22.6 is an outlier for which of the following sets of data?
A. 22.6, 21.5, 23.7, 22.6, 28.9, 22.6, 20.9
B. 2.4, 5.3, 3.5, 22.6, 1.8, 2.1, 4.6, 1.9
C. 20.5, 20.8, 21.6, 22.6, 23.7, 24.5, 25.1
D. 13.6, 31.7, 25.8, 22.6, 18.9, 21.6, 30.5
B is where 22.6 is an outlier.
the set where most of the values are distant from 22.6
Hmm, let me think about this... Ah! I've got it! The clown of the data sets is option B. Why? Because while all the other data sets have repeated values of 22.6, option B stands out with just one 22.6. It's like the odd clown out!
To determine if 22.6 is an outlier for each set of data, we need to calculate the mean and standard deviation.
For set A:
Mean = (22.6 + 21.5 + 23.7 + 22.6 + 28.9 + 22.6 + 20.9) / 7 = 21.99
Standard Deviation = √[ (22.6 - 21.99)² + (21.5 - 21.99)² + (23.7 - 21.99)² + (22.6 - 21.99)² + (28.9 - 21.99)² + (22.6 - 21.99)² + (20.9 - 21.99)² ] / 7 = 2.20
Since 22.6 is within 1 standard deviation of the mean, it is not an outlier for set A.
For set B:
Mean = (2.4 + 5.3 + 3.5 + 22.6 + 1.8 + 2.1 + 4.6 + 1.9) / 8 = 6.10
Standard Deviation = √[ (2.4 - 6.10)² + (5.3 - 6.10)² + (3.5 - 6.10)² + (22.6 - 6.10)² + (1.8 - 6.10)² + (2.1 - 6.10)² + (4.6 - 6.10)² + (1.9 - 6.10)² ] / 8 = 7.54
Since 22.6 is more than 2 standard deviations away from the mean, it is an outlier for set B.
For set C:
Mean = (20.5 + 20.8 + 21.6 + 22.6 + 23.7 + 24.5 + 25.1) / 7 = 22.22
Standard Deviation = √[ (20.5 - 22.22)² + (20.8 - 22.22)² + (21.6 - 22.22)² + (22.6 - 22.22)² + (23.7 - 22.22)² + (24.5 - 22.22)² + (25.1 - 22.22)² ] / 7 = 1.32
Since 22.6 is within 1 standard deviation of the mean, it is not an outlier for set C.
For set D:
Mean = (13.6 + 31.7 + 25.8 + 22.6 + 18.9 + 21.6 + 30.5) / 7 = 23.06
Standard Deviation = √[ (13.6 - 23.06)² + (31.7 - 23.06)² + (25.8 - 23.06)² + (22.6 - 23.06)² + (18.9 - 23.06)² + (21.6 - 23.06)² + (30.5 - 23.06)² ] / 7 = 5.94
Since 22.6 is within 1 standard deviation of the mean, it is not an outlier for set D.
Therefore, 22.6 is an outlier only for set B.