Which is considered an outlier?

x:1,2,3,4,5,6,7,8,9 y:0.8,2.1,3.1,3.8,5.2,8.4,6.9,8.1,9

(2,2.1)
(4, 3.8)
(6, 8.4)
(9.9)

Can someone help, I think it's C or D?

I think C is farther out than D.

To determine whether a data point is an outlier, we typically look at the overall pattern of the data and identify values that deviate significantly from that pattern. One common method for identifying outliers is by using the concept of z-scores.

To calculate the z-score, you need to follow these steps:

1. Calculate the mean (average) of the data set.
2. Calculate the standard deviation of the data set.
3. For each data point, subtract the mean from the data point and divide the result by the standard deviation. This gives you the z-score.

The z-score indicates how many standard deviations a particular data point is away from the mean. Generally, a z-score greater than 3 or less than -3 is considered an outlier.

Now, let's calculate the z-scores for the given data points:

For x = 2, z-score = (2 - mean(x)) / standard deviation(x)
= (2 - (1+2+3+4+5+6+7+8+9) / 9) / sqrt(variance(x))
= (2 - 5) / 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)
= (2 - 5) / sqrt((-14) / 9)
= (2 - 5) / sqrt(-14/9)
= (2 - 5) / sqrt(-2)
Here, the denominator is imaginary, so we cannot calculate the z-score for this data point.

For x = 4, z-score = (4 - mean(x)) / standard deviation(x)
= (4 - (1+2+3+4+5+6+7+8+9) / 9) / sqrt(variance(x))
= (4 - 5) / 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)
= (4 - 5) / sqrt((-14) / 9)
= (4 - 5) / sqrt(-14/9)
= (4 - 5) / sqrt(-2)
Here, the denominator is imaginary, so we cannot calculate the z-score for this data point.

For x = 6, z-score = (6 - mean(x)) / standard deviation(x)
= (6 - (1+2+3+4+5+6+7+8+9) / 9) / sqrt(variance(x))
= (6 - 5) / 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)
= (6 - 5) / sqrt((-14) / 9)
= (6 - 5) / sqrt(-14/9)
= (6 - 5) / sqrt(-2)
Here, the denominator is imaginary, so we cannot calculate the z-score for this data point.

For x = 9, z-score = (9 - mean(x)) / standard deviation(x)
= (9 - (1+2+3+4+5+6+7+8+9) / 9) / sqrt(variance(x))
= (9 - 5) / 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)
= (9 - 5) / sqrt((-14) / 9)
= (9 - 5) / sqrt(-14/9)
= (9 - 5) / sqrt(-2)
Here, the denominator is imaginary, so we cannot calculate the z-score for this data point.

Based on the z-score calculation, none of the given data points have a valid z-score. Therefore, none of the options (C or D) can be considered outliers in this case.