Question

a data set is shown.
10, 10, 11, 12, 13, 13, 13, 13, 14, 14, 15, 15, 15, 16, 17, 17, 17, 35
move the options to the blanks to describe the effect removing the outlier, 35, will have on the data set.
the mean of the data set will __.
the median of the data set will __.
the range of the data set will __.
decrease
not change

Explanation:

Step1: Calculate the range before and after removing the outlier

• Before removing 35:

The range is calculated as \( \text{Range}=\text{Max}-\text{Min} \). Here, \( \text{Max} = 35 \) and \( \text{Min}=10 \), so \( \text{Range}=35 - 10=25 \).

• After removing 35:

The new \( \text{Max}=17 \) and \( \text{Min} = 10 \), so \( \text{Range}=17-10 = 7 \). The range decreases.

Step2: Calculate the median before and after removing the outlier

• Before removing 35:

There are \( n = 20 \) data points. The median is the average of the \( \frac{n}{2}=10^{th} \) and \( (\frac{n}{2}+1)=11^{th} \) ordered values. The \( 10^{th} \) value is \( 14 \) and the \( 11^{th} \) value is \( 14 \), so \( \text{Median}=\frac{14 + 14}{2}=14 \).

• After removing 35:

There are \( n=19 \) data points. The median is the \( \frac{n + 1}{2}=10^{th} \) ordered value. The \( 10^{th} \) value is still \( 14 \). The median does not change.

Step3: Calculate the mean before and after removing the outlier

• Before removing 35:

The mean \( \bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n} \). \( \sum_{i=1}^{20}x_{i}=10+10+11+12+13+13+13+14+14+15+15+15+16+17+17+17+35\)
\( \sum_{i = 1}^{20}x_{i}=280 \), \( \bar{x}=\frac{280}{20}=14 \).

• After removing 35:

\( \sum_{i=1}^{19}x_{i}=280 - 35=245 \), \( \bar{x}=\frac{245}{19}\approx12.89 \). The mean decreases.

Answer:

The mean of the data set will decrease. The median of the data set will not change. The range of the data set will decrease.