Question

Which data set contains an outlier?

{30, 31, 31, 32, 34, 37}

{11, 12, 12, 12, 13, 13, 14, 14}

{6.1, 6.8, 7.3, 8, 9.1, 9.5}

{14, 15, 16, 20, 22, 25, 51}

Answer 1

Answer:

The last one {14,15,16,20,22,25, and 51}

Step-by-step explanation:

This set shows an outlier because 51 is a huge jump from 25 which causes it to be the outlier.

Answer 2

The data set {14,15,16,20,22,25,51} contains an outlier. So, option D is correct.

What is an outlier:

• An outlier is an extreme value in a data set that is either much larger or much smaller than all the other values.
• A data point is an outlier if it is over 1.5 times the IQR below the first quartile $(Q_1)$ or 1.5 times the IQR above the third quartile $(Q_3)$.

How to calculate an outlier:

• To calculate an outlier for a given data, arrange the data in the ascending order
• Separate the data into two sections or groups
• Find the medians for each group. The median for the first group is said to be the first quartile $(Q_1)$ and the median for the second group is said to be the third quartile $(Q_3)$
• Find the difference between them by $(Q_3)$ minus $(Q_1)$ which is called the interquartile range (IQR).

• Then, if the data points are over 1.5 times the IQR below the first quartile $(Q_1)$ or 1.5 times the IQR above the third quartile $(Q_3)$ is said to have an outlier.

Calculating the outlier for the given data sets:

Option A:

{30, 31, 31, 32, 34, 37}

There is an even number of data points.

Diving into 2 groups- {30,31,31} and {32,34,37}

So, $Q_1$=31 and $Q_3$=34

Then $Q_3$ - $Q_1$=34-31=3

Then 1.5×3=4.5

$Q_1$-4.5=31-4.5=26.5

$Q_3$+4.5=34+4.5=38.5

So, for this data set, there are no outliers either below or above.

Option B:

{11,12,12,12,13,13,14,14}

There is an even number of data points.

Diving into 2 groups- {11,12,12,12} and {13,13,14,14}

So, $Q_1$=$\frac{12+12}{2}=12$ and $Q_3$=$\frac{13+14}{2}=13.5$

Then $Q_3$ - $Q_1$=13.5-12=1.5

Then 1.5×1.5=2.25

$Q_1$-2.25=12-2.25=9.75

$Q_3$+2.25=13.5+2.25=15.75

So, for this data set, there are no outliers either below or above.

Option C:

{6.1,6.8,7.3,8,9.1,9.5}

There is an even number of data points.

Diving into 2 groups- {6.1,6.8,7.3} and {8,9.1,9.5}

So, $Q_1$=6.8 and $Q_3$=9.1

Then $Q_3$ - $Q_1$=9.1-6.8=2.3

Then 1.5×2.3=3.45

$Q_1$-3.45=6.8-3.45=3.35

$Q_3$+3.45=9.1+3.45=12.55

So, for this data set, there are no outliers either below or above.

Option D:

{14,15,16,20,22,25,51}

There is an odd number of data points.

Diving into 2 groups- {14,15,16,20} and {20,22,25,51}

So, $Q_1$=$\frac{15+16}{2}=15.5$ and $Q_3$=$\frac{22+25}{2}=23.5$

Then $Q_3$ - $Q_1$=23.5-15.5=8

Then 1.5×8=12

$Q_1$-8=15.5-8=3.5

$Q_3$+8=23.5+8=35.5

In this data set, some values are above the value of 35.5 (35.5<51).

So, this data set is said to have outliers.

Therefore, Option D is correct.

The required data set is {14,15,16,20,22,25,51}.

Learn more about outliers here:

https://brainly.com/question/24195133

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