Question

The table below shows data from a survey about the amount of time, in hours, high school students spent reading and the amount of time spent watching videos each week (without reading):
Reading Video
3 1
3 2
4 3
5 4
6 6
7 7
10 8
12 8
14 9
30 15

Which response best describes outliers in these data sets?
A, Reading has a suspected outlier in the 30-hour value.
B. Reading has a suspected outlier in the 30-hour value, and video has a possible outlier in the 15-hour value.
C. Neither data set has suspected outliers.
D. The range of data is too small to identify outliers.

Answer 1

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. 

Answer 2

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:

$\dfrac{3+3+4+5+6+7+10+12+14+30}{10}=\dfrac{94}{10}=9.4$

Also the mean after omitting the odd value i.e. 30 we get:

$\dfrac{3+3+4+5+6+7+10+12+14}{9}=\dfrac{64}{9}=7.11$

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:

$\dfrac{1+2+3+4+6+7+8+8+9+15}{10}=\dfrac{63}{10}=6.3$

Also the mean after omitting the odd value i.e. 15 we get:

$\dfrac{1+2+3+4+6+7+8+8+9}{9}=\dfrac{48}{9}=5.33$

Hence, there is a change in its mean hence 15 is an outlier.

Hence option A and B are true.