Given:
The data set is
10, 10, 11, 12, 13, 13, 13, 14, 14, 15, 15, 15, 16, 17, 17, 17, 35
To find:
Effect on mean, median, range after removing the outlier, 35.
Solution:
We have, the data set
10, 10, 11, 12, 13, 13, 13, 14, 14, 15, 15, 15, 16, 17, 17, 17, 35
[tex]\text{Mean}=\dfrac{\text{Sum of observations}}{\text{Number of observations}}[/tex]
[tex]\text{Mean}=\dfrac{10+10+11+12+13+13+13+14+14+15+15+15+16+17+17+17+35}{17}[/tex]
[tex]\text{Mean}=\dfrac{257}{17}[/tex]
[tex]\text{Mean}=15.117647[/tex]
[tex]\text{Mean}\approx 15.12[/tex]
Mean of the data set is 15.12.
[tex]Median=\dfrac{n+1}{2}\text{th term}[/tex] because n=17, which is odd.
[tex]Median=\dfrac{17+1}{2}\text{th term}[/tex]
[tex]Median=\dfrac{18}{2}\text{th term}[/tex]
[tex]Median=9\text{th term}[/tex]
[tex]Median=14[/tex]
Median is 14.
[tex]Range=Maximum-Minimum[/tex]
[tex]Range=35-10[/tex]
[tex]Range=25[/tex]
Range is 25.
After removing the outlier, 35, the data set is
10, 10, 11, 12, 13, 13, 13, 14, 14, 15, 15, 15, 16, 17, 17, 17
[tex]\text{Mean}=\dfrac{10+10+11+12+13+13+13+14+14+15+15+15+16+17+17+17}{16}[/tex]
[tex]\text{Mean}=\dfrac{222}{16}[/tex]
[tex]\text{Mean}=13.875[/tex]
Mean of the new data set is 13.875, which is less than 15.12.
[tex]Median=\dfrac{\dfrac{n}{2}\text{th term}+(\dfrac{n}{2}+1)\text{th term}}{2}[/tex] because n=16, which is even.
[tex]Median=\dfrac{\dfrac{16}{2}\text{th term}+(\dfrac{16}{2}+1)\text{th term}}{2}[/tex]
[tex]Median=\dfrac{8\text{th term}+9\text{th term}}{2}[/tex]
[tex]Median=\dfrac{14+14}{2}[/tex]
[tex]Median=\dfrac{28}{2}[/tex]
[tex]Median=14[/tex]
New median is 14, which is same as original median.
[tex]Range=17-10[/tex]
[tex]Range=7[/tex]
So, the new range is 7 which is less than 25.
Therefore, the mean of the data set will decrease, the median of the data set will not change, the Range of the data set will decrease.
