Chapter 11.6, Problem 11PPSE

Solution

Toidentify:How to compare the temperature if table shows the average monthly temperature of two cities.

	J	F	M	A	M	J	J	A	S	O	N	D
Jacksonville, Florida	52.4	55.2	61.1	67	73.4	79.1	81.6	81.2	78.1	69.8	61.9	55.1
Austin, Texas	48.8	52.8	61.5	69.9	75.6	81.3	84.5	84.8	80.2	71.1	60.9	51.6

The range and the inter quartile range shows the temperatures varying less at Jacksonville comparing that the Austin temperature.

Given information:

The given table is:

	J	F	M	A	M	J	J	A	S	O	N	D
Jacksonville, Florida	52.4	55.2	61.1	67	73.4	79.1	81.6	81.2	78.1	69.8	61.9	55.1
Austin, Texas	48.8	52.8	61.5	69.9	75.6	81.3	84.5	84.8	80.2	71.1	60.9	51.6

Explanation:

Consider the given table.

For Jacksonville, Florida

Mean:

  $\begin{array}{c}mean=\frac{52.4+55.2+61.1+67+73.4+79.1+81.6+81.2+78.1+69.8+61.9+55.1}{12}\\ \approx 67.99\end{array}$

Median:

The given data has 12 observations which is even so the median will be the average of the middle two numbers if data arranged in ascending order.

52.4, 55.1, 55.2, 61.1, 61.9, 67.0, 69.8, 73.4, 78.1, 79.1, 81.2, 81.6

  $\begin{array}{c}median=\frac{67+69.8}{2}\\ =68.4\end{array}$

Mode is the highest frequency value of the given data set. Since no value is repeated so mode is None.

From the data set the minimum and maximum values are 52.4 and 81.6 respectively.

  $\begin{array}{c}Range=\left(\max -\min \right)\\ =81.6-52.4\\ =29.2\end{array}$

There is total 12 observations so lower part contains first 6 observation and upper part contains last 6 observations.

Median of lower part

  $\begin{array}{c}{Q}_1=\frac{55.2+61.1}{2}\\ =58.15\end{array}$

Median of upper part

  $\begin{array}{c}{Q}_3=\frac{78.1+79.1}{2}\\ =78.6\end{array}$

Inter quartile range:

  $\begin{array}{c}{Q}_3-Q_1=78.6-58.15\\ =20.45\end{array}$

For Austin, Texas

Mean:

  $\begin{array}{c}mean=\frac{48.8+52.8+61.5+69.9+75.6+81.3+84.5+84.8+80.2+71.1+60.9+51.6}{12}\\ \approx 68.53\end{array}$

Median:

The given data has 12 observations which is even so the median will be the average of the middle two numbers if data arranged in ascending order.

48.8, 51.6, 52.8, 60.9, 61.5, 69.9, 71.1, 75.6, 80.2, 81.3, 84.5, 84.8

  $\begin{array}{c}median=\frac{69.9+71.1}{2}\\ =70.5\end{array}$

Mode is the highest frequency value of the given data set. Since no value is repeated so mode is None.

From the data set the minimum and maximum values are 48.8 and 84.8 respectively.

  $\begin{array}{c}Range=\left(\max -\min \right)\\ =84.8-48.8\\ =36\end{array}$

There is total 12 observations so lower part contains first 6 observation and upper part contains last 6 observations.

Median of lower part

  $\begin{array}{c}{Q}_1=\frac{52.8+60.9}{2}\\ =56.85\end{array}$

Median of upper part

  $\begin{array}{c}{Q}_3=\frac{80.2+81.3}{2}\\ =80.75\end{array}$

Inter quartile range:

  $\begin{array}{c}{Q}_3-Q_1=80.75-56.85\\ =23.9\end{array}$

The range and the inter quartile range shows the temperatures varying less at Jacksonville comparing that the Austin temperature.