Answer to Question #90381 in Statistics and Probability for Sharon Tan

Question

Monthly Expenses (S$) Frequency
0 and less than 150 5
150 and less than 300 8
300 and less than 450 15
450 and less than 600 9
600 and less than 750 17
750 and less than 900 6

Total 60

(a) Explain with the aid of a diagram, the skewness of the monthly expenses distribution.
(b) Which measure of central tendency (mean, median or mode) would you choose for the
above grouped data? Explain.

Answer 1

$$\def\arraystretch{0.5} \begin{array}{c:c:c:c:c} \begin{matrix} Monthly \\ Expenses (S \ \$) \end{matrix} & \begin{matrix} Frequency \\ (f) \end{matrix} &\begin{matrix} Midpoint \\ (x) \end{matrix} & \begin{matrix} \ \ \ \ f\cdot x \ \ \ \ \\ \end{matrix} \\ \hline \\ \end{array}$$ $$\ \ \ \ \ \ \ \ \ 0\ - 150 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 5 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 75 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 375$$ $$\ \ \ \ \ 15 0\ - 300 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 8 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 225 \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1800$$ $$\ \ \ \ \ 300\ - 450 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 15 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 375 \ \ \ \ \ \ \ \ \ \ \ \ \ \ 5625$$ $$\ \ \ \ \ 450\ - 600 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 9 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 525 \ \ \ \ \ \ \ \ \ \ \ \ \ \ 4725$$ $$\ \ \ \ \ 600\ - 750 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 17 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 675 \ \ \ \ \ \ \ \ \ \ \ \ 11475$$ $$\ \ \ \ \ 750\ - 900 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 6 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 825 \ \ \ \ \ \ \ \ \ \ \ \ \ \ 4950$$

$\sum f\cdot x=$

$=375+1800+5625+4725+11475+4950=28950$

$\overline{x}=\dfrac{\sum f\cdot x}{n}=\dfrac{28950}{60}=482.5$

To find Median Class 

$$value\ of\ ({n \over 2})^{th}\ observation=$$ $$=value\ of\ ({60 \over 2})^{th}\ observation=$$ $$=value\ of\ (30)^{th}\ observation$$

From the column of frequency $f$, we find that the $30^{th}$ observation lies in the class 450 −600.

The median class is 450 −600.

Cumulative frequency of the class preceding the median class is 5+8+15=28.

Frequency of the median class is 9.

Class length of median class is 150.

$$median=450+{30-28 \over 9}\cdot 150\approx 483.3333$$

To find Mode Class 

Here, maximum frequency is 17.

The mode class is 600 −750.

Lower boundary point of mode class =600.

Frequency of the mode class =17.

Frequency of the preceding class =9.

Frequency of the succedding class =6.

Class length of mode class =150.

$$mode=600+{17 - 9 \over 2\cdot 17-9-6}\cdot 150=663.1579$$

$$mean<median<mode$$

The monthly expenses distribution is A left-skewed distribution. But the mean and median are close together. They are approximately equal. 

The mode is greater than mean and then median.

The form of diagram does not help me well.

I may suppose that the data are approximately symmetric.

The mean has one main disadvantage: it is particularly susceptible to the influence of outliers. These are values that are unusual compared to the rest of the data set by being especially small or large in numerical value. 

The median is the middle score for a set of data that has been arranged in order of magnitude. The median is less affected by outliers and skewed data.

Normally, the mode is used for categorical data where we wish to know which is the most common category

I think, I would choose the median for the above grouped data.