Question

1. 4, 2, 8, 5, 6, 20, 7, 9

with outlier
mean: ______
median: ______
mode: ______
range: ______
stdevp: ______
without outlier
mean: ______
median: ______
mode: ______
range: ______
stdevp: ______
lower bound: ______
upper bound: ______

Explanation:

Step1: Identify the outlier

First, we check the data set \(4, 2, 8, 5, 6, 20, 7, 9\). The value \(20\) is much larger than the other values, so it is the outlier.

Step2: Calculate with outlier (all data points: \(n = 8\))

Mean:

The formula for the mean \(\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}\)
\(\sum_{i = 1}^{8}x_{i}=4 + 2+8 + 5+6 + 20+7 + 9=61\)
\(\bar{x}=\frac{61}{8}=7.625\)

Median:

First, sort the data: \(2, 4, 5, 6, 7, 8, 9, 20\)
Since \(n = 8\) (even), the median is the average of the \(\frac{n}{2}\)-th and \((\frac{n}{2}+ 1)\)-th terms.
\(\frac{n}{2}=4\), \(\frac{n}{2}+ 1 = 5\)
The 4th term is \(6\), the 5th term is \(7\)
Median \(=\frac{6 + 7}{2}=6.5\)

Mode:

All values appear once, so there is no mode (or we can say all values are modes, but typically we say no mode in this case)

Range:

Range \(=\text{Max}-\text{Min}=20 - 2=18\)

Stdevp (Population Standard Deviation):

The formula for population standard deviation \(\sigma=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n}}\)
First, calculate \((x_{i}-\bar{x})^{2}\) for each \(x_{i}\):

• \( (2 - 7.625)^{2}=(- 5.625)^{2}=31.640625\)
• \( (4 - 7.625)^{2}=(-3.625)^{2}=13.140625\)
• \( (5 - 7.625)^{2}=(-2.625)^{2}=6.890625\)
• \( (6 - 7.625)^{2}=(-1.625)^{2}=2.640625\)
• \( (7 - 7.625)^{2}=(-0.625)^{2}=0.390625\)
• \( (8 - 7.625)^{2}=(0.375)^{2}=0.140625\)
• \( (9 - 7.625)^{2}=(1.375)^{2}=1.890625\)
• \( (20 - 7.625)^{2}=(12.375)^{2}=153.140625\)

\(\sum_{i = 1}^{8}(x_{i}-\bar{x})^{2}=31.640625+13.140625 + 6.890625+2.640625+0.390625+0.140625+1.890625+153.140625 = 209.875\)
\(\sigma=\sqrt{\frac{209.875}{8}}=\sqrt{26.234375}\approx5.122\)

Step3: Calculate without outlier (remove \(20\), \(n = 7\))

Mean:

\(\sum_{i = 1}^{7}x_{i}=4 + 2+8 + 5+6 + 7+9 = 41\)
\(\bar{x}=\frac{41}{7}\approx5.857\)

Median:

Sort the data (without \(20\)): \(2, 4, 5, 6, 7, 8, 9\)
Since \(n = 7\) (odd), the median is the \(\frac{n + 1}{2}=4\)-th term, which is \(6\)

Mode:

All values appear once, so no mode.

Range:

Range \(=\text{Max}-\text{Min}=9 - 2 = 7\)

Stdevp (Population Standard Deviation):

First, calculate \((x_{i}-\bar{x})^{2}\) for each \(x_{i}\) (where \(\bar{x}=\frac{41}{7}\approx5.857\))

• \( (2 - 5.857)^{2}=(-3.857)^{2}\approx14.876\)
• \( (4 - 5.857)^{2}=(-1.857)^{2}\approx3.448\)
• \( (5 - 5.857)^{2}=(-0.857)^{2}\approx0.734\)
• \( (6 - 5.857)^{2}=(0.143)^{2}\approx0.020\)
• \( (7 - 5.857)^{2}=(1.143)^{2}\approx1.306\)
• \( (8 - 5.857)^{2}=(2.143)^{2}\approx4.592\)
• \( (9 - 5.857)^{2}=(3.143)^{2}\approx9.878\)

\(\sum_{i = 1}^{7}(x_{i}-\bar{x})^{2}\approx14.876+3.448+0.734 + 0.020+1.306+4.592+9.878=34.854\)
\(\sigma=\sqrt{\frac{34.854}{7}}\approx\sqrt{4.979}\approx2.231\)

Step4: Calculate Lower and Upper Bounds (for outlier detection, using IQR method)

First, find Q1 and Q3 for the data without outlier (sorted: \(2, 4, 5, 6, 7, 8, 9\))

• Q1 (25th percentile): The median of the lower half (\(2, 4, 5\)) is \(4\)
• Q3 (75th percentile): The median of the upper half (\(7, 8, 9\)) is \(8\)
• IQR \(=Q3 - Q1=8 - 4 = 4\)
• Lower Bound \(=Q1-1.5\times\text{IQR}=4-1.5\times4=4 - 6=- 2\)
• Upper Bound \(=Q3 + 1.5\times\text{IQR}=8+1.5\times4=8 + 6 = 14\) (this shows why \(20\) is an outlier, since \(20>14\))

Answer:

With Outlier:

• Mean: \(7.625\)
• Median: \(6.5\)
• Mode: No mode (or all values)
• Range: \(18\)
• Stdevp: \(\approx5.122\)

Without Outlier:

• Mean: \(\approx5.857\)
• Median: \(6\)
• Mode: No mode (or all values)
• Range: \(7\)
• Stdevp: \(\approx2.231\)
• Lower Bound: \(-2\)
• Upper Bound: \(14\)