Monthly starting salaries for a sample of 12 business school graduates are given below.
Calculate the mean, standard deviation, variance and coefficient of variation.
Then calculate the three quartiles and identify which quartile represents the median.
Detailed Step-by-Step Solution
Given salary data:
\(3450,\;3550,\;3650,\;3480,\;3355,\;3310,\;3490,\;3730,\;3540,\;3925,\;3520,\;3480\)
\( n=12 \)
\( \sum x=42480 \)
\( \sum x^2=150681050 \)
Sample mean:
\[
\bar{x}=\frac{\sum x}{n}
\]
\[
\bar{x}=\frac{42480}{12}
\]
\[
\bar{x}=3540.00
\]
Sample variance:
\[
s^2=\frac{\sum x^2-\frac{(\sum x)^2}{n}}{n-1}
\]
\[
s^2=\frac{150681050-\frac{42480^2}{12}}{11}
\]
\[
s^2=27440.91
\]
Sample standard deviation:
\[
s=\sqrt{s^2}
\]
\[
s=\sqrt{27440.91}
\]
\[
s=165.65
\]
Coefficient of variation:
\[
CV=\frac{s}{\bar{x}}\times 100
\]
\[
CV=\frac{165.65}{3540.00}\times 100
\]
\[
CV=4.68\%
\]
Arrange the data in ascending order:
\(3310,\;3355,\;3450,\;3480,\;3480,\;3490,\;3520,\;3540,\;3550,\;3650,\;3730,\;3925\)
First quartile \(Q_1=P_{25}\):
\[
i=\frac{25}{100}\times 12
\]
\[
i=3
\]
Since \(i\) is an integer, average the 3rd and 4th values.
\[
Q_1=\frac{3450+3480}{2}
\]
\[
Q_1=3465
\]
Second quartile \(Q_2=P_{50}\):
\[
i=\frac{50}{100}\times 12
\]
\[
i=6
\]
Since \(i\) is an integer, average the 6th and 7th values.
\[
Q_2=\frac{3490+3520}{2}
\]
\[
Q_2=3505
\]
\(Q_2\) represents the median.
Third quartile \(Q_3=P_{75}\):
\[
i=\frac{75}{100}\times 12
\]
\[
i=9
\]
Since \(i\) is an integer, average the 9th and 10th values.
\[
Q_3=\frac{3550+3650}{2}
\]
\[
Q_3=3600
\]
Five number summary:
\[
\text{Minimum}=3310
\]
\[
Q_1=3465
\]
\[
\text{Median}=3505
\]
\[
Q_3=3600
\]
\[
\text{Maximum}=3925
\]
Box plot idea:
Min: 3310
Q1: 3465
Median: 3505
Q3: 3600
Max: 3925
Final Answer
Mean \(=3540.00\)
Variance \(=27440.91\)
SD \(=165.65\)
CV \(=4.68\%\)
\(Q_1=3465,\; Q_2=3505,\; Q_3=3600\)
Median is represented by \(Q_2\).
Five number summary: \(3310,\;3465,\;3505,\;3600,\;3925\)