Respuesta :
Reading: 3, 3, 4, 5, 6, 7, 10, 12, 14, 30
Video: 1, 2, 3, 4, 6, 7, 8, 8, 9, 15
Reading Video
Minimum 3 1
1st quartile 4 3
Median 6.5 6.5
3rd quartile 12 8
Maximum 30 15
IQR 8 5
Outliers 30 no outliers
A, Reading has a suspected outlier in the 30-hour value.
Video: 1, 2, 3, 4, 6, 7, 8, 8, 9, 15
Reading Video
Minimum 3 1
1st quartile 4 3
Median 6.5 6.5
3rd quartile 12 8
Maximum 30 15
IQR 8 5
Outliers 30 no outliers
A, Reading has a suspected outlier in the 30-hour value.
Answer:
The correct option is: A,B
Step-by-step explanation:
" An outlier is a odd value which on removing from the data leads to the less mean as calculated mean on including that value ".
There are total 10 data points.
- The table for reading is given as:
3 3 4 5 6 7 10 12 14 30
The mean of this data is given by:
[tex]\dfrac{3+3+4+5+6+7+10+12+14+30}{10}=\dfrac{94}{10}=9.4[/tex]
Also the mean after omitting the odd value i.e. 30 we get:
[tex]\dfrac{3+3+4+5+6+7+10+12+14}{9}=\dfrac{64}{9}=7.11[/tex]
Hence, there is a change in its mean;
Hence the outlier for reading is 30.
- The table for Video is given by:
1 2 3 4 6 7 8 8 9 15
The possible maximum value here is 15 so we will check whether it is an outlier or not.
The mean of this data is given by:
[tex]\dfrac{1+2+3+4+6+7+8+8+9+15}{10}=\dfrac{63}{10}=6.3[/tex]
Also the mean after omitting the odd value i.e. 15 we get:
[tex]\dfrac{1+2+3+4+6+7+8+8+9}{9}=\dfrac{48}{9}=5.33[/tex]
Hence, there is a change in its mean hence 15 is an outlier.
Hence option A and B are true.