Chapter 13.3, Problem 9PPS

(a)

To determine

To Find:City has greater range of temperature

Answer to Problem 9PPS

Antelope has higher range than Augusta

Explanation of Solution

Given:The table shows the average monthly temperatures.

  

Which city has greater range of temperature?

Finding range

For Antelope MT

Highest vale = $84$

Lowest value = $21$

Range = Highest value − lowest value

  $\begin{array}{l}=84-21\\ =53\end{array}$

For Augusta

Highest vale = $80$

Lowest value = $28$

Range = Highest value − lowest value

  $\begin{array}{l}=80-28\\ =52\end{array}$

Hence, Antelope has higher range than Augusta

(b)

To determine

To Find:Measure of variation for each city

Answer to Problem 9PPS

For Antelope

Median = $58$

Lower quartile $=33.5$

Upper quartile $=74.5$

Interquartile range $=41$

For Augusta

Median = $55.5$

Lower quartile $=37.5$

Upper quartile $=72.5$

Interquartile range $=35$

Explanation of Solution

Given:The table shows the average monthly temperatures.

  

Find the measure of variation for each city.

For Antelope After arranging the data in ascending order

Median = $58$

Lower quartile $=\frac{30+37}{2}=33.5$

Upper quartile $=\frac{70+79}{2}=74.5$

Interquartile range = Upper quartile − lower quartile

  $\begin{array}{l}74.5-33.5\\ =41\end{array}$

For Augusta After arranging the data in ascending order

Median = $55.5$

Lower quartile $=\frac{34+31}{2}=37.5$

Upper quartile $=\frac{70+75}{2}=72.5$

Interquartile range = Upper quartile − lower quartile

  $\begin{array}{l}72.5-37.5\\ =35\end{array}$

(c)

To determine

To Compare:Medians and the interquartile ranges of average temperatures

Answer to Problem 9PPS

Median and interquartile range both of Antelope is greater than Augusta

Explanation of Solution

Given:The table shows the average monthly temperatures.

  

Compare the medians and the interquartile ranges of average temperatures.

For Antelope

Median = $58$

Interquartile range $=41$

For Augusta

Median = $55.5$

Interquartile range $=35$

It is concluded that Median and interquartile range both of Antelope is greater than Augusta

(d)

To determine

To Find:The appropriate measure of central tendency to describe the average high temperature for Augusta.

Answer to Problem 9PPS

Mean

Explanation of Solution

Given: The table shows the average monthly temperatures.

  

Select the appropriate measure of central tendency to describe the average high temperature for Augusta. Justify your response.

Mean is theappropriate measure of central tendency to describe the average high temperature for Augusta.

Because the range of values are large.

(e)

To determine

To Describe:Average temperatures of Antelope and Augusta.

Answer to Problem 9PPS

The average temperature of Antelope MT is higher than Augusta ME

Explanation of Solution

Given:The table shows the average monthly temperatures.

  

Describe the average temperatures of Antelope and Augusta using both the measures of central tendency and variation.

Measures of Central tendency

For Antelope

Mean

  $\frac{21+24+30+\mathrm{...}+84}{12}=54.92$

Median= $58$

Mode = $58\text{ and 84}$

For Augusta

Mean

  $\frac{28+32+34+\mathrm{...}+80}{12}=55.17$

Median= $55.5$

No Mode

It is concluded that the average temperature of Antelope MT is higher than Augusta ME.

Measures of variation

For Antelope

Median = $58$

Range= $42$

Interquartile range $=41$

For Augusta

Median = $55.5$

Range $=35$

Interquartile range $=35$

In the measures of variation the average temperature in Antelope is higher than Augusta.