Chapter 12.3, Problem 23PPE

a.

To determine

To find: The mean, median, mode, and range of each data set.

Answer to Problem 23PPE

  $\begin{array}{l}\text{Manufacturing plant A: Mean}=5.79,\text{Median}=6.5,\text{Mode}=5.4,\text{Range}=1.2\\ \text{Manufacturing plant B: Mean}=5.55,\text{Median}=5.45,\text{Mode}=\text{None},\text{Range}=2.9\end{array}$

Explanation of Solution

Given information:

Two data set is given in the below back to back stem and leaf plot.

  

Calculation:

Given data set is −

  

  $\begin{array}{l}\text{For manufacturing plant A:}\\ \text{Arranging in ascending order: 5}\text{.2 5}\text{.4 5}\text{.4 5}\text{.7 5}\text{.8 6}\text{.1 6}\text{.3 6}\text{.4}\\ \text{Here, mean}=\frac{5.2+5.4+5.4+5.7+5.8+6.1+6.3+6.4}{8}=5.7875\\ \text{Median}=\frac{5.7+5.8}{2}=5.75\\ \text{Mode}=5.4\\ \text{Range}=6.4-5.2=1.2\end{array}$

  $\begin{array}{l}\text{For manufacturing plant B:}\\ \text{Arranging in ascending order: 4}\text{.3 4}\text{.5 4}\text{.9 5}\text{.2 5}\text{.7 6}\text{.3 6}\text{.4 7}\text{.2}\\ \text{Here, mean}=\frac{4.3+4.5+4.9+5.2+5.7+6.3+6.4+7.2}{8}=5.55\\ \text{Median}=\frac{5.2+5.7}{2}=5.45\\ \text{Mode}=\text{None}\\ \text{Range}=7.2-4.3=2.9\end{array}$

Hence, for the given data set −

  $\begin{array}{l}\text{Manufacturing plant A: Mean}=5.79,\text{Median}=6.5,\text{Mode}=5.4,\text{Range}=1.2\\ \text{Manufacturing plant B: Mean}=5.55,\text{Median}=5.45,\text{Mode}=\text{None},\text{Range}=2.9\end{array}$

b.

To determine

To find: Which measure of central tendency best describes the given data set.

Answer to Problem 23PPE

Mean is the best measure to describe the central tendency.

Explanation of Solution

Given information:

Same as part $a$ of this exercise.

Calculation:

From part $a$ of this exercise,

  $\begin{array}{l}\text{Manufacturing plant A: Mean}=5.79,\text{Median}=6.5,\text{Mode}=5.4,\text{Range}=1.2\\ \text{Manufacturing plant B: Mean}=5.55,\text{Median}=5.45,\text{Mode}=\text{None},\text{Range}=2.9\end{array}$

Since both data sets do not have any outliers. Hence, mean is the best measure to describe the central tendency.

c.

To determine

To find: How to determine which data set has greater mean value using shape of back to back stem and leaf plot.

Explanation of Solution

Given information:

Same as part $a$ of this exercise.

Calculation:

From part $a$ of this exercise,

  $\begin{array}{l}\text{Manufacturing plant A: Mean}=5.79,\text{Median}=6.5,\text{Mode}=5.4,\text{Range}=1.2\\ \text{Manufacturing plant B: Mean}=5.55,\text{Median}=5.45,\text{Mode}=\text{None},\text{Range}=2.9\end{array}$

  $\begin{array}{l}\text{The mean value of - }\\ \text{Right shifted graph }>\text{symmetrical graph }>\text{Left shifted graph}\text{.}\end{array}$

Therefore, data set for manufacturing plant A has greater mean value because plot for manufacturing plant B is left skewed.