Question

the following were calculated from the times (in minutes) that 10 individuals reported spending on the internet each day. mean: 87 median: 72 mode: 72 range: 75 iqr: 30 there was one individual in this group who reported spending 150 minutes per day on the internet. if you removed this individuals time from the data set, which two statements are most likely true? press to hear a reminder or hint for this problem. outliers affect which measures of center or dispersion the most? a the range will decrease. b the median will increase. c the iqr will increase. d the mean will decrease.

Explanation:

Step1: Analyze the effect on range

Range is calculated as \( \text{Range}=\text{Maximum}-\text{Minimum} \). Since \( 150 \) is the maximum value (because range \( = 75 \), if \( \text{Minimum}=x \), then \( 150 - x=75\), so \( x = 75\)), removing \( 150 \) (the maximum) will make the new range \( \text{New Range}=\text{New Maximum}-\text{Minimum}\), and \( \text{New Maximum}<150 \). So the range will decrease.

Step2: Analyze the effect on median

The median is the middle - value (for \( n = 10 \), the average of the 5th and 6th ordered values). Since \( 150 \) is an extreme value (an outlier) and not in the middle of the data set (the median is \( 72 \)), removing it will not increase the median. In fact, if the data set was ordered, and \( 150 \) was at the end, the middle values (5th and 6th) may stay the same or change slightly, but not increase in a way that is significant for this multiple - choice.

Step3: Analyze the effect on IQR

IQR (Inter - Quartile Range) is \( Q_{3}-Q_{1} \). Since \( 150 \) is an outlier (not in the middle \( 50\%\) of the data, because median \( = 72\) and \( 150\) is much larger), removing it will not increase the IQR. The IQR is based on the middle \( 50\%\) of the data.

Step4: Analyze the effect on mean

The mean is \( \bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n} \). Let \( \sum_{i = 1}^{10}x_{i}=S\), then \( \bar{x}=\frac{S}{10}=87\), so \( S = 870 \). If we remove \( x = 150 \), the new sum \( S'=S - 150=870-150 = 720 \), and the new mean \( \bar{x}'=\frac{720}{9}=80<87 \). So the mean will decrease.

Answer:

A. The range will decrease, D. The mean will decrease.