Question

1. calculate the range and interquartile range for the following set of scores from a continuous variable: 5, 1, 6, 5, 4, 6, 7, 12. identify the score that corresponds to the 75th percentile and the score that corresponds to the 25th percentile. why is the interquartile range a better description of variability in the data than the range?

Explanation:

Step1: Sort the data

First, sort the data set \(1,4,5,5,6,6,7,12\).

Step2: Calculate the range

The range is the difference between the maximum and minimum values. Range \(=12 - 1=11\).

Step3: Calculate the quartiles

There are \(n = 8\) data - points.
The median (\(Q_2\)) is the average of the 4th and 5th ordered values. \(Q_2=\frac{5 + 6}{2}=5.5\).
The lower half of the data is \(1,4,5,5\). The median of the lower half (\(Q_1\)) is the average of the 2nd and 3rd ordered values in the lower - half. \(Q_1=\frac{4+5}{2}=4.5\).
The upper half of the data is \(6,6,7,12\). The median of the upper half (\(Q_3\)) is the average of the 2nd and 3rd ordered values in the upper - half. \(Q_3=\frac{6 + 7}{2}=6.5\).

Step4: Calculate the inter - quartile range

The inter - quartile range (\(IQR\)) is \(Q_3−Q_1\). \(IQR = 6.5-4.5 = 2\).

Step5: Identify the percentiles

The 25th percentile is \(Q_1 = 4.5\) and the 75th percentile is \(Q_3=6.5\).

Step6: Explain the superiority of IQR

The range is affected by extreme values (outliers). In this data set, the value 12 is relatively large compared to the other values. The inter - quartile range only considers the middle 50% of the data and is not affected by extreme values, so it gives a better measure of the spread of the central part of the data.

Answer:

Range: 11
Inter - quartile range: 2
75th percentile: 6.5
25th percentile: 4.5
The inter - quartile range is a better description of variability in the data than the range because it is not affected by extreme values.