Design and conduct a mini-project involving statistical investigations based on measures of central tendency and measures of dispersion and hence interpret and communicate research findings.

Measures of Dispersion

7.2

Measures of Dispersion

Some formula for a grouped data

\(\text{range}=\text{midpoint of the highest class} - \text{midpoint of the lowest class}\)

\(\text{Interquartile range, IQR}=Q_3-Q_1\)

\(\begin{aligned} \text{Mean, }\bar{x}=\frac{\sum fx}{\sum f} \end{aligned}\)

\(\begin{aligned} \text{Variance, }\sigma^2=\frac{\sum fx^2}{\sum f}-\bar{x}^2 \end{aligned}\)

\(\begin{aligned} \text{Standard deviation, }\sigma=\sqrt{\frac{\sum fx^2}{\sum f}-\bar{x}} \end{aligned}\)

where \(\begin{aligned} &x=\text{midpoint of the class interval}\\ &f=\text{frequency}\\ \end{aligned}\)

Example 5

The frequency table below shows the volumes of water to the nearest litres, used daily by a group of families in a housing area. Calculate the variance and standard deviation of the data.

Solution:

\(\begin{aligned} \text{Mean, }\bar{x}&=\frac{\sum fx}{\sum f} \\ &=\frac{17417.5}{95}\\ &=183.34 \end{aligned}\)

\(\begin{aligned} \text{Variance, }\sigma^2&=\frac{\sum fx^2}{\sum f}-\bar{x}\\ &=\frac{3 215 133.75}{95}-\bigg(\frac{17 417.5}{95}\bigg)^2\\ &=229.1856 \end{aligned}\)

\(\begin{aligned} \text{Standard deviation, }\sigma&=\sqrt{\frac{\sum fx^2}{\sum f}-\bar{x}}\\ &=\sqrt{229.1856} \\ &=15.1389 \end{aligned}\)

What is a boxplot?

A box plot is a method to display a group of numerical data graphically based on the five number summary of data. They are the minimum value, first quartile, median, third quartile and maximum value. Similar to the histrogram and frequency polygon, the shape of a distribution can also be identified through the box plot.

Example 6

The ogive below shows the masses in g, of 90 starfruits.

a) Construct a box plot based on the ogive.

b) Hence, state the distribution shape of the data.

Solution:

a) \(\begin{aligned} &\text{Minimum value} = 80\\\\ &\text{Maximum value} = 150\\\\ &Q_1=116\\\\ &Q_2=123\\\\ &Q_3=128 \end{aligned}\)

b) The distribution of the data is skewed to the left because the left side of the box plot is longer than the right side of the box plot.

Measures of Dispersion

7.2

Measures of Dispersion

Some formula for a grouped data

\(\text{range}=\text{midpoint of the highest class} - \text{midpoint of the lowest class}\)

\(\text{Interquartile range, IQR}=Q_3-Q_1\)

\(\begin{aligned} \text{Mean, }\bar{x}=\frac{\sum fx}{\sum f} \end{aligned}\)

\(\begin{aligned} \text{Variance, }\sigma^2=\frac{\sum fx^2}{\sum f}-\bar{x}^2 \end{aligned}\)

\(\begin{aligned} \text{Standard deviation, }\sigma=\sqrt{\frac{\sum fx^2}{\sum f}-\bar{x}} \end{aligned}\)

where \(\begin{aligned} &x=\text{midpoint of the class interval}\\ &f=\text{frequency}\\ \end{aligned}\)

Example 5

The frequency table below shows the volumes of water to the nearest litres, used daily by a group of families in a housing area. Calculate the variance and standard deviation of the data.

Solution:

\(\begin{aligned} \text{Mean, }\bar{x}&=\frac{\sum fx}{\sum f} \\ &=\frac{17417.5}{95}\\ &=183.34 \end{aligned}\)

\(\begin{aligned} \text{Variance, }\sigma^2&=\frac{\sum fx^2}{\sum f}-\bar{x}\\ &=\frac{3 215 133.75}{95}-\bigg(\frac{17 417.5}{95}\bigg)^2\\ &=229.1856 \end{aligned}\)

\(\begin{aligned} \text{Standard deviation, }\sigma&=\sqrt{\frac{\sum fx^2}{\sum f}-\bar{x}}\\ &=\sqrt{229.1856} \\ &=15.1389 \end{aligned}\)