate the median of the data provided.

Answer to Problem 2ICA

The median of the data provided is 6 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.

Arrange the given data in ascending order as follows.

$1\text{psi},1\text{psi},2\text{psi},3\text{psi},6\text{psi},10\text{psi},14\text{psi},15\text{psi},18\text{psi}$

The center value in the series is 6 psi, so the median of given data series is 6 psi.

Conclusion:

Thus, the median of the data provided is 6 psi.

(c)

To determine

Calculate the variance of the data provided.

Answer to Problem 2ICA

The variance of the data provided is 43.94 psi².

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 7.78 psi for Mean, 1 for ${\text{Data}}_1$, 14 for ${\text{Data}}_2$, 2 for ${\text{Data}}_3$, 15 for ${\text{Data}}_4$, 10 for ${\text{Data}}_5$, 6 for ${\text{Data}}_6$, 3 for ${\text{Data}}_7$, 1 for ${\text{Data}}_8$ and 18 for ${\text{Data}}_9$ to calculate the variance.

$\begin{array}{c}\text{Variance}=\frac{1}{\text{8}}\times \left(\begin{array}{l}{\left(\text{7}\text{.78}-1\right)}^2+{\left(\text{7}\text{.78}-14\right)}^2+{\left(\text{7}\text{.78}-2\right)}^2+\\ {\left(\text{7}\text{.78}-15\right)}^2+{\left(\text{7}\text{.78}-10\right)}^2+{\left(\text{7}\text{.78}-6\right)}^2\\ +{\left(\text{7}\text{.78}-3\right)}^2+{\left(\text{7}\text{.78}-1\right)}^2+{\left(\text{7}\text{.78}-18\right)}^2\end{array}\right)\\ =\frac{1}{8}\times \left(45.96+38.68+33.40+52.12+4.92+3.16+22.84+45.98+104.44\right)\\ =43.937{\text{psi}}^2\\ \cong 43.94{\text{psi}}^2\end{array}$

Conclusion:

Thus, the variance of the data provided is 43.94 psi².

(d)

To determine

Calculate the standard deviation of the data provided.

Explanation of Solution

Formula used:

Write the expression for standard deviation.

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

Calculation:

Substitute 43.94 psi² for Variance in equation (3) to obtain the value of standard deviation.

$\begin{array}{c}\text{Standard deviation}=\sqrt{\text{43}\text{.94}{\text{psi}}^2}\\ =6.6287\text{psi}\\ \cong 6.629\text{psi}\end{array}$

Conclusion:

Thus, the standard deviation of the data provided is 6.629 psi.