Need help Given the following values, which point would be considered an outlier?
x1 ,2 ,3 ,4 ,5,6, 7 ,8 ,9
y1.2,4.3,2.8,5.2, 6,7.1,7.9,9.4
It's (6,6) I believe or in other words D.
To determine if there is an outlier, we need to calculate the z-score for each data point and check if any of them fall outside a certain threshold. In this case, an outlier would be considered a point that falls outside the range of the mean plus or minus 2 standard deviations.
First, let's calculate the mean and standard deviation for each set of values:
For x:
Mean (x̄) = (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9) / 9 = 5
Standard Deviation (σ) = sqrt(((1-5)² + (2-5)² + (3-5)² + (4-5)² + (5-5)² + (6-5)² + (7-5)² + (8-5)² + (9-5)²) / 9)
= sqrt((16 + 9 + 4 + 1 + 0 + 1 + 4 + 9 + 16) / 9)
= sqrt(60 / 9)
≈ sqrt(6.67)
≈ 2.58
For y:
Mean (ȳ) = (1.2 + 4.3 + 2.8 + 5.2 + 6 + 7.1 + 7.9 + 9.4) / 8 ≈ 5.86
Standard Deviation (σ) = sqrt(((1.2-5.86)² + (4.3-5.86)² + (2.8-5.86)² + (5.2-5.86)² + (6-5.86)² + (7.1-5.86)² + (7.9-5.86)² + (9.4-5.86)²) / 8)
= sqrt((19.1436 + 2.4644 + 13.2436 + 0.4224 + 0.0224 + 1.8404 + 2.1244 + 11.4736) / 8)
= sqrt(50.7244 / 8)
≈ sqrt(6.34)
≈ 2.52
Now let's calculate the z-scores for each point:
For x:
z-score (x1) = (1 - 5) / 2.58 ≈ -1.55
z-score (x2) = (2 - 5) / 2.58 ≈ -1.16
z-score (x3) = (3 - 5) / 2.58 ≈ -0.78
z-score (x4) = (4 - 5) / 2.58 ≈ -0.39
z-score (x5) = (5 - 5) / 2.58 ≈ 0
z-score (x6) = (6 - 5) / 2.58 ≈ 0.39
z-score (x7) = (7 - 5) / 2.58 ≈ 0.78
z-score (x8) = (8 - 5) / 2.58 ≈ 1.16
z-score (x9) = (9 - 5) / 2.58 ≈ 1.55
For y:
z-score (y1) = (1.2 - 5.86) / 2.52 ≈ -1.86
z-score (y2) = (4.3 - 5.86) / 2.52 ≈ -0.62
z-score (y3) = (2.8 - 5.86) / 2.52 ≈ -1.21
z-score (y4) = (5.2 - 5.86) / 2.52 ≈ -0.26
z-score (y5) = (6 - 5.86) / 2.52 ≈ 0.06
z-score (y6) = (7.1 - 5.86) / 2.52 ≈ 0.49
z-score (y7) = (7.9 - 5.86) / 2.52 ≈ 0.81
z-score (y8) = (9.4 - 5.86) / 2.52 ≈ 1.40
No point in either set falls outside the range of the mean plus or minus 2 standard deviations. Therefore, there are no outliers in this data set.
To determine if a point is an outlier, you can use the concept of "outlier boundaries" based on the interquartile range (IQR) method.
First, you need to find the IQR for the dataset.
1. Sort the values in ascending order for each variable:
x: 1, 2, 3, 4, 5, 6, 7, 8, 9
y: 1.2, 2.8, 4.3, 5.2, 6, 7.1, 7.9, 9.4
2. Calculate the first quartile (Q1) and third quartile (Q3) for each variable:
Q1x = 2, Q3x = 8
Q1y ≈ 3.55, Q3y ≈ 7.5
3. Calculate the IQR for each variable:
IQRx = Q3x - Q1x = 8 - 2 = 6
IQry = Q3y - Q1y ≈ 7.5 - 3.55 ≈ 3.95
4. Define the lower and upper outlier boundaries for each variable:
Lower boundary for x: Q1x - 1.5 * IQRx ≈ 2 - 1.5 * 6 = -7
Upper boundary for x: Q3x + 1.5 * IQRx ≈ 8 + 1.5 * 6 = 17
Lower boundary for y: Q1y - 1.5 * IQry ≈ 3.55 - 1.5 * 3.95 ≈ -2.18
Upper boundary for y: Q3y + 1.5 * IQry ≈ 7.5 + 1.5 * 3.95 ≈ 13.62
5. Check if any point lies outside the outlier boundaries.
Given the values:
x: 1, 2, 3, 4, 5, 6, 7, 8, 9
y: 1.2, 4.3, 2.8, 5.2, 6, 7.1, 7.9, 9.4
Looking at the values, there are no points that fall outside the outlier boundaries. Therefore, there are no outliers in this dataset.