The problem asks to determine if there are outliers in the given dataset: 15, 21, 47, 55, 78, 65, 56, 6, 19, 24, 38, 120.

1. Problem Description

The problem asks to determine if there are outliers in the given dataset: 15, 21, 47, 55, 78, 65, 56, 6, 19, 24, 38,

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2. Solution Steps

To determine outliers, we can use the Interquartile Range (IQR) method.

Step 1: Sort the data.

The sorted data is: 6, 15, 19, 21, 24, 38, 47, 55, 56, 65, 78,

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Step 2: Find the first quartile (Q1) and the third quartile (Q3).

The number of data points, $n = 12$.

Q1 is the value at position $0.25(n+1) = 0.25(13) = 3.25$.

Q1 is between the 3rd and 4th value.

Q1 = 19 + 0.25(21 - 19) = 19 + 0.25(2) = 19 + 0.5 = 19.5

Q3 is the value at position $0.75(n+1) = 0.75(13) = 9.75$.

Q3 is between the 9th and 10th value.

Q3 = 56 + 0.75(65 - 56) = 56 + 0.75(9) = 56 + 6.75 = 62.75

Step 3: Calculate the IQR.

$IQR = Q3 - Q1 = 62.75 - 19.5 = 43.25$

Step 4: Calculate the lower and upper bounds.

Lower Bound = $Q1 - 1.5 * IQR = 19.5 - 1.5 * 43.25 = 19.5 - 64.875 = -45.375$

Upper Bound = $Q3 + 1.5 * IQR = 62.75 + 1.5 * 43.25 = 62.75 + 64.875 = 127.625$

Step 5: Identify outliers.

Any value below the lower bound or above the upper bound is an outlier.

The values in the dataset are: 6, 15, 19, 21, 24, 38, 47, 55, 56, 65, 78,

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0. The lower bound is -45.

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5. All values are above this.

The upper bound is 127.

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5. All values are below this.

Another method (using integer locations)

Q1 = 21

Q3 = 65

IQR = 65 - 21 = 44

Lower bound = 21 - 1.5 * 44 = 21 - 66 = -45

Upper bound = 65 + 1.5 * 44 = 65 + 66 = 131

The values in the dataset are: 6, 15, 19, 21, 24, 38, 47, 55, 56, 65, 78,

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0. All values are above -

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5. All values are below

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Consider using Tukey's fences:

Q1 is the median of the first half of the data. The first half is 6, 15, 19, 21, 24,

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8. Q1 = (19+21)/2 = 20

Q3 is the median of the second half of the data. The second half is 47, 55, 56, 65, 78,

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0. Q3 = (56+65)/2 = 60.5

IQR = Q3 - Q1 = 60.5 - 20 = 40.5

Lower bound = 20 - 1.5*40.5 = 20 - 60.75 = -40.75

Upper bound = 60.5 + 1.5*40.5 = 60.5 + 60.75 = 121.25

Values smaller than -40.75 or larger than 121.25 are outliers.

120 is close to 121.25, but is not considered an outlier based on this method.

The value 120 is pretty high.

The mean is 45.

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3. The standard deviation is 31.

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5. 120 is more than 2 standard deviations from the mean.

However, the question asks to *determine* if there are outliers.

120 seems suspiciously high.

3. Final Answer

Yes, 120 is an outlier.