Answer to Question #90332 in Statistics and Probability for Sharon Tan

Question

XYZ, a retail store, operates in a highly competitive market with many competitors. The manager has been tasked to do some statistical analysis on the monthly expenses over the last five years. Table 1.1 shows the data that was collected.

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) For the above grouped data, calculate:
(i) Mean
(ii) Median
(iii) Mode
(iv) Sample variance
(v) Coefficient of Variation

Answer 1

$\def\arraystretch{0.5} \begin{array}{c:c:c:c:c} \begin{matrix} Monthly \\ Expenses (S \ \$) \end{matrix} & \begin{matrix} Frequency \\ (f) \end{matrix} &\begin{matrix} Midpoint \\ (x) \end{matrix} & \begin{matrix} \ \ \ \ f\cdot x \ \ \ \ \\ \end{matrix} \\ \hline \\ \end{array}$

$\ \ \ \ \ \ \ \ \ 0\ - 150 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 5 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 75 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 375$

$\ \ \ \ \ 15 0\ - 300 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 8 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 225 \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1800$

$\ \ \ \ \ 300\ - 450 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 15 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 375 \ \ \ \ \ \ \ \ \ \ \ \ \ \ 5625$

$\ \ \ \ \ 450\ - 600 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 9 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 525 \ \ \ \ \ \ \ \ \ \ \ \ \ \ 4725$

$\ \ \ \ \ 600\ - 750 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 17 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 675 \ \ \ \ \ \ \ \ \ \ \ \ 11475$

$\ \ \ \ \ 750\ - 900 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 6 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 825 \ \ \ \ \ \ \ \ \ \ \ \ \ \ 4950$

$\sum f\cdot x=$

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

(i)

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

(ii)

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 $150.$

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

(iii)

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 $=150.$

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

(iv)

Sample Variance 

$$s^2={\sum f_i x_i^2-\Big \lbrack {\dfrac{(\sum f_i\cdot x_i)^2}{n}}\Big \rbrack\over \ n -1}$$

$$\sum f_i x_i^2=5(75)^2+8(225)^2+15(375)^2+9(525)^2+$$ $$+17(675)^2+6(825)^2=16852500$$

$$s^2={16852500-\Big \lbrack {\dfrac{(28950)^2}{60}}\Big \rbrack\over \ 60 -1}\approx 48883.4746$$

(v)

Coefficient of Variation (Sample) 

$${s \over \overline{x}}\cdot 100\%\approx{\sqrt {48883.4746} \over \ 482.5}\cdot 100\%\approx45.823 \%$$