Respuesta :
Given: Sets of data below
[tex]\begin{gathered} data1:4,15,6,12,68,12 \\ data2:0,54,62,64,55,5,54,62 \end{gathered}[/tex]To Determine: The mean, median and mode
Solution
Data1; Calculate the mean as below
[tex]\begin{gathered} Mean=\frac{\Sigma x}{n} \\ Mean=\frac{4+15+6+12+68+12}{6} \\ Mean=\frac{117}{6}=19.5 \end{gathered}[/tex]Calculate the median
[tex]\begin{gathered} Re-arrange\text{ the data} \\ 4,6,12,12,15,68 \end{gathered}[/tex]The middle numbers are
[tex]12,12[/tex]The average of the two middle numbers is
[tex]\begin{gathered} Median=\frac{12+12}{2} \\ Median=\frac{24}{2} \\ Median=12 \end{gathered}[/tex]The mode is the number that occur the most
From the data 1 given, all the number appeared once except for 12 that appeared twice. Hence, the mode is 12
Data 2:
The mean is as calculated below
[tex]Mean=\frac{0+54+62+64+55+5+54+62}{8}[/tex][tex]\begin{gathered} Mean=\frac{356}{8} \\ Mean=44.5 \end{gathered}[/tex]Calculate the median as shown below
[tex]\begin{gathered} Re-arrange \\ 0,5,54,54,55,62,62,64 \end{gathered}[/tex]The middle numbers are
[tex]54,55[/tex]The median is the average of the two middle numbers
[tex]Median=\frac{54+55}{2}=\frac{109}{2}=54.5[/tex]Calculate the mode
It can be observed that 54 and 62 both appeared twice while other numbers appeared once. The mode is 54 and 62
Hence, the data is bi-modal, that it has two modes, 54 and 62
Let us calculate without outliers
Data 1
The outlier in data 1 is 68
[tex]\begin{gathered} Mean=\frac{4+6+12+12+15}{5} \\ Mean=\frac{49}{5} \\ Mean=9.8 \end{gathered}[/tex][tex]\begin{gathered} Re-arrange \\ 4,6,12,12,15 \\ Median=12 \\ Mode=12 \end{gathered}[/tex]Data 2
The outliers are 0 and 5
[tex]\begin{gathered} Mean=\frac{54+54+55+62+62+64}{6} \\ Mean=\frac{351}{6} \\ Mean=58.5 \end{gathered}[/tex][tex]\begin{gathered} Re-arrange \\ 54,54,55,62,62,64 \\ Median=\frac{55+62}{2}=58.5 \\ Mode=54,and,62 \end{gathered}[/tex]The outliers had major effect on the mean compare to the median and the mode
