Chapter 4, Problem 50E

To determine

To Explain: the appropriate graphical displays and summary statistics.

Explanation of Solution

Given:

  

Calculation:

Median: First ascending the data

-1.6, -1, -0.8, -0.3, 0, 1.1, 1.4, 1.9, 2, 2.2, 2.3, 2.5, 3.3, 3.5, 4.4, 5.5, 5.6, 6, 6.5, 7.9, 11.9

These are 21 values so median is 11^th value. So median will be

Median = 2.3%

Mean: Sum of these 21 values is 64.3% so mean will be

  $\text{mean=}\frac{64.3}{21}=3.1\%$

So standard deviation would be

  $\begin{array}{c}S.D=\sqrt{\frac{\left(x-\overline{x}\right)}{n-1}}\\ =\sqrt{\frac{220.98}{20}}\\ =\sqrt{11.04}\\ =3.3\%\end{array}$

So standard deviation is 3.3%

IQR

To find IQR let us first find $Q_1$ and $Q_3$ . Lower quartile is middle value of lower half.

Lower half of data has 11 values (including median). Lower half of information

-1.6, -1, -0.8, -0.3, 0, 1.1, 1.4, 1.9, 2, 2.2, 2.3

The lower quartile is 6^th value. So

  $Q_1=1.1$

Upper half of data has 25 values. upper half of data is

2.3, 2.5, 3.3, 3.5, 4.4, 5.5, 5.6, 6, 6.5, 7.9, 11.9

The upper quartile is 6^th value. So

  $Q_3=5.5$

Therefore IQR is

  $\begin{array}{c}\text{IQR}=Q_3-Q_1\\ =5.5-1.1\\ =4.4\%\end{array}$

Graph:

  

Median and mean of the data is 2.3% and 3.1% respectively. Distribution is skewed to right. So median works as a good measure of centre of distribution in comparison of mean. There is an outlier also means in one state percent change is around 11.9% which is higher than any other state.